# 7.3: Using the Central Limit Theorem

It is important for you to understand when to use the central limit theorem. 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.

# 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 $\displaystyle\overline{{x}}$ of the sample tends to get closer and closer to the population mean μ.

The formula for the standard deviation of variable $\overline{x}$ is $\frac{\sigma}{\sqrt{n}}$. If n is getting larger, then $\frac{\sigma}{\sqrt{n}}$ is getting smaller. Indirectly, the sample mean $\overline{x}$ will be closed to the population mean $\mu$

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

### Example 1

A study involving stress is conducted among the students on a college campus.
The stress scores follow a uniform distribution with the lowest stress score = 1 and the highest score =5.
Using a sample of 75 students, find

a. The probability that the mean stress score for the 75 students is less than two.
b. The 90th percentile for the mean stress score for the 75 students.
c. The probability that the total of the 75 stress scores is less than 200.
d. The 90th percentile for the total stress score for the 75 students.

#### Solution

Let X = one stress score. The sample size n = 75.

We are looking for a probability or a percentile for a mean score in problem a and b.
We are looking for a probability or a percentile for a total or sum of score in problem c and d.

Since the individual stress scores follow a uniform distribution, X ~ U(1, 5) where lowest score = 1 and highest score = 5. $\displaystyle{\mu}_{X}=\frac{{a+b}}{{2}}=\frac{{1+5}}{{2}}={3}$ $\displaystyle{\sigma}_{X}=\sqrt{\frac{{{(b-a)}^{2}}}{{12}}}=\sqrt{\frac{{({5-1)}^{2}}}{{12}}}$ = 1.15

For problems a and b, let $\displaystyle\overline{X}$ = the mean stress score for the 75 students.
Then, $\displaystyle\overline{X}\sim{N}({3},\frac{{1.15}}{{\sqrt{75}}})\text{ where } {n}={75}$.

### Solution

a. Find the probability that the mean stress score for the 75 students is less than two.
We are asked to find P( $\displaystyle\overline{x}{<}{2}$). By plotting the graph, We will use TI-83/84 to solve for part (a).

TI-Calculator: normalcdf $\displaystyle{({1},{2},{3},\frac{{1.15}}{\sqrt{{75}}})}={0}$

 Remember that the smallest stress score is one.

b. Find the 90th percentile for the mean stress score for the 75 students.
We are asked to find the 90th percentile for the mean of 75 stress scores. By plotting a graph, Let k = the 90th percentile. Find the value of k where P( $\overline{x}$ < k) = 0.90.
TI-Calculator: invNorm (0.90, 3, $\frac{1.15}{\sqrt{75}}$)

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 that 10% are at least 3.2.

 For problems c and d, let ΣX = the sum of the 75 stress scores. Then, $\displaystyle\sum{X}{\sim}{N}{[{({75})}{({3})},{(\sqrt{{75}})}{({1.15})}]}$ The mean of the sum of 75 stress scores is (75)(3) = 225. The standard deviation of the sum of 75 stress scores is $\displaystyle{(\sqrt{{75}})}$(1.15) = 9.96

c. Find the probability that the total of the 75 stress scores is less than 200.
We are asked to find P(Σx < 200). By plotting the graph, TI-Calculator: normalcdf (75, 200, (75)(3), ( $\sqrt{75}$)(1.15))
Therefore, P(Σx < 200) = 0

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

 Remember, since the smallest single score is 1, it is possible that we draw the smallest score of 1 for 75 times theoretically. The smallest total of 75 stress scores is 75.

d.  Find the 90th percentile for the total stress score for the 75 students.
We are asked to solve for the 90th percentile for the total of 75 stress scores. By plotting the graph, Let k= the 90th percentile. Find the value of k where P(Σx < k) = 0.90.
TI-Calculator: invNorm(0.90,(75)(3), $\displaystyle{(\sqrt{{75}})}$(1.15))

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.= 237.8

### Try It

A study involving stress is conducted among the students on a college campus.
The stress scores follow a uniform distribution with the lowest stress score = 1 and the highest score =5.
Use a sample size of 55 to answer the following questions.

