OverviewOverview
61 Discrete Random Variables
62 Binomial Probability Distribution
63 Continuous Random Variables and the Normal Probability Distribution
64 Standard Normal Distribution
65 Applications of the Normal Distribution
61 Discrete Random Variables61 Discrete Random Variables
ObjectivesBy the end of this section I will beable tohellip
1) Identify random variables2)Explain what a discrete probability
distribution is and construct probability distribution tables and graphs
3)Calculate the mean variance and standard deviation of a discrete random variable
Random VariablesRandom Variables
A variable whose values are determined by chance
Chance in the definition of a random variable is crucial
Example 62 - Notation Example 62 - Notation for random variablesfor random variables
Suppose our experiment is to toss a single fair die and we are interested in the numberrolled We define our random variable X to be the outcome of a single die rolla Why is the variable X a random variableb What are the possible values that the random variable X can takec What is the notation used for rolling a 5d Use random variable notation to express the
probability of rolling a 5
Example 62 continuedExample 62 continuedSolution
a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment
b) Possible values for X 1 2 3 4 5 and 6
c) When a 5 is rolled then X equals the outcome 5 or X = 5
d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16
Main types of random variablesMain types of random variables
Discrete random variable - a finite or a countable number of values
Continuous random variable can take infinitely many values
Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables
For the following random variables (i) determine
whether they are discrete or continuousand (ii) indicate the possible values they can
takea The number of automobiles owned by a
familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs
Example 63 continuedExample 63 continued
Solution
aSince the possible number of automobiles
owned by a family is finite and may be written as a list of numbers it represents a discrete random variable
The possible values are 0 1 2 3 4
Example 63 continuedExample 63 continued
Solution
bWidth is something that must be measured not counted Width can take infinitely many different
possible values with these values forming an interval on the number line
Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft
Example 63 continuedExample 63 continued
Solution
c The number of games played in the next
World Series can be counted and thus represents a discrete random variable
The possible values are 4 5 6 7
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
61 Discrete Random Variables61 Discrete Random Variables
ObjectivesBy the end of this section I will beable tohellip
1) Identify random variables2)Explain what a discrete probability
distribution is and construct probability distribution tables and graphs
3)Calculate the mean variance and standard deviation of a discrete random variable
Random VariablesRandom Variables
A variable whose values are determined by chance
Chance in the definition of a random variable is crucial
Example 62 - Notation Example 62 - Notation for random variablesfor random variables
Suppose our experiment is to toss a single fair die and we are interested in the numberrolled We define our random variable X to be the outcome of a single die rolla Why is the variable X a random variableb What are the possible values that the random variable X can takec What is the notation used for rolling a 5d Use random variable notation to express the
probability of rolling a 5
Example 62 continuedExample 62 continuedSolution
a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment
b) Possible values for X 1 2 3 4 5 and 6
c) When a 5 is rolled then X equals the outcome 5 or X = 5
d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16
Main types of random variablesMain types of random variables
Discrete random variable - a finite or a countable number of values
Continuous random variable can take infinitely many values
Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables
For the following random variables (i) determine
whether they are discrete or continuousand (ii) indicate the possible values they can
takea The number of automobiles owned by a
familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs
Example 63 continuedExample 63 continued
Solution
aSince the possible number of automobiles
owned by a family is finite and may be written as a list of numbers it represents a discrete random variable
The possible values are 0 1 2 3 4
Example 63 continuedExample 63 continued
Solution
bWidth is something that must be measured not counted Width can take infinitely many different
possible values with these values forming an interval on the number line
Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft
Example 63 continuedExample 63 continued
Solution
c The number of games played in the next
World Series can be counted and thus represents a discrete random variable
The possible values are 4 5 6 7
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Random VariablesRandom Variables
A variable whose values are determined by chance
Chance in the definition of a random variable is crucial
Example 62 - Notation Example 62 - Notation for random variablesfor random variables
Suppose our experiment is to toss a single fair die and we are interested in the numberrolled We define our random variable X to be the outcome of a single die rolla Why is the variable X a random variableb What are the possible values that the random variable X can takec What is the notation used for rolling a 5d Use random variable notation to express the
probability of rolling a 5
Example 62 continuedExample 62 continuedSolution
a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment
b) Possible values for X 1 2 3 4 5 and 6
