Collaborative Statistics by Barbara Illowsky, Ph.D. and Susan Dean - HTML preview

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Chapter 7

The Central Limit Theorem

7.1 The Central Limit Theorem1

7.1.1 Student Learning Outcomes

By the end of this chapter, the student should be able to:

• Recognize the Central Limit Theorem problems.

• Classify continuous word problems by their distributions.

• Apply and interpret the Central Limit Theorem for Means.

• Apply and interpret the Central Limit Theorem for Sums.

7.1.2 Introduction

Why are we so concerned with means? Two reasons are that they give us a middle ground for comparison

and they are easy to calculate. In this chapter, you will study means and the Central Limit Theorem.

The Central Limit Theorem (CLT for short) is one of the most powerful and useful ideas in all of statistics.

Both alternatives are concerned with drawing finite samples of size n from a population with a known

mean, µ, and a known standard deviation, σ. The first alternative says that if we collect samples of size

n and n is "large enough," calculate each sample’s mean, and create a histogram of those means, then the

resulting histogram will tend to have an approximate normal bell shape. The second alternative says that

if we again collect samples of size n that are "large enough," calculate the sum of each sample and create a

histogram, then the resulting histogram will again tend to have a normal bell-shape.

In either case, it does not matter what the distribution of the original population is, or whether you even

need to know it. The important fact is that the sample means and the sums tend to follow the normal

distribution. And, the rest you will learn in this chapter.

The size of the sample, n, that is required in order to be to be ’large enough’ depends on the original

population from which the samples are drawn. If the original population is far from normal then more

observations are needed for the sample means or the sample sums to be normal. Sampling is done with

replacement.

Optional Collaborative Classroom Activity

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CHAPTER 7. THE CENTRAL LIMIT THEOREM

Do the following example in class: Suppose 8 of you roll 1 fair die 10 times, 7 of you roll 2 fair dice 10

times, 9 of you roll 5 fair dice 10 times, and 11 of you roll 10 fair dice 10 times.

Each time a person rolls more than one die, he/she calculates the sample mean of the faces showing. For

example, one person might roll 5 fair dice and get a 2, 2, 3, 4, 6 on one roll.

The mean is

2+2+3+4+6 = 3.4.

The 3.4 is one mean when 5 fair dice are rolled. This same person would

5

roll the 5 dice 9 more times and calculate 9 more means for a total of 10 means.

Your instructor will pass out the dice to several people as described above. Roll your dice 10 times. For

each roll, record the faces and find the mean. Round to the nearest 0.5.

Your instructor (and possibly you) will produce one graph (it might be a histogram) for 1 die, one graph for

2 dice, one graph for 5 dice, and one graph for 10 dice. Since the "mean" when you roll one die, is just the

face on the die, what distribution do these means appear to be representing?

Draw the graph for the means using 2 dice. Do the sample means show any kind of pattern?

Draw the graph for the means using 5 dice. Do you see any pattern emerging?

Finally, draw the graph for the means using 10 dice. Do you see any pattern to the graph? What can you

conclude as you increase the number of dice?

As the number of dice rolled increases from 1 to 2 to 5 to 10, the following is happening:

1. The mean of the sample means remains approximately the same.

2. The spread of the sample means (the standard deviation of the sample means) gets smaller.

3. The graph appears steeper and thinner.

You have just demonstrated the Central Limit Theorem (CLT).

The Central Limit Theorem tells you that as you increase the number of dice, the sample means tend

toward a normal distribution (the sampling distribution).

7.2 The Central Limit Theorem for Sample Means (Averages)2

Suppose X is a random variable with a distribution that may be known or unknown (it can be any distri-

bution). Using a subscript that matches the random variable, suppose:

a. µ X = the mean of X

b. σ X = the standard deviation of X

If you draw random samples of size n, then as n increases, the random variable X which consists of sample

means, tends to be normally distributed and

X ∼ N µ X, σ X

√n

The Central Limit Theorem for Sample Means says that if you keep drawing larger and larger samples

(like rolling 1, 2, 5, and, finally, 10 dice) and calculating their means the sample means form their own

normal distribution (the sampling distribution). The normal distribution has the same mean as the

original distribution and a variance that equals the original variance divided by n, the sample size. n is the

number of values that are averaged together not the number of times the experiment is done.