1. Find $\displaystyle{P}{(\overline{{x}}{<}{7})}$.
2. Find $\displaystyle{P}{(\sum{x}{<}{170})}$.
3. Find $\displaystyle{P}{(\sum{x}{>}{170})}$.
4. Find the 80th percentile for the mean of 55 scores.
5. Find the 85th percentile for the sum of 55 scores.

[practice-area rows=”2″][/practice-area]
1. TI-Calculator: normalcdf $\displaystyle{({1},{7},{3},\frac{{1.15}}{\sqrt{{55}}})}={0}$ = 1
2. TI-Calculator: normalcdf (55, 170, (55)(3), ( $\sqrt{55}$)(1.15)) = 0.7211
3. 1 – 0.7211 = 0.2789
4. TI-Calculator: invNorm (0.80, 3, $\frac{1.15}{\sqrt{55}}$) = 3.13
5. TI-Calculator: invNorm (0.85,(55)(3), $\displaystyle{(\sqrt{{55}})}[/latex (1.15)) = 173.84[/hidden-answer]

### Example 2

Suppose that a market research analyst for a cell phone company conducts a study of their customers 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. [latex]\displaystyle{X}{\sim}{E}{x}{p}{(\frac{{1}}{{22}})}$
. From previous chapters, we know that μ = 22 and σ = 22.

Let $\displaystyle\overline{{X}}$ = the mean excess time used by a sample of n = 80 customers who exceed their contracted time allowance. $\displaystyle\overline{{X}}{\sim}{N}{({22},\frac{{22}}{{\sqrt{{80}}}})}$ by the central limit theorem for sample means.

Using the clt to find probability,

1. 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 $\displaystyle{P}{(\overline{{x}}{>}{20})}$. Draw the graph.
2. 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).
3. Explain why the probabilities in parts 1 and 2 are different.
4. 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.

### Solution

1. Find: $\displaystyle{P}{(\overline{{x}}{>}{20})}$ $\displaystyle{P}{(\overline{{x}}{>}{20})}={0.79199}$ using normalcdf $\displaystyle{({20},{1}\text{E99},{22},\frac{{22}}{\sqrt{{80}}})}$
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. Remember, 1E99 = 1099 and –1E99 = –1099. Press the EE key for E. Or just use 1099 instead of 1E99.
2. Find P(x > 20). Remember to use the exponential distribution for an individual: X~ $\frac{1}{22})$. $\displaystyle{P}{({x}{>}{20})}={e}^{{{(-{({122})}{({20})})}}}{\quad\text{or}\quad}{e}^{{{(-{0.04545}{({20})})}}}={0.4029}$
3. $\displaystyle{P}{({x}{>}{20})}={0.4029}\text{ but } {P}{(\overline{{x}}{>}{20})}={0.7919}$.  (1) The probabilities are not equal because we use different distributions to calculate the probability for individuals and for means. (2)  When asked to find the probability of an individual value, use the stated distribution of its random variable; do not use the clt. (3) Use the clt with the normal distribution when you are being asked to find the probability for a mean.
4. Let k = the 95th percentile. Find k where $\displaystyle{P}{(\overline{{x}}{<}{k})}={0.95}$
k = 26.0 using invNorm = 26.0 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.Ninety five percent of such samples would have means under 26 minutes; only five percent of such samples would have means above 26 minutes.

### Try It

Use the information in Example 2, but change the sample size to 144.

1. Find $\displaystyle{P}{({20}{<}\overline{{x}}{<}{30})}$.
2. Find P(Σx is at least 3,000).
3. Find the 75th percentile for the sample mean excess time of 144 customers.
4. Find the 85th percentile for the sum of 144 excess times used by customers.