c) When a 5 is rolled then X equals the outcome 5 or X = 5
d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16
Main types of random variablesMain types of random variables
Discrete random variable - a finite or a countable number of values
Continuous random variable can take infinitely many values
Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables
For the following random variables (i) determine
whether they are discrete or continuousand (ii) indicate the possible values they can
takea The number of automobiles owned by a
familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs
Example 63 continuedExample 63 continued
Solution
aSince the possible number of automobiles
owned by a family is finite and may be written as a list of numbers it represents a discrete random variable
The possible values are 0 1 2 3 4
Example 63 continuedExample 63 continued
Solution
bWidth is something that must be measured not counted Width can take infinitely many different
possible values with these values forming an interval on the number line
Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft
Example 63 continuedExample 63 continued
Solution
c The number of games played in the next
World Series can be counted and thus represents a discrete random variable
The possible values are 4 5 6 7
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 62 - Notation Example 62 - Notation for random variablesfor random variables
Suppose our experiment is to toss a single fair die and we are interested in the numberrolled We define our random variable X to be the outcome of a single die rolla Why is the variable X a random variableb What are the possible values that the random variable X can takec What is the notation used for rolling a 5d Use random variable notation to express the
probability of rolling a 5
Example 62 continuedExample 62 continuedSolution
a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment
b) Possible values for X 1 2 3 4 5 and 6
c) When a 5 is rolled then X equals the outcome 5 or X = 5
d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16
Main types of random variablesMain types of random variables
Discrete random variable - a finite or a countable number of values
Continuous random variable can take infinitely many values
Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables
For the following random variables (i) determine
whether they are discrete or continuousand (ii) indicate the possible values they can
takea The number of automobiles owned by a
familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs
Example 63 continuedExample 63 continued
Solution
aSince the possible number of automobiles
owned by a family is finite and may be written as a list of numbers it represents a discrete random variable
The possible values are 0 1 2 3 4
Example 63 continuedExample 63 continued
Solution
bWidth is something that must be measured not counted Width can take infinitely many different
possible values with these values forming an interval on the number line
Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft
Example 63 continuedExample 63 continued
Solution
c The number of games played in the next
World Series can be counted and thus represents a discrete random variable
The possible values are 4 5 6 7
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 62 continuedExample 62 continuedSolution
a) We donrsquot know the value of X before we toss the die which introduces an element of chance into the experiment
b) Possible values for X 1 2 3 4 5 and 6
c) When a 5 is rolled then X equals the outcome 5 or X = 5
d) Probability of rolling a 5 for a fair die is 16 thus P(X = 5) = 16
Main types of random variablesMain types of random variables
Discrete random variable - a finite or a countable number of values
Continuous random variable can take infinitely many values
Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables
For the following random variables (i) determine
whether they are discrete or continuousand (ii) indicate the possible values they can
takea The number of automobiles owned by a
familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs
Example 63 continuedExample 63 continued
Solution
aSince the possible number of automobiles
owned by a family is finite and may be written as a list of numbers it represents a discrete random variable
The possible values are 0 1 2 3 4
Example 63 continuedExample 63 continued
Solution
bWidth is something that must be measured not counted Width can take infinitely many different
possible values with these values forming an interval on the number line
Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft
Example 63 continuedExample 63 continued
Solution
c The number of games played in the next
World Series can be counted and thus represents a discrete random variable
The possible values are 4 5 6 7
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Main types of random variablesMain types of random variables
Discrete random variable - a finite or a countable number of values
Continuous random variable can take infinitely many values
Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables
For the following random variables (i) determine
whether they are discrete or continuousand (ii) indicate the possible values they can
takea The number of automobiles owned by a
familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs
Example 63 continuedExample 63 continued
Solution
aSince the possible number of automobiles
owned by a family is finite and may be written as a list of numbers it represents a discrete random variable
The possible values are 0 1 2 3 4
Example 63 continuedExample 63 continued
Solution
bWidth is something that must be measured not counted Width can take infinitely many different
possible values with these values forming an interval on the number line
Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft
Example 63 continuedExample 63 continued
