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Chapter 3
Chapter
Chapter Outline
5
Normal Probability
Distributions
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Section 5.4 Objectives
• How to find sampling distributions and verify their
properties
• How to interpret the Central Limit Theorem
• How to apply the Central Limit Theorem to find the
probability of a sample mean
Section 5.4
Sampling Distributions and the
Central Limit Theorem
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Sampling Distributions
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Sampling Distribution of Sample Means
Sampling distribution
• The probability distribution of a sample statistic.
• Formed when samples of size n are repeatedly taken
from a population.
• e.g. Sampling distribution of sample means
Population with μ, σ
Sample 5
Sample 3
x
Sample 1
x1
Sample 2
x
2
3
Sample 4
x5
x4
The sampling distribution consists of the values of the
sample means, x 1 , x 2 , x 3 , x 4 , x 5 , . . .
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Chapter 3
Properties of Sampling Distributions of
Sample Means
Example: Sampling Distribution of
Sample Means
1. The mean of the sample means, m x , is equal to the
population mean μ.
mx = m
The population values {1, 3, 5, 7} are written on slips of
paper and put in a box. Two slips of paper are randomly
selected, with replacement.
a. Find the mean, variance, and standard deviation of
the population.
2. The standard deviation of the sample means, s x , is
equal to the population standard deviation, σ
divided by the square root of the sample size, n.
sx =
Solution:
s
Mean:
n
S( x - m )2
=5
N
Standard Deviation: s = 5 » 2.236
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Example: Sampling Distribution of
Sample Means
c. List all the possible samples of size n = 2 and
calculate the mean of each sample.
Solution:
Solution:
Sample
1, 1
1, 3
1, 5
1, 7
3, 1
3, 3
3, 5
3, 7
Probability Histogram of
Population of x
0.25
Probability
All values have the
same probability of
being selected (uniform
distribution)
x
1
3
5
7
Population values
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Example: Sampling Distribution of
Sample Means
1
0.0625
2
3
4
5
2
3
4
3
0.1250
0.1875
0.2500
0.1875
6
7
2
1
0.1250
0.0625
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Larson/Farber 6th ed
x
1
2
3
4
2
3
4
5
Sample
5, 1
5, 3
5, 5
5, 7
7, 1
7, 3
7, 5
7, 7
x
3
4
5
6
4
5
6
7
These means
form the
sampling
distribution of
sample means.
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Example: Sampling Distribution of
Sample Means
d. Construct the probability distribution of the sample
means.
Solution:
f
Probability
x x f Probability
1
8
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Example: Sampling Distribution of
Sample Means
b. Graph the probability histogram for the population
values.
P(x)
Sx
=4
N
Variance: s 2 =
• Called the standard error of the mean.
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m=
e. Find the mean, variance, and standard deviation of
the sampling distribution of the sample means.
Solution:
The mean, variance, and standard deviation of the
16 sample means are:
mx = 4
s x2 =
5
= 2.5
2
s x = 2.5 » 1.581
These results satisfy the properties of sampling
distributions of sample means.
mx = m = 4
11
s x = s = 5 » 2.236 » 1.581
n
2
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Chapter 3
The Central Limit Theorem
Example: Sampling Distribution of
Sample Means
1. If samples of size n ³ 30, are drawn from any
population with mean = m and standard deviation = s,
f. Graph the probability histogram for the sampling
distribution of the sample means.
Solution:
P(x)
Probability
0.25
Probability Histogram of
Sampling Distribution of x
0.20
0.15
0.10
0.05
x
2
3
4
5
6
x
m
then the sampling distribution of the sample means
approximates a normal distribution. The greater the
sample size, the better the approximation.
The shape of the
graph is symmetric
and bell shaped. It
approximates a
normal distribution.
xx
x x
x x x
x x x x x
7
m
Sample mean
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The Central Limit Theorem
x
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The Central Limit Theorem
2. If the population itself is normally distributed,
• In either case, the sampling distribution of sample
means has a mean equal to the population mean.
mx = m
• The sampling distribution of sample means has a
variance equal to 1/n times the variance of the
population and a standard deviation equal to the
population standard deviation divided by the square
root of n.
2
x
m
the sampling distribution of the sample means is
normally distribution for any sample size n.
xx
x x
x x x
x x x x x
s x2 = s
n
x
m
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sx =
15
Any Population Distribution
Distribution of Sample Means,
n ≥ 30
2.
Normal Population Distribution
Larson/Farber 6th ed
Standard deviation (standard
error of the mean)
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Cellular phone bills for residents of a city have a mean
of $63 and a standard deviation of $11. Random
samples of 100 cellular phone bills are drawn from this
population and the mean of each sample is determined.
Find the mean and standard error of the mean of the
sampling distribution. Then sketch a graph of the
sampling distribution of sample means.