[practice-area rows="2"][/practice-area]

1. 0.8623
2. 0.7377
3. 23.2

## Example 3

In the United States, someone is sexually assaulted every two minutes, on average, according to a number of studies. Suppose the standard deviation is 0.5 minutes and the sample size is 100.

1. Find the median, the first quartile, and the third quartile for the sample mean time of sexual assaults in the United States.
2. Find the median, the first quartile, and the third quartile for the sum of sample times of sexual assaults in the United States.
3. Find the probability that a sexual assault occurs on the average between 1.75 and 1.85 minutes.
4. Find the value that is two standard deviations above the sample mean.
5. Find the IQR for the sum of the sample times.

### Solution

1. We have $\displaystyle{\mu}_{x}={\mu}={2}$ and ${\sigma}_{x}=\frac{{\sigma}}{{\sqrt{n}}}=\frac{{0.5}}{{10}}=0.05$.
Therefore:
50th percentile = μx = μ = 2             TI-83/84: invNorm(0.5,2,0.05) ,
25th percentile = 1.97                      TI-83/84: invNorm(0.25,2,0.05),
75th percentile = 2.03                      TI-83/84: invNorm(0.75,2,0.05).
2. We have $\displaystyle{\mu}_{\sum{x}}=n{\mu}_{x}={100}({2})= 200$ and ${\sigma}_{\sum{x}}={\sqrt{100}}(0.5)=5$.
Therefore:
50th percentile = $\displaystyle{\mu}_{\sum{x}}$ = 200        TI-83/84: invNorm(0.50, 200, 0.05) ,
25th percentile = 196.63                TI-83/84: invNorm(0.25, 200, 0.05),
75th percentile = 203.37                TI-83/84: invNorm(0.75, 200, 0.05)
3. $\displaystyle{P}{(1.75{<}{\overline{x}}{<}{1.85})}$ = 0.0013
TI-83/84: normalcdf(1.75,1.85,2,0.05)
4. Using the z-score equation, $\displaystyle{z}=\frac{{\overline{x}-{\mu}_{\overline{x}}}}{{{\sigma}_{\overline{x}}}}$.
To solve for x, z = 2, ${\mu}_{x}=2$, ${\sigma}_{x}=2$
Then we have x = 2(0.05) + 2 = 2.1
5. The IQR of $\sum{x}$ is 75th percentile – 25th percentile = 203.37 – 196.63 = 6.74

### Try It

Based on data from the National Health Survey, women between the ages of 18 and 24 have an average systolic blood pressures (in mm Hg) of 114.8 with a standard deviation of 13.1. Systolic blood pressure for women between the ages of 18 to 24 follow a normal distribution.

1. If one woman from this population is randomly selected, find the probability that her systolic blood pressure is greater than 120.
2. If 40 women from this population are randomly selected, find the probability that their mean systolic blood pressure is greater than 120.
3. If the sample were four women between the ages of 18 to 24 and we did not know the original distribution, could the central limit theorem be used?

1. P(x > 120) = 0.0272.
TI-Calculator: normalcdf(120,99,114.8,13.1)
There is about a 3%, that the randomly selected woman will have systolics blood pressure greater than 120.
2. Since we are selecting 40 women randomly, P( x > 120) = 0.006.
TI-83/84: normalcdf(120,99,114.8, $\frac{13.1}{\sqrt{40}}$)
There is only a 0.6% chance that the average systolic blood pressure for the randomly selected group is greater than 120.

3. The central limit theorem could not be used if the sample size were four and we did not know the original distribution was normal. The sample size would be too small.[/hidden-answer]

### Example 4

A study was done about violence against prostitutes and the symptoms of the post-traumatic stress that they developed. The age range of the prostitutes was 14 to 61. The mean age was 30.9 years with a standard deviation of nine years.

