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Copyright © 2013, 2010 and 2007 Pearson Education, Inc.

Chapter

The Normal Probability Distribution

7

Chap 2 2

3

Jury Selection Problem

There are 60 women and 40 men waiting outside the courtroom in the jury pool.

Assuming the probability of being

selected for the jury is the same for each person, what is the probability that the 12 person jury selected will consist of 7 men and 5 women?


Section

Properties of the Normal Distribution

7.1

A probability density function (pdf) is an equation used to compute probabilities of

continuous random variables.

The total area under the graph of the equation over all possible values of the random variable must equal 1.

The area under the graph of the density function over an interval represents the probability of observing a value

of the random variable in that interval.

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7-6

“Probability Density Function” is the equation of the blue curve. This equation gives the “y” values of the curve as

“x” moves from left to right.

The “area under the curve” is the area between the curve (blue line) and the horizontal.

7-7

Relative frequency histograms that are symmetric /bell-shaped are said to have the

shape of a normal curve.

7-8

If a continuous random variable is “normally distributed”, or has a “normal probability

distribution”, then a rel freq histogram of that variable has the shape of a “normal” or symmetric

bell-shaped curve

Properties of the Normal Curve

1. It is symmetric about its mean, μ.2. Mean = median = mode. The curve has a

single maximum point at x = μ.3. It has inflection points at μ ± σ 4. The area under the curve is 1.5. The area under the curve to the right of μ,

as well as to the left of μ, equals 1/2.

5-9

6. As “x” increases without bound (gets further from the mean), the graph approaches, but never reaches, the horizontal axis.

7. The Empirical Rule (68-95-99.7): Approximately 68% of the area under the normal curve lies between μ ± 1 σ; approx. 95% of the area lies between μ ± 2σ; and approx. 99.7% of the area lies between μ ± 3σ .

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7-13

The data on the next slide represent the heights (in) of a random sample of 50 two-year old males.

(a) Draw a histogram of the data using first class boundaries of 31.5 and 32.5 in.

(b) Do you think that the variable: “x” = height of 2-year old males

is normally distributed?

A Normal Continuous Random Variable

7-14

36.0 36.2 34.8 36.0 34.6 38.4 35.4 36.834.7 33.4 37.4 38.2 31.5 37.7 36.9 34.034.4 35.7 37.9 39.3 34.0 36.9 35.1 37.033.2 36.1 35.2 35.6 33.0 36.8 33.5 35.035.1 35.2 34.4 36.7 36.0 36.0 35.7 35.738.3 33.6 39.8 37.0 37.2 34.8 35.7 38.937.2 39.3

HEIGHT (IN) OF TWO-YR OLD MALES

7-16

In the next slide, we have a normal density curve superimposed over the histogram.

How does the area of the rectangle corresponding to a height between 34.5 and

35.5 inches relate to the area under the curve between these two heights?

Area under a Normal Curve

Suppose that a random variable “x” is normally distributed with mean μ and standard deviation σ.

The area under the normal curve for any interval of values of “x” (say, 34.5 – 35.5 in) represents either:

1. the proportion of the population with that interval (range) of values or:

2. the probability that a randomly selected individual from the population will have that range of values.

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7-19

EXAMPLE: Area Under a Normal Curve

Suppose, the weights of adult African giraffes are normally distributed with mean μ = 2200

pounds and standard deviation σ = 200 pounds.

(a) Shade the area under the normal curve to the left of x = 2100 pounds.

(b) Suppose that the area under the normal curve to the left of x = 2100 pounds is 0.3085.

Provide two (Proportion/Probability) interpretations of this result.

7-20

EXAMPLE Area Under a Normal Curve

Adult Giraffe: μ = 2200 pounds and σ = 200 pounds

1) The proportion of giraffes whose weight is less than 2100 pounds is 0.3085 or 30.85% of the giraffe population.

2) The probability that a randomly selected giraffe weighs less than 2100 pounds is 0.3085.


Section

Applications of the Normal Distribution

7.2

Standardizing a Normal Random Variable

Suppose that the random variable X is normally distributed with mean μ and standard deviation σ. Then the random variable

is normally distributed with mean μ = 0 and standard deviation σ = 1.The random variable Z has a “standard” normal distribution.

5-22

Z X

7-23

“Standard” Normal Distribution

7-24

IQ scores can be modeled by a normal distribution with μ = 100 and σ = 15.

An IQ score of 120, is 1.33 standard deviations above the mean.

z x

120 100

151.33

25

Normal Probability Accuracy

For these probabilities, we require 4-decimal accuracy, so set your

calculator to:

Mode: Float: 4

7-26

Table IV: the area under the standard normal curve to the left of z = 1.33 is 0.9082.

7-27

Find the area to the right of z = 1.33.

7-28

Areas Under the Standard Normal Curve Empirical Rule: 68-95-99.7

7-29

Find the area under the standard normal curve to the left of z = – 0.38.

Area to the left of z = – 0.38 is 0.3520.

