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Chapter 09

OMIS 2010 Chapter Notes - Chapter 09: Sampling Distribution, Binomial Distribution, Random Variable


Department
Operations Management and Information System
Course Code
OMIS 2010
Professor
Alan Marshall
Chapter
09

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Chapter 9: Sampling Distributions
9.1 Introduction
This chapter connects the material in Chapters 4 through 8 (numerical descriptive statistics,
sampling, and probability distributions, in particular) with statistical inference, which is introduced in
Chapter 10.
At the completion of this chapter, you are expected to know the following:
1. How the sampling distribution of the mean is created and the shape and parameters of the
distribution.
2. How to calculate probabilities using the sampling distribution of the mean.
3. Understand how the normal distribution can be used to approximate the binomial distribution.
4. How to calculate probabilities associated with a sample proportion.
5. How to calculate probabilities associated with the difference between two sample means.
9.2 Sampling Distribution of the Mean
The most important thing to learn from this section is that if we repeatedly draw samples from any
population, the values of
x
calculated in each sample will vary. This new random variable created by
sampling will have three important characteristics:
1.
x
is approximately normally.
2. The mean of
x
will equal the mean of the original random variable. That is, µx
x.
3. The variance of
x
will equal the variance of the original random variable divided by n. That is,
σx
2
x
2/n.
The sampling distribution of
x
allows us to make probability statements about
x
based on knowing the
values of the sample size n and the population parameters µ and
σ
2.
Example 9.1
A random variable possesses the following probability distribution:
x p(x)
1 .2
2 .5
3 .3
106

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a) Find all possible samples of size 2 that can be drawn from this population.
b) Using the results in part a), find the sampling distribution of
x
.
c) Confirm that
µx
x and σx
2
=
σ
x
2/n.
Solution
a) There are nine possible samples of size 2. They are (1,1), (1,2), (1,3), (2,1), (2,2), (2,3), (3,1),
(3,2), and (3,3).
b) The samples, the values of
x
, and the probability of each sample outcome are shown below:
Sample
x
Probability
(1,1) 1.0 (.2)(.2) = .04
(1,2) 1.5 (.2)(.5) = .10
(1,3) 2.0 (.2)(.3) = .06
(2,1) 1.5 (.5)(.2) = .10
(2,2) 2.0 (.5)(.5) = .25
(2,3) 2.5 (.5)(.3) = .15
(3,1) 2.0 (.3)(.2) = .06
(3,2) 2.5 (.3)(.5) = .15
(3,3) 3.0 (.3)(.3) = .09
The sampling distribution of
x
follows:
x
p(
x
)
1.0 .04
1.5 .20
2.0
2.5
3.0
.37
.30
.09
c) Using our definitions of expected value and variance, we find the mean and variance of the
random variable x:
µx=E(x)=xp(x)
=
1(.2)
+
2(.5)
+
3(.3)
=
2.1
σx
2=(x−µ)2
p(x)=(1 2.1)2(.2) +(2 2.1)2(.5) +(3 2.1)2(.3)
=0.49
107

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The mean and variance of the random variable
x
are computed as follows:
µx =E(x )=x p(x )
=1.0(.04) +1.5(.20) +2.0(.37) +2.5(.30) +3.0(.09)
=2.1
σx
2=x −µ
x
()
2
p(x )
=(1.0 2.1)2(.04) +(1.5 2.1)2(.20) +(2.0 2.1)2(.37)
+(2.5 2.1)2(.30) +(3.0 2.1)2(.09)
=0.245
As you can see,
µx
x=2.1
and
σx
2
x
2/n=0.49/2
=
0.245
Example 9.2
Suppose a random sample of 100 observations is drawn from a normal population whose mean is
600 and whose variance is 2,500. Find the following probabilities:
a) P(590 < x < 610)
b) P(590 <
x
< 610)
c) P(x > 650)
d) P(
x
> 650)
Solution
a) X is normally distributed with mean
µ
x = 600 and variance
σ
x
2 = 2,500. We standardize x by
subtracting µ = 600 and dividing by
x
σ
x = 50. Therefore,
P(590 < x < 610) = P590 600
50 <x
µ
x
σx<610 600
50
= P(–.2 < z < .2) = .1586
b) We know that
x
is normally distributed with
µ
x
=
µ
x
=
600 and σx
2
x
2/n
=
Thus, σ2, 500/100 =25. x
=
5. Hence,
P(590 <
x
< 610) = P590 600
5<x
µ
x
σx <610 600
5
= P(–2 < z < 2) = .9544
108
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