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The Central Limit Theorem for Sample Means (Averages)

Module by: Susan Dean, Barbara Illowsky, Ph.D.. E-mail the authors

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 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 nn, the sample size. nn is the number of values that are averaged together not the number of times the experiment is done.

To put it more formally, if you draw random samples of size nn,the distribution of the random variable X¯ 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 nn, the sample size, increases.

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

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

Let X¯ = X = the mean 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 or 84: normalcdf(lower value, upper value, mean, standard error of the mean)

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

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

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 ) μ X +(#ofSTDEVs)( σ X n )

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

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

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 time, in hours, it takes to play one soccer match.

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

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

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 mean time 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 standard deviation σσ. We are sampling with size n and we are interested in two new RVs - the sample mean, X¯X¯, and the sample sum, ΣXΣX. If the size nn of the sample is sufficiently large, then X¯X¯ size 12{ { bar {X}}} {} N μ σ n N μ σ n and ΣXΣX size 12{X} {}N ( , n σ )N(,nσ). 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. The mean of the sample means will equal the population mean and the mean of the sample sums will equal n times the population mean. The standard deviation of the distribution of the sample means, σ n σ n , is called the standard error of the mean.
Normal Distribution:
A continuous random variable (RV) with pdf f(x)=1σe(xμ)2/2f(x)=1σe(xμ)2/2 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 the standard deviation. Notation: XX ~ N μ σ N μ σ . If μ=0μ=0 and σ=1σ=1, the RV is called the standard normal distribution.
Standard Error of the Mean:
The standard deviation of the distribution of the sample means, σ n σ n .

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