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Chapter 09

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

by OC9945

School

York UniversityDepartment

Operations Management and Information SystemCourse Code

OMIS 2010Professor

Alan MarshallChapter

09This

**preview**shows pages 1-3. to view the full**9 pages of the document.**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 =µ

x.

3. The variance of

x

will equal the variance of the original random variable divided by n. That is,

σx

2=σ

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

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

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 =µ

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

Only pages 1-3 are available for preview. Some parts have been intentionally blurred.

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 =µ

x=2.1

and

σx

2=σ

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=σ

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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