Solution
c The number of games played in the next
World Series can be counted and thus represents a discrete random variable
The possible values are 4 5 6 7
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 63 - Discrete and Example 63 - Discrete and continuous random variablescontinuous random variables
For the following random variables (i) determine
whether they are discrete or continuousand (ii) indicate the possible values they can
takea The number of automobiles owned by a
familyb The width of your desk in this classroomc The number of games played in the next World Seriesd The weight of model year 2007 SUVs
Example 63 continuedExample 63 continued
Solution
aSince the possible number of automobiles
owned by a family is finite and may be written as a list of numbers it represents a discrete random variable
The possible values are 0 1 2 3 4
Example 63 continuedExample 63 continued
Solution
bWidth is something that must be measured not counted Width can take infinitely many different
possible values with these values forming an interval on the number line
Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft
Example 63 continuedExample 63 continued
Solution
c The number of games played in the next
World Series can be counted and thus represents a discrete random variable
The possible values are 4 5 6 7
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 63 continuedExample 63 continued
Solution
aSince the possible number of automobiles
owned by a family is finite and may be written as a list of numbers it represents a discrete random variable
The possible values are 0 1 2 3 4
Example 63 continuedExample 63 continued
Solution
bWidth is something that must be measured not counted Width can take infinitely many different
possible values with these values forming an interval on the number line
Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft
Example 63 continuedExample 63 continued
Solution
c The number of games played in the next
World Series can be counted and thus represents a discrete random variable
The possible values are 4 5 6 7
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 63 continuedExample 63 continued
Solution
bWidth is something that must be measured not counted Width can take infinitely many different
possible values with these values forming an interval on the number line
Thus the width of your desk is a continuous random variable The possible values might be 1 ft le W le 10 ft
Example 63 continuedExample 63 continued
Solution
c The number of games played in the next
World Series can be counted and thus represents a discrete random variable
The possible values are 4 5 6 7
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 63 continuedExample 63 continued
Solution
c The number of games played in the next
World Series can be counted and thus represents a discrete random variable
The possible values are 4 5 6 7
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 63 continuedExample 63 continued
Solution
d The weight of model year 2007 SUVs must
be measured not counted and so represents a continuous random variable
Weight can take infinitely many different possible values in an interval possible values 2500 lb le Y le 7000 lb
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Discrete Probability DistributionsDiscrete Probability Distributions
Provides all the possible values that the random variable can assume
Together with the probability associated with each value
Can take the form of a table graph or formula
Describe populations not samples
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 65 - Probability Example 65 - Probability distribution tabledistribution table
Construct the probability distribution table of the number of heads observed whentossing a fair coin twice
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 65 continuedExample 65 continued
Solution
The probability distribution table given in Table 62 uses probabilities we found in Example 64 page 261
Table 62 Probability distribution of number of heads on two fair coin tosses
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
RequirementsRequirements
Probability Distribution of a Discrete Random Variable
The sum of the probabilities of all the possible values of a discrete random variable must equal 1
That is ΣP(X) = 1
The probability of each value of X must be between 0 and 1 inclusive
That is 0 le P(X ) le 1
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Discrete Probability Distribution Discrete Probability Distribution as a Graphas a GraphGraphs show all the information contained in
probability distribution tables
Identify patterns more quickly
FIGURE 61 Graph of probability distribution for Kristinrsquos financial gain
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Mean of a Discrete Random Mean of a Discrete Random VariableVariable
The mean μ of a discrete random variable represents the mean result when the experiment is repeated an indefinitely large number of times
Also called the expected value or expectation of the random variable X
Denoted as E(X )
Holds for discrete and continuous random variables
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Finding the Mean of a Discrete Finding the Mean of a Discrete Random VariableRandom Variable
Multiply each possible value of X by its probability
Add the resulting products
X P X
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 68 - Calculating the Example 68 - Calculating the mean of a discrete random mean of a discrete random variablevariable
Find the mean age of the mother for the babies born to teenagers aged 15-18 in 2004from Example 67 page 265
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 68 continuedExample 68 continued
Solution
Multiply each possible outcome (value of X) by its probability P(X)
Multiply the value X = 15 by its probability P(X) = 007 the value X = 16 by its probability P(X) = 017 and so on
