© 2011 Pearson Education, Inc
Statistics for Business and
Economics
Chapter 4
Random Variables &
Probability Distributions
© 2011 Pearson Education, Inc
Content
1. Two Types of Random Variables
2. Probability Distributions for Discrete Random Variables
3. The Binomial Distribution
4. Poisson and Hypergeometric Distributions
5. Probability Distributions for Continuous Random Variables
6. The Normal Distribution
© 2011 Pearson Education, Inc
Content (continued)
7. Approximating a Binomial Distribution with a Normal Distribution
8. Sampling Distributions
9. The Sampling Distribution of a Sample Mean and the Central Limit Theorem
© 2011 Pearson Education, Inc
Learning Objectives
1. Develop the notion of a random variable
2. Learn that numerical data are observed values of either discrete or continuous random variables
3. Study two important types of random variables and their probability models: the binomial and normal model
4. Define a sampling distribution as the probability of a sample statistic
5. Learn that the sampling distribution of follows a normal model
x
© 2011 Pearson Education, Inc
Thinking Challenge
• You’re taking a 33 question
multiple choice test. Each
question has 4 choices.
Clueless on 1 question, you
decide to guess. What’s the
chance you’ll get it right?
• If you guessed on all 33
questions, what would be your
grade? Would you pass?
© 2011 Pearson Education, Inc
Random Variable
A random variable is a variable that assumes
numerical values associated with the random
outcomes of an experiment, where one (and only
one) numerical value is assigned to each sample
point.
© 2011 Pearson Education, Inc
Discrete
Random Variable
Random variables that can assume a countable
number (finite or infinite) of values are called
discrete.
© 2011 Pearson Education, Inc
Discrete Random Variable
Examples
Experiment Random
Variable
Possible
Values
Count Cars at Toll
Between 11:00 & 1:00
# Cars
Arriving 0, 1, 2, ..., ∞
Make 100 Sales Calls # Sales 0, 1, 2, ..., 100
Inspect 70 Radios # Defective 0, 1, 2, ..., 70
Answer 33 Questions # Correct 0, 1, 2, ..., 33
© 2011 Pearson Education, Inc
Continuous
Random Variable
Random variables that can assume values
corresponding to any of the points contained in
one or more intervals (i.e., values that are
infinite and uncountable) are called continuous.
© 2011 Pearson Education, Inc
Continuous Random Variable
Examples
Measure Time
Between Arrivals
Inter-Arrival
Time
0, 1.3, 2.78, ...
Experiment Random
Variable
Possible
Values
Weight 100 People Weight 45.1, 78, ...
Measure Part Life Hours 900, 875.9, ...
Amount spent on food $ amount 54.12, 42, ...
© 2011 Pearson Education, Inc
Discrete
Probability Distribution
The probability distribution of a discrete
random variable is a graph, table, or formula
that specifies the probability associated with each
possible value the random variable can assume.
© 2011 Pearson Education, Inc
Requirements for the
Probability Distribution of a
Discrete Random Variable x
1. p(x) ≥ 0 for all values of x
2. p(x) = 1
where the summation of p(x) is over all possible
values of x.
© 2011 Pearson Education, Inc
Discrete Probability
Distribution Example
Probability Distribution
Values, x Probabilities, p(x)
0 1/4 = .25
1 2/4 = .50
2 1/4 = .25
Experiment: Toss 2 coins. Count number of
tails.
© 1984-1994 T/Maker Co.
© 2011 Pearson Education, Inc
Visualizing Discrete
Probability Distributions
Listing Table
Formula
# Tails f(x)
Count p(x)
0 1 .25
1 2 .50
2 1 .25
p x n
x!(n – x)! ( )
! = px(1 – p)n – x
Graph
.00
.25
.50
0 1 2
x
p(x)
{ (0, .25), (1, .50), (2, .25) }
© 2011 Pearson Education, Inc
Summary Measures
1. Expected Value (Mean of probability distribution)
• Weighted average of all possible values
• = E(x) = x p(x)
2. Variance
• Weighted average of squared deviation about
mean
• 2 = E[(x 2(x 2p(x)
3. Standard Deviation
2 ●
© 2011 Pearson Education, Inc
Summary Measures
Calculation Table
x p(x) x p(x) x –
Total (x 2p(x)
(x – 2 (x – 2p(x)
xp(x)
© 2011 Pearson Education, Inc
Thinking Challenge
You toss 2 coins. You’re
interested in the number of
tails. What are the expected
value, variance, and
standard deviation of this
random variable, number of
tails? © 1984-1994 T/Maker Co.
