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Central Limit Theorem: Central Limit Theorem for Sample Means (Averages)

Module by: Dr. Barbara Illowsky, Susan Dean

Suppose XX is a random variable with a distribution that may be known or unknown (it can be any distribution). Using a subscript that matches the random variable, suppose:

  • a. μXμX = the mean of XX
  • b. σXσX = the standard deviation of XX
If you draw random samples of size nn, then as nn increases, the random variable X¯ X which consists of sample means, tends to be normally distributed and

X¯ X ~ N ( μ X , σ X n ) N( μ X , σ X n )

The Central Limit Theorem for Sample Means (Averages) 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 (averages) form their own normal distribution. This distribution has the same mean as the original distribution and a variance that equals the original variance divided by nn, the sample size. nn is the number of values that are averaged together not the number of times the experiment is done.

The random variable X¯ X has a different z-score associated with it. x¯ x is one sample

z = x¯ - μ X ( σ X n ) z= x - μ X ( σ X n ) (1)

μXμX is both the average of XX and of X¯ X .

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

Example 1

An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of size nn = 25 are drawn randomly from the population.

Problem 1

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

Solution 1

Let XX = one value from the original unknown population. The probability question asks you to find a probability for the sample mean (or average).

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

then X¯ X ~ N ( 90 , 15 25 ) N(90, 15 25 )

Find P ( 85 < X¯ < 92 ) P ( 85 X 92 ) Draw a graph.

P ( 85 < X¯ < 92 ) = 0.6997 P ( 85 X 92 ) =0.6997

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

Normal distribution curve from -∞ to ∞ and an x-axis with the values of 85, 90, and 92. The x-axis is equal to the mean of a sample size of 25. A vertical upward line extends from points 85 and 92 to the curve. The probability area is between 85 and 92.

TI-83: normalcdf(lower value, upper value, mean for averages, stdev for averages)

stdev = standard deviation

The parameter list is abbreviated (lower, upper, μμ, σ n σ n )

normalcdf(85,92,90, 15 25 ) = 0.6997 (85,92,90, 15 25 ) = 0.6997

Problem 2

Find the average value that is 2 standard deviations above the the mean of the averages.

Solution 2

To find the average value that is 2 standard deviations above the mean of the averages, use the formula

value = μ X + (#ofSTDEVs) ( σ X n ) μ X +(#ofSTDEVs)( σ X n )

value = 90 + 2 ⋅ 15 25 = 96 90+2⋅ 15 25 =96

So, the average value that is 2 standard deviations above the mean of the averages is 96.

Example 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 nn = 50 is drawn randomly from the population.

Problem 1

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

Solution 1

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

The probability question asks you to find a probability for the sample mean or average time, in hours, it takes to play one soccer match.

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

Problem 2

If μ X = μ X = _________, σ X = σ X = __________, and n=n= ___________, then X¯ ~ N (______, ______) X ~N(______, ______) by the Central Limit Theorem for Averages of Sample Means.

Solution 2

μ X = μ X = 2, σ X = σ X = 0.5, n=n= 50, and X ~ N ( 2  ,  0.5 50 ) X~N(2 ,  0.5 50 )

Find P ( 1.8 < X¯ < 2.3 ) P ( 1.8 X 2.3 ) . Draw a graph.

P ( 1.8 < X¯ < 2.3 ) = 0.9977 P ( 1.8 X 2.3 ) =0.9977

normalcdf(1.8,2.3,2, .5 50 ) = 0.9977 (1.8,2.3,2, .5 50 ) = 0.9977

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

Glossary

Average:
A number that describes the central tendency of the data. There are a number of specialized averages, including the arithmetic mean, weighted mean, median, mode, and geometric mean.
Central Limit Theorem:
Given a random variable (RV) with known mean μμ and known variance σσ 22 size 12{ {} rSup { size 8{2} } } {}, we are sampling with size n and we are interested in two new RV - sample mean, XˉXˉ size 12{ { bar {X}}} {},and sample sum,ΣΣ XX size 12{X} {}. If the size n of the sample is sufficiently large, then XˉXˉ size 12{ { bar {X}}} {}∼ N nμ σ 2 n N nμ σ 2 n and ΣXΣX size 12{X} {} ∼ N nμ n σ 2 N nμ n σ 2 . In words, if the size n of the sample is sufficiently large, then the distribution of the sample means and the distribution of the sample sums will approximate a normal distribution regardless of the shape of the population. And even more, the mean of the sampling distribution will equal the population mean and mean of sampling sums will equal n times the population mean. The standard deviation of the distribution of the sample means, σ n σ n , is called standard error of the mean.
Normal Distribution:
A continuous random variable (RV) with pdf=1σ2πe−(x−μ)2/2σ2pdf=1σ2πe−(x−μ)2/2σ2 size 12{ ital "pdf"= { {1} over {σ sqrt {2π} } } e rSup { size 8{ - \( x - μ \) rSup { size 6{2} } /2σ rSup { size 6{2} } } } } {}, where μμ is the mean of the distribution and σσ is its standard deviation. Notation: XX ~ N μ σ 2 N μ σ 2 . If μ=0μ=0 and σ=1σ=1, the RV is called standard normal distribution, or z-score.
Standard Error of the Mean:
The standard deviation of the distribution of the sample means, σ n σ n .

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