Distribution of Sample Means,
(any n)
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n
Example: Interpreting the Central Limit
Theorem
The Central Limit Theorem
1.
s
Variance
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Chapter 3
Solution: Interpreting the Central Limit
Theorem
Solution: Interpreting the Central Limit
Theorem
• The mean of the sampling distribution is equal to the
population mean
m x = m = 63
• Since the sample size is greater than 30, the sampling
distribution can be approximated by a normal
distribution with
s x = $1.10
m x = $63
• The standard error of the mean is equal to the
population standard deviation divided by the square
root of n.
s x = s = 11 = 1.1
n
100
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Example: Interpreting the Central Limit
Theorem
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Solution: Interpreting the Central Limit
Theorem
The training heart rates of all 20-years old athletes are
normally distributed, with a mean of 135 beats per
minute and standard deviation of 18 beats per minute.
Random samples of size 4 are drawn from this
population, and the mean of each sample is determined.
Find the mean and standard error of the mean of the
sampling distribution. Then sketch a graph of the
sampling distribution of sample means.
• The mean of the sampling distribution is equal to the
population mean
m x = m = 135
• The standard error of the mean is equal to the
population standard deviation divided by the square
root of n.
s x = s = 18 = 9
n
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Solution: Interpreting the Central Limit
Theorem
Larson/Farber 6th ed
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Probability and the Central Limit
Theorem
• Since the population is normally distributed, the
sampling distribution of the sample means is also
normally distributed.
sx = 9
m x = 135
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• To transform x to a z-score
z=
23 of 104
23
x - mx x - m
Value-Mean
=
=
s
Standard Error
sx
n
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Chapter 3
Example: Probabilities for Sampling
Distributions
Solution: Probabilities for Sampling
Distributions
The graph shows the length of
time people spend driving each
day. You randomly select 50
drivers age 15 to 19. What is the
probability that the mean time
they spend driving each day is
between 24.7 and 25.5 minutes?
Assume that σ = 1.5 minutes.
From the Central Limit Theorem (sample size is greater
than 30), the sampling distribution of sample means is
approximately normal with
m x = m = 25
25
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Solution: Probabilities for Sampling
Distributions
Normal Distribution
μ = 25 σ = 0.21213
24.7 - 25
z1 =
=
= -1.41
s
1 .5
n
50
P(24.7 < x < 25.5)
z2 =
x-m
s
n
=
25
25.5
z
0
2.36
P(24 < x < 54) = P(-1.41 < z < 2.36)
= 0.9909 – 0.0793 = 0.9116
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Solution: Probabilities for x and x
Normal Distribution
μ = 3173 σ = 1120
z=
P(x < 2700)
s
=
2700 - 3173
» -0.42
1120
P(z < -0.42)
0.3372
x
2500 2870
z
-0.42
0
Larson/Farber 6th ed
An education finance corporation claims that the
average credit card debts carried by undergraduates are
normally distributed, with a mean of $3173 and a
standard deviation of $1120. (Adapted from Sallie Mae)
Solution:
You are asked to find the probability associated with
a certain value of the random variable x.
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2. You randomly select 25 undergraduates who are
credit card holders. What is the probability that
their mean credit card balance is less than $2700?
Solution:
You are asked to find the probability associated with
a sample mean x.
m x = m = 3173
P( x < 2700) = P(z < 0.42) = 0.3372
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Example: Probabilities for x and x
Standard Normal Distribution
μ=0 σ=1
x-m
1.5
» 0.21213
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1. What is the probability that a randomly selected
undergraduate, who is a credit card holder, has a
credit card balance less than $2700?
0.9909
0.0793
-1.41
=
P(-1.41 < z < 2.36)
25.5 - 25
= 2.36
1.5
50
x
24.7
n
Example: Probabilities for x and x
Standard Normal Distribution
μ=0 σ=1
x-m
s
sx =
29
s x = s = 1120 = 224
n
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Chapter 3
Solution: Probabilities for x and x
Normal Distribution
μ = 3173 σ = 1120
z=
Standard Normal Distribution
μ=0 σ=1
x-m
s
Solution: Probabilities for x and x
n
=
2700 - 3173 -473
=
» -2.11
1120
224
25
P(z < -2.11)
P(x < 2700)
0.0174
x
2500
2870
z
-2.11
0
• There is a 34% chance that an undergraduate will
have a balance less than $2700.
• There is only a 2% chance that the mean of a sample
of 25 will have a balance less than $2700 (unusual
event).
• It is possible that the sample is unusual or it is
possible that the corporation’s claim that the mean is
$3173 is incorrect.
P( x < 2700) = P(z < -2.11) = 0.0174
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Section 5.4 Summary
• Found sampling distributions and verify their
properties
• Interpreted the Central Limit Theorem
• Applied the Central Limit Theorem to find the
probability of a sample mean
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Larson/Farber 6th ed
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