© 2011 Pearson Education, Inc
Expected Value & Variance
Solution*
0 .25 –1.00 1.00
1 .50 0 0
2 .25 1.00 1.00
0
.50
.50
= 1.0
x p(x) x p(x) x – (x – 2 (x – 2p(x)
.25
0
.25
2.50
.71
© 2011 Pearson Education, Inc
Probability Rules for Discrete
Random Variables
Let x be a discrete random variable with probability
distribution p(x), mean µ, and standard deviation .
Then, depending on the shape of p(x), the following
probability statements can be made:
Chebyshev’s Rule Empirical Rule
P x x µ 0 .68
P x 2 x µ 2 34
.95
P x 3 x µ 3 89
1.00
© 2011 Pearson Education, Inc
Binomial Distribution
Number of ‘successes’ in a sample of n
observations (trials)
• Number of reds in 15 spins of roulette wheel
• Number of defective items in a batch of 5 items
• Number correct on a 33 question exam
• Number of customers who purchase out of 100
customers who enter store (each customer is
equally likely to pyrchase)
© 2011 Pearson Education, Inc
Binomial Probability
Characteristics of a Binomial Experiment
1. The experiment consists of n identical trials.
2. There are only two possible outcomes on each trial. We
will denote one outcome by S (for success) and the other
by F (for failure).
3. The probability of S remains the same from trial to trial.
This probability is denoted by p, and the probability of
F is denoted by q. Note that q = 1 – p.
4. The trials are independent.
5. The binomial random variable x is the number of S’s in
n trials.
© 2011 Pearson Education, Inc
Binomial Probability
Distribution !
( ) (1 )! ( )!
x n x x n xn n
p x p q p px x n x
p(x) = Probability of x ‘Successes’
p = Probability of a ‘Success’ on a single trial
q = 1 – p
n = Number of trials
x = Number of ‘Successes’ in n trials
(x = 0, 1, 2, ..., n)
n – x = Number of failures in n trials
© 2011 Pearson Education, Inc
Binomial Probability
Distribution Example
3 5 3
!( ) (1 )
!( )!
5!(3) .5 (1 .5)
3!(5 3)!
.3125
x n xnp x p p
x n x
p
Experiment: Toss 1 coin 5 times in a row. Note
number of tails. What’s the probability of 3 tails?
© 1984-1994 T/Maker Co.
© 2011 Pearson Education, Inc
Binomial Probability Table
(Portion) n = 5 p
k .01 … 0.50 … .99
0 .951 … .031 … .000
1 .999 … .188 … .000
2 1.000 … .500 … .000
3 1.000 … .812 … .001
4 1.000 … .969 … .049
Cumulative Probabilities
p(x ≤ 3) – p(x ≤ 2) = .812 – .500 = .312
© 2011 Pearson Education, Inc
Binomial Distribution
Characteristics
.0
.5
1.0
0 1 2 3 4 5
X
P(X)
.0
.2
.4
.6
0 1 2 3 4 5
X
P(X)
n = 5 p = 0.1
n = 5 p = 0.5
E(x) np
Mean
Standard Deviation
npq
© 2011 Pearson Education, Inc
Binomial Distribution
Thinking Challenge
You’re a telemarketer selling service
contracts for Macy’s. You’ve sold 20
in your last 100 calls (p = .20). If you
call 12 people tonight, what’s the
probability of
A. No sales?
B. Exactly 2 sales?
C. At most 2 sales?
D. At least 2 sales?
© 2011 Pearson Education, Inc
Binomial Distribution Solution*
n = 12, p = .20
A. p(0) = .0687
B. p(2) = .2835
C. p(at most 2) = p(0) + p(1) + p(2) = .0687 + .2062 + .2835 = .5584
D. p(at least 2) = p(2) + p(3)...+ p(12) = 1 – [p(0) + p(1)] = 1 – .0687 – .2062 = .7251
© 2011 Pearson Education, Inc
Poisson Distribution
1. Number of events that occur in an interval
• events per unit
— Time, Length, Area, Space
2. Examples
• Number of customers arriving in 20 minutes
• Number of strikes per year in the U.S.