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To put it more formally, if you draw random samples of size n,the distribution of the random vari-

able X, which consists of sample means, is called the sampling distribution of the mean. The sampling

distribution of the mean approaches a normal distribution as n, the sample size, increases.

The random variable X has a different z-score associated with it than the random variable X. x is the value

of X in one sample.

x − µ

z =

X

(7.1)

σ X

√n

µ X is both the average of X and of X.

σ

= σ X

= standard deviation of X and is called the standard error of the mean.

X

n

Example 7.1

An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size n = 25

are drawn randomly from the population.

Problem 1

Find the probability that the sample mean is between 85 and 92.

Solution

Let X = one value from the original unknown population. The probability question asks you to

find a probability for the sample mean.

Let X = the mean of a sample of size 25. Since µ X = 90, σ X = 15, and n = 25;

then X ∼ N 90, 15

√25

Find P (85 < x < 92)

Draw a graph.

P (85 < x < 92) = 0.6997

The probability that the sample mean is between 85 and 92 is 0.6997.

TI-83 or 84: ♥♦r♠❛❧❝❞❢(lower value, upper value, mean, standard error of the mean)

The parameter list is abbreviated (lower value, upper value, µ, σ

√ )

n

♥♦r♠❛❧❝❞❢ 85, 92, 90, 15

= 0.6997

25

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CHAPTER 7. THE CENTRAL LIMIT THEOREM

Problem 2

Find the value that is 2 standard deviations above the expected value (it is 90) of the sample mean.

Solution

To find the value that is 2 standard deviations above the expected value 90, use the formula

value = µ X + (#ofSTDEVs) σ X

√n

value = 90 + 2 · 15

= 96

25

So, the value that is 2 standard deviations above the expected value is 96.

Example 7.2

The length of time, in hours, it takes an "over 40" group of people to play one soccer match is

normally distributed with a mean of 2 hours and a standard deviation of 0.5 hours. A sample of

size n = 50 is drawn randomly from the population.

Problem

Find the probability that the sample mean is between 1.8 hours and 2.3 hours.

Solution

Let X = the time, in hours, it takes to play one soccer match.

The probability question asks you to find a probability for the sample mean time, in hours, it

takes to play one soccer match.

Let X = the mean time, in hours, it takes to play one soccer match.

If µ X = _________, σ X = __________, and n = ___________, then X ∼ N(______, ______)

by the Central Limit Theorem for Means.

µ X = 2, σ X = 0.5, n = 50, and X ∼ N 2 , 0.5

√50

Find P (1.8 < x < 2.3).

Draw a graph.

P (1.8 < x < 2.3) = 0.9977

♥♦r♠❛❧❝❞❢ 1.8, 2.3, 2, .5

= 0.9977

50

The probability that the mean time is between 1.8 hours and 2.3 hours is ______.

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7.3 The Central Limit Theorem for Sums3

Suppose X is a random variable with a distribution that may be known or unknown (it can be any distri-

bution) and suppose:

a. µ X = the mean of X

b. σ X = the standard deviation of X

If you draw random samples of size n, then as n increases, the random variable ΣX which consists of sums

tends to be normally distributed and

ΣX ∼ N n · µ X, n · σ X

The Central Limit Theorem for Sums says that if you keep drawing larger and larger samples and taking

their sums, the sums form their own normal distribution (the sampling distribution) which approaches a

normal distribution as the sample size increases. The normal distribution has a mean equal to the original

mean multiplied by the sample size and a standard deviation equal to the original standard deviation

multiplied by the square root of the sample size.

The random variable ΣX has the following z-score associated with it:

a. Σx is one sum.

Σ

b. z = x−n· µ X

√n· σ X

a. n · µ X = the mean of ΣX

b.

n · σ X = standard deviation of ΣX

Example 7.3

An unknown distribution has a mean of 90 and a standard deviation of 15. A sample of size 80 is

drawn randomly from the population.

Problem

a. Find the probability that the sum of the 80 values (or the total of the 80 values) is more than

7500.

b. Find the sum that is 1.5 standard deviations above the mean of the sums.

