Module 7: The Central Limit Theorem
Using the Central Limit Theorem
Barbara Illowsky & OpenStax et al.
Learning Objectives
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 [latex]displaystyleoverline{{x}}[/latex] 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
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 equal to one and the highest equal to five. Using a sample of 75 students, find:
- The probability that the mean stress score for the 75 students is less than two.
- The 90th percentile for the mean stress score for the 75 students.
- The probability that the total of the 75 stress scores is less than 200.
- The 90th percentile for the total stress score for the 75 students.
Let X = one stress score.
Problems a and b ask you to find a probability or a percentile for a mean. Problems c and d 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.
[latex]displaystyle{mu}_{X}=frac{{a+b}}{{2}}=frac{{1+5}}{{2}}={3}[/latex][latex]displaystyle{sigma}_{X}=sqrt{frac{{{(b-a)}^{2}}}{{12}}}=sqrt{frac{{({5-1)}^{2}}}{{12}}}[/latex]= 1.15
For problems 1. and 2., let [latex]displaystyleoverline{X}[/latex] = the mean stress score for the 75 students. Then,
[latex]displaystyleoverline{X}sim{N}({3},frac{{1.15}}{{sqrt{75}}})text{ where } {n}={75}[/latex].
Example
- Find P([latex]displaystyleoverline{x}{<}{2}[/latex]). Draw the graph.
- Find the 90th percentile for the mean of 75 stress scores. Draw a graph.
- Find P(Σx < 200). Draw the graph.
- Find the 90th percentile for the total of 75 stress scores. Draw a graph.
Solution:
- P( < 2) = 0 The probability that the mean stress score is less than two is about zero.
normalcdf
[latex]displaystyle{({1},{2},{3},frac{{1.15}}{sqrt{{75}}})}={0}[/latex]
Remember that the smallest stress score is one.
2. Let k = the 90th percentile.Find
k, where P( < 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 that 10% are at least