• Number of defects per lot (group) of DVD’s
© 2011 Pearson Education, Inc
Characteristics of a Poisson
Random Variable
1. Consists of counting number of times an event occurs during a given unit of time or in a given area or volume (any unit of measurement).
2. The probability that an event occurs in a given unit of time, area, or volume is the same for all units.
3. The number of events that occur in one unit of time, area, or volume is independent of the number that occur in any other mutually exclusive unit.
4. The mean number of events in each unit is denoted by .
© 2011 Pearson Education, Inc
Poisson Probability
Distribution Function
2
p(x) = Probability of x given
= Mean (expected) number of events in unit
e = 2.71828 . . . (base of natural logarithm)
x = Number of events per unit
p x x
( ) !
x
e
–
(x = 0, 1, 2, 3, . . .)
© 2011 Pearson Education, Inc
Poisson Probability
Distribution Function
.0
.2
.4
.6
.8
0 1 2 3 4 5
X
P(X)
.0
.1
.2
.3
0 2 4 6 8 10
X
P(X)
= 0.5
= 6
Mean
Standard Deviation
E(x)
© 2011 Pearson Education, Inc
Poisson Distribution Example
Customers arrive at a
rate of 72 per hour.
What is the probability
of 4 customers arriving
in 3 minutes?
© 1995 Corel Corp.
© 2011 Pearson Education, Inc
Poisson Distribution Solution
72 Per Hr. = 1.2 Per Min. = 3.6 Per 3 Min. Interval
-
4 -3.6
( )!
3.6(4) .1912
4!
x ep x
x
ep
© 2011 Pearson Education, Inc
Poisson Probability Table
(Portion) x
0 … 3 4 … 9
.02 .980 …
: : : : : : :
3.4 .033 … .558 .744 … .997
3.6 .027 … .515 .706 … .996
3.8 .022 … .473 .668 … .994
: : : : : : :
Cumulative Probabilities
p(x ≤ 4) – p(x ≤ 3) = .706 – .515 = .191
© 2011 Pearson Education, Inc
Thinking Challenge
You work in Quality Assurance
for an investment firm. A clerk
enters 75 words per minute
with 6 errors per hour. What is
the probability of 0 errors in a
255-word bond transaction?
© 1984-1994 T/Maker Co.
© 2011 Pearson Education, Inc
Poisson Distribution Solution:
Finding * • 75 words/min = (75 words/min)(60 min/hr)
= 4500 words/hr
• 6 errors/hr = 6 errors/4500 words
= .00133 errors/word
• In a 255-word transaction (interval):
= (.00133 errors/word )(255 words)
= .34 errors/255-word transaction
© 2011 Pearson Education, Inc
Poisson Distribution Solution:
Finding p(0)*
-
0 -.34
( )!
.34(0) .7118
0!
x ep x
x
ep
© 2011 Pearson Education, Inc
Characteristics of a
Hypergeometric
Random Variable
1. The experiment consists of randomly drawing n elements without replacement from a set of N elements, r of which are S’s (for success) and (N – r) of which are F’s (for failure).
2. The hypergeometric random variable x is the number of S’s in the draw of n elements.
© 2011 Pearson Education, Inc
Hypergeometric Probability
Distribution Function
where . . .
[x = Maximum [0, n – (N – r), …,
Minimum (r, n)] p x
r
x
N r
n x
N
n
µ nr
N 2
r N r n N n N 2 N 1
© 2011 Pearson Education, Inc
Hypergeometric Probability
Distribution Function
N = Total number of elements
r = Number of S’s in the N elements
n = Number of elements drawn
x = Number of S’s drawn in the n elements
© 2011 Pearson Education, Inc
Continuous Probability
Density Function
The graphical form of the probability distribution for a
continuous random variable x is a smooth curve
© 2011 Pearson Education, Inc
Continuous Probability
Density Function
This curve, a function of x, is denoted by the symbol f(x)
and is variously called a probability density function
(pdf), a frequency function, or a probability
distribution.
The areas under a probability
distribution correspond to
probabilities for x. The area A
beneath the curve between two
points a and b is the probability
that x assumes a value between a and b.