7-30

Find the area under the standard normal curve to the right of z = 1.25

Area right of 1.25 = 1 – area left of 1.25= 1 – 0.8944 = 0.1056

7-31

Find the area under the standard normal curve between z = –1.02 and z = 2.94

Area between z = –1.02 and 2.94= (Area left of z = 2.94) – (area left of z = –1.02)= 0.9984 – 0.1539= 0.8445

TI-84 2:Distr: normcdf… (lo, up, mn, stddev)

(using z-score): normcdf (-1,1) = 0.6827 normcdf (-3.49,-0.38) = 0.3517

normcdf (-1E99,-0.38) = 0.3520

TI-84 2:Distr: normcdf… (lo, up, mn, stddev)

If you wanted to know the probability of getting 120 or less on an IQ test whose mean is 100 and stddev is 15…

(using x-score) normcdf (0,120,100,15) = 0.9088 (using z-score): normcdf (-1E99,1.333) = 0.9087 (using lo bd of z= -3.49 gives 0.9085)

invNorm (area/prob , mean, stddev)

Find the raw score/z-score that corresponds to on the standard IQ Test (100;15)

(using x-score) InvNorm (0.85 ,100,15)

= 115.55

(using z-score) InvNorm (0.85) = 1.0364

x

85P

TI-84 sample problems…normcdf (-1.5,1.25) → 0.8275

normcdf (0,133,100,15) → 0.9861 (89,133,100,15) → 0.7544

invNorm (0.9861) → 2.2001invNorm (0.9861,100,15) → 133.0015

normcdf (-1E99,1.25) → 0.8944

7-36

The combined (verbal + quantitative) score on the GRE (Graduate Record Exam (GRE ~ to

get accepted into graduate school) is normally distributed with mean=1049 and stddev =189

Finding the Value of a Normal Variable

What is the raw test score for a student whose rank is at the 85th percentile?

7-37

Value of a Normal Random Variable

The z-score that corresponds to the 85th percentile is the z-score which

has .8500 area to its left.

TI-84 or Table V : InvNorm (0.8500) = 1.0364

So, this z-score is 1.0364.x = µ + zσ = 1049 + 1.0364(189) = 1244.88 or 1245

A person who scores 1245 on the GRE would rank in the 85th percentile, or the prob a person scores 1245 on

the GRE is 0.850

7-38

Karlos, Inc. manufactures construction-grade steel rebar rods. Their length is normally distributed with a mean of

100 cm (1 m) and a standard deviation of 0.45 cm.

Suppose Quality Control wants to accept 90% of all rods manufactured, but rebuild all rods outside that tolerance.

Determine the min/max length of these rods that make up the middle 90% of all rods manufactured.


7-39


Area = 0.05

Area = 0.05

InvNorm(0.05) = - 1.645z1 = –1.645 and z2 = 1.645x1 = µ + z1σ = 100 + (–1.645)(0.45) = 99.2598 cm

ORInvNorm(0.05,100,0.45)= 99.2598 = 99.3 cm

The steel rods that make up the middle 90% would have lengths between 99.3 cm and

100.7 cm.


Section

Assessing Normality

(Skip)

7.3


Section

The Normal Approximation to the Binomial

Probability Distribution

7.4

Criteria for a Binomial Probability Experiment

An experiment is said to be a binomial experiment if:

1. The experiment is performed “n” independent times. Each repetition is called a trial. Independence means that the outcome of one trial will not affect the outcome of the other trials.

2. For each trial, there are only two possible disjoint outcomes: Success / Failure (T/F; Yes/No, H/T, etc.)

3. The probability of success, “p” and the prob of failure “q”, are the same for each trial of the experiment.

Note: p + q =1 and q = 1- p.

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7-43

For a fixed prob of success “p”, as the number of trials “n” in a binomial experiment increases,

the probability distribution of “x” becomes more nearly symmetric and bell-shaped.

As a rule of thumb, if the variance = npq > 10, then the probability distribution will be approx normal (bell-shaped) and we can use “z” scores to predict probability of experiment outcomes.

The Normal Approximation to the Binomial Probability Distribution

If npq ≥ 10, (stddev = ≥ 3.2) , the binomial random variable “x” is approximately normally distributed with :

Mean: μX = np

Std Dev:

Note: “q” = (1- p)

5-44

X np 1 p .

npq

7-45

P(X = 18) ≈ P(17.5 < X < 18.5)

7-46

P(X < 18) ≈ P(X < 18.5)

7-47

Exact Probability Using Binomial: P(X ≤ a)

Approximate ProbabilityUsing Normal: P(X ≤ a + 0.5)

7-48

Exact Probability Using Binomial: P(X ≥ a)

Approximate ProbabilityUsing Normal: P(X ≥ a – 0.5)

7-49

Exact ProbabilityUsing Binomial: P(a ≤ X ≤ b)

Approximate ProbabilityUsing Normal: P(a=0.5 ≤ X ≤ b+0.5)

7-50

Using the Binomial Probability Distribution Function

35% of all car-owning households have three or more cars.

(a) In a sample of 400 car-owning households, what is the probability that fewer than 150 have three or more cars?(1 ) 400(0.35)(1 0.35)

91 10

np p

400(0.35)

140X

400(0.35)(1 0.35)

9.54X

P(x<150) ≈ P(x<150.5)

8428.0

)006.1(

)54.9

1405.150(

zP

xzP

7-51

Using the Binomial Probability Distribution Function

35% of all car-owning households have three or more cars.

(b) In a sample of 400 car-owning households, what is the probability that at least 160 have three or more cars?

P(X 160) P(X 159.5)

P Z 159.5 140

9.54

0.0207

Chap 2 52

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