Solution

Let X = one value from the original unknown population. The probability question asks you to

find a probability for the sum (or total of) 80 values.

ΣX = the sum or total of 80 values. Since µ X = 90, σ X = 15, and n = 80, then

ΣX ∼ N 80 · 90, 80 · 15

•. mean of the sums = n · µ X = (80) (90) = 7200

•. standard deviation of the sums =

n · σ X =

80 · 15

•. sum of 80 values = Σx = 7500

a: Find P (Σx > 7500)

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CHAPTER 7. THE CENTRAL LIMIT THEOREM

P (Σx > 7500) = 0.0127

♥♦r♠❛❧❝❞❢(lower value, upper value, mean of sums, st❞❡✈ of sums)

The parameter list is abbreviated (lower, upper, n · µ X, n · σ X)

♥♦r♠❛❧❝❞❢(7500,1E99, 80 · 90, 80 · 15 = 0.0127

Reminder: 1E99 = 1099. Press the ❊❊ key for E.

b: Find Σx where z = 1.5:

Σx = n · µ X + z · n · σ X = (80)(90) + (1.5)( 80) (15) = 7401.2

7.4 Using the Central Limit Theorem4

It is important for you to understand when to use the CLT. If you are being asked to find the probability of

the mean, use the CLT for the mean. If you are being asked to find the probability of a sum or total, use the

CLT for sums. This also applies to percentiles for means and sums.

NOTE: If you are being asked to find the probability of an individual value, do not use the CLT.

Use the distribution of its random variable.

7.4.1 Examples of the Central Limit Theorem

Law of Large Numbers

The Law of Large Numbers says that if you take samples of larger and larger size from any population,

then the mean x of the sample tends to get closer and closer to µ. From the Central Limit Theorem, we

know that as n gets larger and larger, the sample means follow a normal distribution. The larger n gets, the

smaller the standard deviation gets. (Remember that the standard deviation for X is σ

.) This means that

n

the sample mean x must be close to the population mean µ. We can say that µ is the value that the sample

means approach as n gets larger. The Central Limit Theorem illustrates the Law of Large Numbers.

Central Limit Theorem for the Mean and Sum Examples

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Example 7.4

A study involving stress is done on a college campus among the students. The stress scores follow

a uniform distribution with the lowest stress score equal to 1 and the highest equal to 5. Using a

sample of 75 students, find:

1. The probability that the mean stress score for the 75 students is less than 2.

2. The 90th percentile for the mean stress score for the 75 students.

3. The probability that the total of the 75 stress scores is less than 200.

4. The 90th percentile for the total stress score for the 75 students.

Let X = one stress score.

Problems 1. and 2. ask you to find a probability or a percentile for a mean. Problems 3 and 4 ask

you to find a probability or a percentile for a total or sum. The sample size, n, is equal to 75.

Since the individual stress scores follow a uniform distribution, X ∼ U (1, 5) where a = 1 and

b = 5 (See Continuous Random Variables (Section 5.1) for the uniform).

µ X = a+b = 1+5 = 3

2

2

σ X =

(b−a)2 =

(5−1)2 = 1.15

12

12

For problems 1. and 2., let X = the mean stress score for the 75 students. Then,

X ∼ N 3, 1.15

where n = 75.

75

Problem 1

Find P (x < 2).

Draw the graph.

Solution

P (x < 2) = 0

The probability that the mean stress score is less than 2 is about 0.

♥♦r♠❛❧❝❞❢ 1, 2, 3, 1.15

= 0

75

REMINDER: The smallest stress score is 1. Therefore, the smallest mean for 75 stress scores is 1.

Problem 2

Find the 90th percentile for the mean of 75 stress scores. Draw a graph.

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CHAPTER 7. THE CENTRAL LIMIT THEOREM

Solution

Let k = the 90th precentile.

Find k where P (x < k) = 0.90.

k = 3.2

The 90th percentile for the mean of 75 scores is about 3.2. This tells us that 90% of all the means of

75 stress scores are at most 3.2 and 10% are at least 3.2.

✐♥✈◆♦r♠ .90, 3, 1.15

= 3.2

75

For problems c and d, let ΣX = the sum of the 75 stress scores.