© 2011 Pearson Education, Inc
Importance of
Normal Distribution
1. Describes many random processes or
continuous phenomena
2. Can be used to approximate discrete
probability distributions
• Example: binomial
3. Basis for classical statistical inference
© 2011 Pearson Education, Inc
Normal Distribution
1. ‘Bell-shaped’ &
symmetrical
2. Mean, median, mode
are equal x
f ( x )
Mean
Median
Mode
© 2011 Pearson Education, Inc
Probability Density Function
where
µ = Mean of the normal random variable x
= Standard deviation
π = 3.1415 . . .
e = 2.71828 . . .
P(x < a) is obtained from a table of normal probabilities
f (x) 1
2e
1
2
x
2
© 2011 Pearson Education, Inc
Normal Distribution
Probability
c d x
f ( x )
Probability is
area under
curve! P(c x d) f (x)
c
d
dx?
© 2011 Pearson Education, Inc
Standard Normal Distribution
The standard normal distribution is a normal
distribution with µ = 0 and = 1. A random variable
with a standard normal distribution, denoted by the
symbol z, is called a standard normal random variable.
© 2011 Pearson Education, Inc
z = 0
= 1
1.96
Z .04 .05
1.8 .4671 .4678 .4686
.4738 .4744
2.0 .4793 .4798 .4803
2.1 .4838 .4842 .4846
The Standard Normal Table:
P(0 < z < 1.96)
.06
1.9 .4750
Standardized Normal
Probability Table (Portion)
Probabilities
.4750
Shaded area
exaggerated
© 2011 Pearson Education, Inc
The Standard Normal Table:
P(–1.26 z 1.26)
z = 0
= 1
–1.26
Standardized Normal Distribution
Shaded area exaggerated
.3962
1.26
.3962 P(–1.26 ≤ z ≤ 1.26)
= .3962 + .3962
= .7924
© 2011 Pearson Education, Inc
The Standard Normal Table:
P(z > 1.26)
z = 0
= 1
Standardized Normal Distribution
1.26
P(z > 1.26)
= .5000 – .3962
= .1038
.3962
.5000
© 2011 Pearson Education, Inc
The Standard Normal Table:
P(–2.78 z –2.00)
= 1
= 0
–2.78 z –2.00
.4973
.4772
Standardized Normal Distribution
Shaded area exaggerated
P(–2.78 ≤ z ≤ –2.00)
= .4973 – .4772
= .0201
© 2011 Pearson Education, Inc
The Standard Normal Table:
P(z > –2.13)
z = 0
= 1
–2.13
Standardized Normal Distribution
Shaded area exaggerated
P(z > –2.13)
= .4834 + .5000
= .9834
.5000 .4834
© 2011 Pearson Education, Inc
x
f(x)
Non-standard Normal
Distribution
Normal distributions differ by
mean & standard deviation.
Each distribution would
require its own table.
That’s an infinite
number of tables!
© 2011 Pearson Education, Inc
Property of Normal Distribution
If x is a normal random variable with mean μ and
standard deviation , then the random variable z,
defined by the formula
has a standard normal distribution. The value z describes
the number of standard deviations between x and µ.
z x µ
© 2011 Pearson Education, Inc
Standardize the
Normal Distribution
Normal
Distribution
x
One table!
= 0
= 1
z
Standardized Normal
Distribution
z x
© 2011 Pearson Education, Inc
Finding a Probability Corresponding
to a Normal Random Variable
1. Sketch normal distribution, indicate mean, and shade
the area corresponding to the probability you want.
2. Convert the boundaries of the shaded area from x
values to standard normal random variable z
z x µ
Show the z values under corresponding x values.
3. Use Table IV in Appendix A to find the areas
corresponding to the z values. Use symmetry when
necessary.