1. In a sample of 25 prostitutes, what is the probability that the mean age of the prostitutes is less than 35?
2. Is it likely that the mean age of the sample group could be more than 50 years? Interpret the results.
3. In a sample of 49 prostitutes, what is the probability that the sum of the ages is no less than 1,600?
4. Is it likely that the sum of the ages of the 49 prostitutes is at most 1,595? Interpret the results.
5. Find the 95th percentile for the sample mean age of 65 prostitutes. Interpret the results.
6. Find the 90th percentile for the sum of the ages of 65 prostitutes. Interpret the results.

### Solution

1. P( $\overline{x}$ < 35) = 0.9886
TI-83/84: normalcdf(-E99,35,30.9,1.8)
2. P $\overline{x}$ > 50) ≈ 0.
TI-83/84: normalcdf(50, E99,30.9,1.8)
For this sample group, it is almost impossible for the group's average age to be more than 50. However, it is still possible for an individual in this group to have an age greater than 50.
3. P(Σx ≥ 1,600) = 0.0864
TI-8/84: normalcdf(1600,E99,1514.10,63)
4. P(Σx ≤ 1,595) = 0.9005.
TI-83/84: normalcdf(-E99,1595,1514.10,63)
This means that there is a 90% chance that the sum of the ages for the sample group n = 49 is at most 1595.
5. The 95th percentile = 32.7.
TI-83/84: invNorm(0.95,30.9,1.1)
This indicates that 95% of the prostitutes in the sample of 65 are younger than 32.7 years, on average.
6. The 90th percentile = 2101.5.
TI-83/84: invNorm(0.90,2008.5,72.56)
This indicates that 90% of the prostitutes in the sample of 65 have a sum of ages less than 2,101.5 years.

### Try It

According to Boeing data, the 757 airliner carries 200 passengers and has doors with a mean height of 72 inches. Assume for a certain population of men we have a mean of 69.0 inches and a standard deviation of 2.8 inches.

1. What mean doorway height would allow 95% of men to enter the aircraft without bending?
2. Assume that half of the 200 passengers are men. What mean doorway height satisfies the condition that there is a 0.95 probability that this height is greater than the mean height of 100 men?
3. For engineers designing the 757, which result is more relevant: the height from part 1 or part 2? Why?

1. We know that μx = μ = 69 and we have σx = 2.8.
The 95th percentile = the height of the doorway that would allow 95th of men to enter the aircraft without bending.
TI-83/84: invNorm(0.95,69,2.8)
The 95th percentile = 73.61

2. We know that μx = μ = 69 and σx = $\frac{2.8}{\sqrt{100}}$.
We are looking the 95th percentile for a random sample of 100 men.
TI-83/84: invNorm(0.95,69, $\frac{2.8}{\sqrt{100}}$)
The mean doorway height that satisfies the condition is 69.49 inches.

3. When designing the doorway heights, we need to incorporate as much variability as possible in order to accommodate as many passengers as possible. Therefore, we need to use the result based on part 1.[/hidden-answer]

## Historical Note: 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 with a small value for n(say, 20) were displayed in a table in a book. To calculate the probabilities with large values of n, you had to use the binomial formula, which could be very complicated. Using the normal approximation to the binomial distribution simplified the process. To compute the normal approximation to the binomial distribution, take a simple random sample from a population. You must meet the conditions for a binomial distribution:

• there are a certain number n of independent trials
• the outcomes of any trial are success or failure
• each trial has the same probability of a success p

Recall that if X is the binomial random variable, then X ~ B(n, p). The shape of the binomial distribution needs to be similar to the shape of the normal distribution. To ensure this, the quantities np and nq must both be greater than five (np > 5 and nq > 5; the approximation is better if they are both greater than or equal to 10). Then the binomial can be approximated by the normal distribution with mean μ = np and standard deviation $\sigma={\sqrt{npq}}$. Remember that q = 1 – p. In order to get the best approximation, add 0.5 to x or subtract 0.5 from x (use x + 0.5 or x – 0.5). The number 0.5 is called the continuity correction factor and is used in the following example.