Then, ΣX ∼

N (75) · (3) ,

75 · 1.15

Problem 3

Find P (Σx < 200).

Draw the graph.

Solution

The mean of the sum of 75 stress scores is 75 · 3 = 225

The standard deviation of the sum of 75 stress scores is

75 · 1.15 = 9.96

P (Σx < 200) = 0

The probability that the total of 75 scores is less than 200 is about 0.

♥♦r♠❛❧❝❞❢ 75, 200, 75 · 3, 75 · 1.15 = 0.

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REMINDER: The smallest total of 75 stress scores is 75 since the smallest single score is 1.

Problem 4

Find the 90th percentile for the total of 75 stress scores. Draw a graph.

Solution

Let k = the 90th percentile.

Find k where P (Σx < k) = 0.90.

k = 237.8

The 90th percentile for the sum of 75 scores is about 237.8. This tells us that 90% of all the sums of

75 scores are no more than 237.8 and 10% are no less than 237.8.

✐♥✈◆♦r♠ .90, 75 · 3, 75 · 1.15 = 237.8

Example 7.5

Suppose that a market research analyst for a cell phone company conducts a study of their cus-

tomers who exceed the time allowance included on their basic cell phone contract; the analyst

finds that for those people who exceed the time included in their basic contract, the excess time

used follows an exponential distribution with a mean of 22 minutes.

Consider a random sample of 80 customers who exceed the time allowance included in their basic

cell phone contract.

Let X = the excess time used by one INDIVIDUAL cell phone customer who exceeds his contracted

time allowance.

X ∼ Exp

1

From Chapter 5, we know that

22

µ = 22 and σ = 22.

Let X = the mean excess time used by a sample of n = 80 customers who exceed their contracted

time allowance.

X ∼ N 22, 22

by the CLT for Sample Means

80

Problem 1

Using the CLT to find Probability:

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CHAPTER 7. THE CENTRAL LIMIT THEOREM

a. Find the probability that the mean excess time used by the 80 customers in the sample is longer

than 20 minutes. This is asking us to find P (x > 20)

Draw the graph.

b. Suppose that one customer who exceeds the time limit for his cell phone contract is randomly

selected. Find the probability that this individual customer’s excess time is longer than 20

minutes. This is asking us to find P (x > 20)

c. Explain why the probabilities in (a) and (b) are different.

Solution

Part a.

Find: P (x > 20)

P (x > 20) = 0.7919 using ♥♦r♠❛❧❝❞❢ 20, 1E99, 22, 22

√80

The probability is 0.7919 that the mean excess time used is more than 20 minutes, for a sample of

80 customers who exceed their contracted time allowance.

REMINDER: 1E99 = 1099 and−1E99 = −1099. Press the ❊❊ key for E. Or just use 10^99 instead of

1E99.

Part b.

Find P(x>20) . Remember to use the exponential distribution for an individual: XExp(1/22).

P(X>20) = e^(–(1/22)*20) or e^(–.04545*20) = 0.4029

Part c. Explain why the probabilities in (a) and (b) are different.

P (x > 20) = 0.4029 but P (x > 20) = 0.7919

The probabilities are not equal because we use different distributions to calculate the probability

for individuals and for means.

When asked to find the probability of an individual value, use the stated distribution of its ran-

dom variable; do not use the CLT. Use the CLT with the normal distribution when you are

being asked to find the probability for an mean.

Problem 2

Using the CLT to find Percentiles:

Find the 95th percentile for the sample mean excess time for samples of 80 customers who exceed

their basic contract time allowances. Draw a graph.

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Solution

Let k = the 95th percentile. Find k where P (x < k) = 0.95

k = 26.0 using ✐♥✈◆♦r♠ .95, 22, 22

= 26.0

80

The 95th percentile for the sample mean excess time used is about 26.0 minutes for random

samples of 80 customers who exceed their contractual allowed time.

95% of such samples would have means under 26 minutes; only 5% of such samples would have

means above 26 minutes.

NOTE: (HISTORICAL): Normal Approximation to the Binomial

Historically, being able to compute binomial probabilities was one of the most important applications of the

Central Limit Theorem. Binomial probabilities were displayed in a table in a book with a small value for n

(say, 20). To calculate the pr