© 2011 Pearson Education, Inc
z = 0
= 1
.12
Standardized Normal
Distribution
Shaded area exaggerated
.0478
Non-standard Normal μ = 5, σ = 10:
P(5 < x < 6.2)
Normal
Distribution
x = 5
= 10
6.2
z
x
6.2 5
10 .12
© 2011 Pearson Education, Inc
z = 0
= 1
-.12
Standardized Normal
Distribution
Non-standard Normal μ = 5, σ = 10:
P(3.8 x 5)
Normal
Distribution
x = 5
= 10
3.8
.0478
Shaded area exaggerated
z
x
3.8 5
10 .12
© 2011 Pearson Education, Inc
0
= 1
-.21 z .21
Standardized Normal
Distribution
Non-standard Normal μ = 5, σ = 10:
P(2.9 x 7.1)
5
= 10
2.9 7.1 x
Normal
Distribution
.1664
.0832 .0832
Shaded area exaggerated
z
x
2.9 5
10 .21
z
x
7.1 5
10 .21
© 2011 Pearson Education, Inc
Non-standard Normal μ = 5, σ = 10:
P(x 8)
x = 5
= 10
8
Normal
Distribution
z = 0 .30
Standardized Normal
Distribution
= 1
.3821
.5000
.1179
Shaded area exaggerated
z
x
8 5
10 .30
© 2011 Pearson Education, Inc
= 0
= 1
.30 z .21
Standardized Normal
Distribution
Non-standard Normal μ = 5, σ = 10:
P(7.1 X 8)
= 5
= 10
8 7.1 x
Normal
Distribution
.1179 .0347
.0832
Shaded area exaggerated
z
x
7.1 5
10 .21
z
x
8 5
10 .30
© 2011 Pearson Education, Inc
Normal Distribution Thinking
Challenge You work in Quality Control for
GE. Light bulb life has a normal
distribution with = 2000 hours
and = 200 hours. What’s the
probability that a bulb will last
A. between 2000 and 2400
hours?
B. less than 1470 hours?
© 2011 Pearson Education, Inc
Standardized Normal
Distribution
z = 0
= 1
2.0
Solution* P(2000 x 2400)
Normal
Distribution
x = 2000
= 200
2400
.4772
z
x
2400 2000
200 2.0
© 2011 Pearson Education, Inc
z = 0
= 1
–2.65
Standardized Normal
Distribution
Solution* P(x 1470)
x = 2000
= 200
1470
Normal
Distribution
.0040 .4960
.5000
z
x
1470 2000
200 2.65
© 2011 Pearson Education, Inc
Finding z-Values
for Known Probabilities
What is Z, given
P(z) = .1217?
Shaded area
exaggerated
z = 0
= 1
?
.1217
Standardized Normal
Probability Table (Portion)
Z .00 0.2
0.0 .0000 .0040 .0080
0.1 .0398 .0438 .0478
0.2 .0793 .0832 .0871
.1179 .1255
.01
0.3 .1217
.31
© 2011 Pearson Education, Inc
Finding x Values
for Known Probabilities
Normal Distribution
x = 5
= 10
?
.1217
Standardized Normal Distribution
Shaded areas exaggerated
z = 0
= 1
.31
.1217
8.1
x z 5 .31 10
© 2011 Pearson Education, Inc
Normal Approximation of
Binomial Distribution
1. Useful because not all
binomial tables exist
2. Requires large sample
size
3. Gives approximate
probability only
4. Need correction for
continuity
n = 10 p = 0.50
.0
.1
.2
.3
0 2 4 6 8 10
x
p(x)
© 2011 Pearson Education, Inc
.0
.1
.2
.3
0 2 4 6 8 10
x
p(x)
Why Probability
Is Approximate
Binomial Probability:
Bar Height
Normal Probability: Area Under
Curve from 3.5 to 4.5
Probability Added
by Normal Curve
Probability Lost by
Normal Curve
© 2011 Pearson Education, Inc
Correction for Continuity
1. A 1/2 unit adjustment to
discrete variable
2. Used when approximating
a discrete distribution
with a continuous
distribution
3. Improves accuracy
4.5
(4 + .5)
3.5
(4 – .5) 4
© 2011 Pearson Education, Inc
Using a Normal Distribution to
Approximate Binomial
Probabilities
1. Determine n and p for the binomial distribution,
then calculate the interval:
If interval lies in the range 0 to n, the normal
distribution will provide a reasonable
approximation to the probabilities of most
binomial events.