Add these four products to find the mean
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 68 continuedExample 68 continued
Solution
μ=15(007) + 16(017) + 17(029) +18(047) = 1716
The mean age of the mother for the babies born to teenagers aged 15ndash18 is 1716 years old which is near 17 years old
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Variability of a Discrete Variability of a Discrete Random VariableRandom VariableFormulas for the Variance and Standard Deviation of a Discrete Random Variable
22
2
Definition Formulas
X P X
X P X
2 2 2
2 2
Computational Formulas
X P X
X P X
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
SummarySummary
Section 61 introduces the idea of random variables a crucial concept that we will use to assess the behavior of variable processes for the remainder of the text
Random variables are variables whose value is determined at least partly by chance
Discrete random variables take values that are either finite or countable and may be put in a list
Continuous random variables take an infinite number of possible values represented by an interval on the number line
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
SummarySummary
Discrete random variables can be described using a probability distribution which specifies the probability of observing each value of the random variable
Such a distribution can take the form of a table graph or formula
Probability distributions describe populations not samples
We can find the mean μ standard deviation σ and variance σ2 of a discrete random variable using formulas
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
62 Binomial Probability 62 Binomial Probability DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Explain what constitutes a binomial experiment
2)Compute probabilities using the binomial probability formula
3)Find probabilities using the binomial tables4)Calculate and interpret the mean variance
and standard deviation of the binomial random variable
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Binomial ExperimentBinomial Experiment
Four Requirements
1)Each trial of the experiment has only two possible outcomes (success or failure)
2)Fixed number of trials
3)Experimental outcomes are independent of each other
4)Probability of observing a success remains the same from trial to trial
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
NotationNotation
Table 66 Notation for binomial experiments and the binomial distribution
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Formula for the Number of Formula for the Number of CombinationsCombinations
The number of combinations of X items chosen from n different items is given by
where n represents n factorial which equals n(n - 1)(n - 2) (2)(1) and 0 is defined to be 1
n X
nC
X n X
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 613 - How many Example 613 - How many team combinations in the team combinations in the intramural volleyball leagueintramural volleyball league
Jeffrey is in charge of drawing up a schedule for his collegersquos intramural volleyballleague This year five teams have been fielded and they must play each other onceHow many games will be held
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 613 continuedExample 613 continued
Solution
Number of combinations of n = 5
Volleyball teams taken X = 2 at a time
Ten games will be held
5 2
5 5 4 3 2 1 12010
2 5 2 2 1 3 2 1 2 6C
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Binomial Probability Binomial Probability Distribution FormulaDistribution Formula
The probability of observing exactly X successes in n trials of a binomial experiment is
P(X) = (nCX) pX (1 - p)n-X
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Binomial Distribution TablesBinomial Distribution Tables
n is the number of trialsX is the number of successesp is the probability of observing a success
FIGURE 67 Excerpt from the binomial tables
See Example 616 on page 278 for more information
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Binomial Mean Variance and Binomial Mean Variance and Standard DeviationStandard Deviation
Mean (or expected value) μ = n p
Variance σ2 = n p (1 - p)
Standard deviation 1n p p
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 618 - Mean variance Example 618 - Mean variance and standard deviation of and standard deviation of left-handed studentsleft-handed students
Suppose we know that the population proportion p of left-handed students is010a In a sample of 200 students how manywould we expect to be left-handed
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 618 continuedExample 618 continued
Solution
The binomial random variable here is X = the number of left-handed students
a Here n = 200 and p = 010
So the expected number of left-handed students in a sample of 200 is
E(X) = μ = n p = (200)(010) = 20
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
SummarySummary
The most important discrete distribution is the binomial distribution where there are two possible outcomes each with probability of success p and n independent trials
The probability of observing a particular number of successes can be calculated using the binomial probability distribution formula
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
SummarySummary
Binomial probabilities can also be found using the binomial tables or using technology
There are formulas for finding the mean variance and standard deviation of a binomial random variable
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
63 Continuous Random 63 Continuous Random Variables and the NormalVariables and the NormalProbability DistributionProbability Distribution
ObjectivesBy the end of this section I will beable tohellip
1) Identify a continuous probability distribution
2)Explain the properties of the normal probability distribution
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