### Example 5

Suppose in a local Kindergarten through 12th grade (K - 12) school district, 53 percent of the population favor a charter school for grades K through 5. A simple random sample of 300 is surveyed.

1. Find the probability that at least 150 favor a charter school.
2. Find the probability that at most 160 favor a charter school.
3. Find the probability that more than 155 favor a charter school.
4. Find the probability that fewer than 147 favor a charter school.
5. Find the probability that exactly 175 favor a charter school.

### Solution

Let X = the number that favor a charter school for grades K trough 5. X ~B(n, p) where n = 300, p = 0.53, and q = 1-p = 1-0.53 = 0.47.
Since np > 5 and nq > 5, use the normal approximation to the binomial.
The formulas for the mean and standard deviation are μ = np and $\sigma={\sqrt{npq}}$.
The mean = 300*0.53 = 159 and the standard deviation = $\sqrt{300*0.53*0.47}$ = 8.6447.
The random variable for the normal distribution is Y. Y ~ N(159, 8.6447).

1.  you include 150 so P(X ≥ 150) has normal approximation P(Y ≥ 149.5) = 0.8641.
TI-83/84: normalcdf(149.5,10^99,159,8.6447).
2. you include 160 so P(X ≤ 160) has normal approximation P(Y ≤ 160.5) = 0.5689.
TI-83/84: normalcdf(0,160.5,159,8.6447) = 0.5689
3. you exclude 155 so P(X > 155) has normal approximation P(y > 155.5) = 0.6572.
TI-83/84: normalcdf(155.5,10^99,159,8.6447) = 0.6572.
4. you exclude 147 so P(X < 147) has normal approximationP(Y < 146.5) = 0.0741.
TI-83/84: normalcdf(0,146.5,159,8.6447) = 0.0741
5. P(X = 175) has normal approximation P(174.5 < Y < 175.5) = 0.0083.
TI-83/84: normalcdf(174.5,175.5,159,8.6447) = 0.0083

Because of calculators and computer software that let you calculate binomial probabilities for large values of n easily, it is not necessary to use the the normal approximation to the binomial distribution, provided that you have access to these technology tools. Most school labs have Microsoft Excel, an example of computer software that calculates binomial probabilities. Many students have access to the TI-83 or 84 series calculators, and they easily calculate probabilities for the binomial distribution. If you type in "binomial probability distribution calculation" in an Internet browser, you can find at least one online calculator for the binomial.

For Example 3, the probabilities are calculated using the following binomial distribution: ( n = 300 and p = 0.53).
Compare the binomial and normal distribution answers.

1. P(X ≥ 150) : 1 - binomialcdf(300,0.53,149) = 0.8641
2. P(X ≤ 160) : binomialcdf(300,0.53,160) = 0.5684
3. P(X > 155) : 1 - binomialcdf(300,0.53,155) = 0.6576
4. P(X < 147) : binomialcdf(300,0.53,146) = 0.0742
5. P(X = 175) : binomialpdf(300,0.53,175) = 0.0083 (You need to use the binomial pdf.)

### Try It

In a city, 46 percent of the population favor the incumbent, Dawn Morgan, for mayor. A simple random sample of 500 is taken. Using the continuity correction factor, find the probability that at least 250 favor Dawn Morgan for mayor.

[practice-area rows="1"][/practice-area]
n = 500, p = 0.46, q = 1 - p = 1 - 0.46 = 0.54
0.0401
TI-Calculator: normalcdf (250, 1E99, 500*0.46, $\sqrt{500*0.46*0.54}$)[/hidden-answer]

## References

Data from the Wall Street Journal.

"National Health and Nutrition Examination Survey." Center for Disease Control and Prevention. Available online at http://www.cdc.gov/nchs/nhanes.htm (accessed May 17, 2013).