3 np 3 np 1 p
© 2011 Pearson Education, Inc
Using a Normal Distribution to
Approximate Binomial
Probabilities 2. Express the binomial probability to be
approximated by the form
For example,
P x a or P x b P x a
P x 3 P x 2
P x 5 1 P x 4
P 7 x 10 P x 10 P x 6
© 2011 Pearson Education, Inc
Using a Normal Distribution to
Approximate Binomial
Probabilities 3. For each value of interest a, the correction for
continuity is (a + .5), and the corresponding
standard normal z-value is
z a .5 µ
© 2011 Pearson Education, Inc
Using a Normal Distribution to
Approximate Binomial
Probabilities 4. Sketch the approximating normal distribution and
shade the area corresponding to the event of
interest. Using Table IV and the z-value (step 3),
find the shaded area.
This is the approximate
probability of the
binomial event.
© 2011 Pearson Education, Inc
.0
.1
.2
.3
0 2 4 6 8 10
x
P(x)
Normal Approximation Example
3.5 4.5
What is the normal approximation of p(x = 4)
given n = 10, and p = 0.5?
© 2011 Pearson Education, Inc
Normal Approximation Solution
1. Calculate the interval:
2. Express binomial probability in form:
P x 4 P x 4 P x 3
np 3 np 1 p 10 0.5 3 10 0.5 1 0.5
5 4.74 0.26, 9.74
Interval lies in range 0 to 10, so normal
approximation can be used
© 2011 Pearson Education, Inc
Normal Approximation Solution
3. Compute standard normal z values:
z a .5 n p
n p 1 p
3.5 10 .5
10 .5 1 .5 0.95
z a .5 n p
n p 1 p
4.5 10 .5
10 .5 1 .5 0.32
© 2011 Pearson Education, Inc
= 0 = 1
-.32 z -.95
Normal Approximation Solution
.1255
.3289
- .1255
.2034
.3289
4. Sketch the approximate normal distribution:
© 2011 Pearson Education, Inc
Normal Approximation Solution
.0
.1
.2
.3
0 2 4 6 8 10
x
p(x)
5. The exact probability from the binomial formula is
0.2051 (versus .2034)
© 2011 Pearson Education, Inc
Parameter & Statistic
A parameter is a numerical descriptive measure
of a population. Because it is based on all the
observations in the population, its value is almost
always unknown.
A sample statistic is a numerical descriptive
measure of a sample. It is calculated from the
observations in the sample.
© 2011 Pearson Education, Inc
Common Statistics & Parameters
Sample Statistic Population Parameter
Variance s2 2
Standard
Deviation s
Mean x
Binomial
Proportion p p ̂
© 2011 Pearson Education, Inc
The sampling distribution of a sample statistic
calculated from a sample of n measurements is
the probability distribution of the statistic.
Sampling Distribution
© 2011 Pearson Education, Inc
Developing
Sampling Distributions
• Population size, N = 4
• Random variable, x
• Values of x: 1, 2, 3, 4
• Uniform distribution
© 1984-1994 T/Maker Co.
Suppose There’s a Population ...
© 2011 Pearson Education, Inc
Population Characteristics
xi
i1
N
N 2.5
Population Distribution Summary Measure
.0
.1
.2
.3
1 2 3 4
P(x)
x
© 2011 Pearson Education, Inc
All Possible Samples
of Size n = 2
Sample with replacement
1.0 1.5 2.0 2.5
1.5 2.0 2.5 3.0
2.0 2.5 3.0 3.5
2.5 3.0 3.5 4.0
16 Samples
1st
Obs
1,1 1,2 1,3 1,4
2,1 2,2 2,3 2,4
3,1 3,2 3,3 3,4
4,1 4,2 4,3 4,4
2nd Observation
1 2 3 4
1
2
3
4
2nd Observation
1 2 3 4
1
2
3
4
1st
Obs
16 Sample Means
© 2011 Pearson Education, Inc
Sampling Distribution
of All Sample Means
1.0 1.5 2.0 2.5
1.5 2.0 2.5 3.0
2.0 2.5 3.0 3.5
2.5 3.0 3.5 4.0
2nd Observation
1 2 3 4
1
2
3
4
1st
Obs
16 Sample Means Sampling Distribution
of the Sample Mean
.0
.1
.2
.3
1.0 1.5 2.0 2.5 3.0 3.5 4.0
P(x)
x
© 2011 Pearson Education, Inc
Summary Measure of
All Sample Means
X
xi
i1
N
N
1.0 1.5 ... 4.0
16 2.5
© 2011 Pearson Education, Inc
Comparison
Population Sampling Distribution
2.5x
.0
.1
.2
.3
1 2 3 4
2.5
.0
.1
.2
.3
1.0 1.5 2.0 2.5 3.0 3.5 4.0
P(x)
x
P(x)
x
© 2011 Pearson Education, Inc
4.11
The Sampling Distribution of
a Sample Mean and the
Central Limit Theorem
© 2011 Pearson Education, Inc
Properties of the Sampling
Distribution of x
1. Mean of the sampling distribution equals mean of
sampled population*, that is,
x E x .