(a) Relatively small sample (n = 100)with large classwidths (05 lb)
FIGURE 615FIGURE 615
(b) Large sample(n = 200) with smaller class widths (02 lb)
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Figure 615 continuedFigure 615 continued
(c) Very large sample (n = 400) with very small class widths (01 lb)
(d) Eventually theoretical histogram of entire population becomes smooth curve with class widths arbitrarily small
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Continuous Probability DistributionsContinuous Probability Distributions
A graph that indicates on the horizontal axis the range of values that the continuous random variable X can take
Density curve is drawn above the horizontal axis
Must follow the Requirements for the Probability Distribution of a Continuous Random Variable
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
RequirementsRequirements
1) The total area under the density curve must equal 1 (this is the Law of Total Probability for Continuous Random Variables)
2) The vertical height of the density curve can never be negative That is the density curve never goes below the horizontal axis
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
ProbabilityProbability
Probability for Continuous Distributions is represented by area under the curve above an interval
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
The Normal Probability DistributionThe Normal Probability DistributionMost important probability distribution in the
world
Population is said to be normally distributed the data values follow a normal probability distribution
Specific population mean μ
Specific population standard deviation σ
μ and σ are parameters of the normal distribution
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
FIGURE 619FIGURE 619
The normal distribution is symmetric about its mean μ
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Properties of the Normal Properties of the Normal Density Curve (Normal Curve)Density Curve (Normal Curve)1) It is symmetric about the mean μ
2) The highest point occurs at X = μ because symmetry implies that the mean equals the median which equals the mode of the distribution
3) It has inflection points at μ-σ and μ+σ
4) The total area under the curve equals 1
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Properties of the Normal Properties of the Normal Density Curve (Normal Curve) Density Curve (Normal Curve) continuedcontinued5) Symmetry also implies that the area under
the curve to the left of μ and the area under the curve to the right of μ are both equal to 05 (Figure 619)
6) The normal distribution is defined for values of X extending indefinitely in both the positive and negative directions As X moves farther from the mean the density curve approaches but never quite touches the horizontal axis
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Drawing a Graph to Solve Drawing a Graph to Solve Normal Probability ProblemsNormal Probability Problems
1 Draw a ldquogenericrdquo bell-shaped curve with a horizontal number line under it that is labeled as the random variable X
Insert the mean μ in the center of the number line
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Steps in Drawing a Graph to Steps in Drawing a Graph to Help You Solve Normal Help You Solve Normal Probability ProblemsProbability Problems
2) Mark on the number line the value of X indicated in the problem
Shade in the desired area under the normal curve to the right or left of this value
3) Proceed to find the desired area or probability
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
SummarySummary
Continuous random variables assume infinitely many possible values with no gap between the values
Probability for continuous random variables consists of area above an interval on the number line and under the distribution curve
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
SummarySummary
The normal distribution is the most important continuous probability distribution
It is symmetric about its mean μ and has standard deviation σ
One should always sketch a picture of a normal probability problem to help solve it
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
64 Standard Normal 64 Standard Normal DistributionDistribution
ObjectivesBy the end of this section I will beable tohellip
1)Find areas under the standard normal curve given a Z-value
2)Find the standard normal Z-value given an area
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
The Standard Normal (Z) The Standard Normal (Z) DistributionDistribution
A normal distribution with
mean μ = 0 and
standard deviation σ = 1
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Steps for finding areas under the standard normal curve
Table 67Table 67
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 629 - Expressing areas Example 629 - Expressing areas under the standard normal under the standard normal curve as probabilitiescurve as probabilities
Reexpress the following areas as probabilitiesa In Example 624 page 299 we found the area under the standard normal curve to the right of Z =- 125 to be 08944b In Example 625 we found the area under the standard normal curve between Z = -1 and Z = 1 to be 06826
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 629 continuedExample 629 continuedSolution
a
The probability that Z is greater than -125 is 08944
That is P(Z gt 125) = 08944
bThe probability that Z is between -1 and 1 is
06826
That is P(1 lt Z lt 1) = 06826
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 630 Example 630
Find the value of Z with area 090 to the left
Find the standard normal Z-value that has area 090 to the left of it
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 630 continuedExample 630 continuedSolution
Step 1Draw the standard normal curve Label the Z-value Z1
Step 2Shade the area to the left of Z1 Remember that we are given an area and
are looking for a value of Z Label the area to the left of Z1 with the given
