Module 7: The Central Limit Theorem
The Central Limit Theorem for Sample Means (Averages)
Barbara Illowsky & OpenStax et al.
Learning Objectives
Suppose X 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:
- μX = the mean of X
- σX = the standard deviation of X
If you draw random samples of size n, then as n increases, the random variable [latex]displaystyleoverline{{X}}[/latex].
The central limit theorem for sample means says that if you keep drawing larger and larger samples (such as rolling one, two, five, and finally, ten 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, the sample size. The variable n 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 n, the distribution of the random variable [latex]displaystyleoverline{{X}}[/latex], 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 [latex]displaystyleoverline{{X}}[/latex] in one sample.
[latex]displaystylefrac{{overline{X}-{mu}_{x}}}{{frac{{sigma{x}}}{{sqrt{n}}}}}[/latex][latex]displaystyle{mu}_{x}[/latex] is the average of both X and [latex]displaystyleoverline{X}[/latex]
[latex]displaystyle{sigma}overline{x} = frac{{overline{X}-{mu}_{x}}}{{frac{{sigma{x}}}{{sqrt{n}}}}}[/latex] = standard deviation of [latex]displaystyleoverline{{X}}[/latex] and is called the standard error of the mean.
To find probabilities for means on the calculator, follow these steps:
- 2nd DISTR
- 2:normalcdf
- normalcdf (Lower value of the area, upper value of the area, mean,[latex]displaystylesqrt{frac{{text{standard deviation}}}{{text{sample size}}}}[/latex]
- where: mean is the mean of the original distribution standard deviation is the standard deviation of the original distribution sample size =n
Example
An unknown distribution has a mean of 90 and a standard deviation of 15. Samples of sizen = 25 are drawn randomly from the population.
- Find the probability that the sample mean is between 85 and 92.
- Find the value that is two standard deviations above the expected value, 90, of the sample mean.
Solution:
normalcdf:
(lower value, upper value, mean, standard error of the mean)
The parameter list is abbreviated (lower value, upper value, μ,[latex]displaystylefrac{{sigma}}{{sqrt{n}}}[/latex]
normalcdf:
(85,92,90, [latex]displaystylefrac{{15}}{{sqrt{25}}}[/latex] = 0.6997
To find the value that is two standard deviations above the expected value 90, use the formula:
- value = [latex]displaystyle{mu}_{x}[/latex] + (# of STDEVs)[latex]displaystyleleft