2. Standard deviation of the sampling distribution equals
Standard deviation of sampled population
Square root of sample size
x
n.That is,
© 2011 Pearson Education, Inc
Standard Error of the Mean
The standard deviation is often referred to
as the standard error of the mean.
x
© 2011 Pearson Education, Inc
Theorem 4.1
If a random sample of n observations is selected from
a population with a normal distribution, the sampling
distribution of will be a normal distribution. x
© 2011 Pearson Education, Inc
n =16x = 2.5
Sampling from
Normal Populations
• Central Tendency
• Dispersion
– Sampling with
replacement
= 50
= 10
x
n = 4x = 5
x
= 50 - x
Sampling Distribution
Population Distribution
x
xn
© 2011 Pearson Education, Inc
Standardizing the Sampling
Distribution of x
Standardized Normal
Distribution
= 0
= 1
z
z x
x
x
x
nSampling
Distribution
x x
x
© 2011 Pearson Education, Inc
Thinking Challenge
You’re an operations analyst
for AT&T. Long-distance
telephone calls are normally
distributed with = 8 min.
and = 2 min. If you select
random samples of 25 calls,
what percentage of the
sample means would be
between 7.8 & 8.2 minutes?
© 1984-1994 T/Maker Co.
© 2011 Pearson Education, Inc
Sampling Distribution Solution*
Sampling
Distribution
8
x
= .4
7.8 8.2 x 0
= 1
–.50 z .50
.3830
Standardized Normal
Distribution
.1915 .1915
z x
n
7.8 8
225
.50
z x
n
8.2 8
225
.50
© 2011 Pearson Education, Inc
Sampling from
Non-Normal Populations
• Central Tendency
• Dispersion
– Sampling with
replacement
Population Distribution
Sampling Distribution
n =30x = 1.8
n = 4x = 5
= 50
= 10
x
x
= 50 - x
x
xn
© 2011 Pearson Education, Inc
Central Limit Theorem
Consider a random sample of n observations selected
from a population (any probability distribution) with
mean μ and standard deviation . Then, when n is
sufficiently large, the sampling distribution of will be
approximately a normal distribution with mean
and standard deviation The larger the
sample size, the better will be the normal approximation
to the sampling distribution of .
x x
x n .
x
© 2011 Pearson Education, Inc
Central Limit Theorem
x
As sample
size gets
large
enough
(n 30) ...
sampling
distribution
becomes almost
normal.
x
xn
© 2011 Pearson Education, Inc
Central Limit Theorem Example
SODA
The amount of soda in cans of a
particular brand has a mean of 12
oz and a standard deviation of .2
oz. If you select random samples
of 50 cans, what percentage of
the sample means would be less
than 11.95 oz?
© 2011 Pearson Education, Inc
Central Limit Theorem Solution*
Sampling
Distribution
12
x = .03
11.95 x 0
= 1
–1.77 z
.0384
Standardized Normal
Distribution
.4616
z x
n
11.9512
.250
1.77
Shaded area exaggerated
© 2011 Pearson Education, Inc
Key Ideas
Properties of Probability Distributions
Discrete Distributions
1. p(x) ≥ 0
2.
Continuous Distributions
1. P(x = a) = 0
2. P(a < x < b) = area under curve between a and b
p x 1all x
© 2011 Pearson Education, Inc
Key Ideas
Normal Approximation to Binomial
x is binomial (n, p)
P x a P z a .5 µ
© 2011 Pearson Education, Inc
Key Ideas
Methods for Assessing Normality
2. Stem-and-leaf display
1 7
2 3389
3 245677
4 19
5 2
© 2011 Pearson Education, Inc
Key Ideas
Methods for Assessing Normality
3. (IQR)/S ≈ 1.3
4. Normal probability plot