area (090) as shown in Figure 635
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 630 continuedExample 630 continued
Step 3Look up 097 on the inside of the Z table The closest area is 09699 Move from 09699 left 18Then move up from 09699 to 008 (see
Figure 638 page 305) Putting these values together Z = 18 + 008 = 188Z-value with area 003 to its right is Z=188
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 630 continuedExample 630 continued
FIGURE 635
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
SummarySummary
The standard normal distribution has mean μ=0 and standard deviation σ=1
This distribution is often called the Z distribution
The Z table and technology can be used to find areas under the standard normal curve
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
SummarySummary
In the Z table the numbers on the outside are values of Z and the numbers inside are areas to the left of values of Z
The Z table and technology can also be used to find a value of Z given a probability or an area under the curve
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
65 Applications of the Normal 65 Applications of the Normal DistributionDistributionObjectivesBy the end of this section I will beable tohellip
1)Compute probabilities for a given value of any normal random variable
2)Find the appropriate value of any normal random variable given an area or probability
3)Calculate binomial probabilities using the normal approximation to the binomial distribution
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Standardizing a Normal Standardizing a Normal Random VariableRandom Variable
Any normal random variable X can be transformed into the standard normal random variable Z by standardizing using the formula
xZ
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 1 Determine the random variable X the mean
μ and the standard deviation σ
Draw the normal curve for X and shade the desired area
Step 2Standardize by using the formula Z = (X - μ )σ to find the values of Z corresponding to the X-values
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Finding Probabilities for Any Finding Probabilities for Any Normal DistributionNormal DistributionStep 3Draw the standard normal curve and shade
the area corresponding to the shaded area in the graph of X
Step4Find the area under the standard normal
curve using either (a) the Z table or (b) technology
This area is equal to the area under the normal curve for X drawn in Step 1
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 1
Determine X μ and σ and draw the normal curve for X
Shade the desired area
Mark the position of X1 the unknown value of X
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified ProbabilitiesStep 2
Find the Z-value corresponding to the desired area
Look up the area you identified in Step 1 on the inside of the Z table
If you do not find the exact value of your area by convention choose the area that is closest
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Finding Normal Data Values for Finding Normal Data Values for Specified ProbabilitiesSpecified Probabilities
Step 3
Transform this value of Z into a value of X which is the solution
Use the formula X1 = Zσ + μ
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 638 - Finding the X-Example 638 - Finding the X-values that mark the values that mark the Boundaries of the middle 95 of X-Boundaries of the middle 95 of X-valuesvalues
Edmundscom reported that the averageamount that people were paying for a 2007Toyota Camry XLE was $23400 Let X = price and assume that price follows a normaldistribution with μ = $23400 and σ = $1000 Find the prices that separate the middle95 of 2007 Toyota Camry XLE prices from the bottom 25 and the top 25
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 638 continuedExample 638 continued
Solution
Step 1 Determine X μ and σ and draw the normal
curve for X Let X = price μ = $23400 and σ = $1000 The middle 95 of prices are delimited by X1
and X2 as shown in Figure 649
FIGURE 649
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 638 continuedExample 638 continued
Solution
Step 2
Find the Z-values corresponding to the desired area
The area to the left of X1 equals 0025 and the area to the left of X2 equals 0975
Looking up area 0025 on the inside of the Z table gives us Z1 = ndash196
Looking up area 0975 on the inside of the Z table gives us Z2 = 196
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
Example 638 continuedExample 638 continuedSolutionStep 3 Transform using the formula X1 = Zσ + μ
X1 = Z1σ + μ
=(ndash196)(1000) + 23400 = 21440 X2 =Z2σ + μ
=(196)(1000) + 23400 = 25360The prices that separate the middle 95 of
2007 Toyota Camry XLE prices from the bottom 25 of prices and the top 25 of prices are $21440 and $25360
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
The Normal Approximation to the The Normal Approximation to the Binomial Probability DistributionBinomial Probability Distribution
For the binomial random variable X with probability of success p and number of trials n
if n p ge 5 and n (1 - p) ge 5 the binomial distribution may be approximated by a normal distribution with mean
μX = np
and standard deviation σX = 1 np p
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
SummarySummary
Section 65 showed how to solve normal probability problems for any conceivable normal random variable by first standardizing and then using the methods of Section 64
Methods for finding probabilities for a given value of the normal random variable X were discussed
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5
SummarySummary
For any normal probability distribution values of X can be found for given probabilities using the ldquobackwardrdquo methods of Section 64 and the formula X1 = Z1σ + μ
The normal distribution can be used to approximate binomial probabilities when np ge 5 and np(1 - p) ge 5