35 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.
NoteIf 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]\displaystyle\overline{{x}}[/latex] of the sample tends to get closer and closer to the population mean μ.
The formula for the standard deviation of variable [latex]\overline{x}[/latex] is [latex]\frac{\sigma}{\sqrt{n}}[/latex]. If n is getting larger, then [latex]\frac{\sigma}{\sqrt{n}}[/latex] is getting smaller. Indirectly, the sample mean [latex]\overline{x}[/latex] will be closed to the population mean [latex]\mu[/latex]
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.
[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 a and b, let [latex]\displaystyle\overline{X}[/latex] = the mean stress score for the 75 students.
Then, [latex]\displaystyle\overline{X}\sim{N}({3},\frac{{1.15}}{{\sqrt{75}}})\text{ where } {n}={75}[/latex].
Solution
a. Find the probability that the mean stress score for the 75 students is less than two.
We are asked to find P([latex]\displaystyle\overline{x}{<}{2}[/latex]). By plotting the graph,
We will use TI-83/84 to solve for part (a).
TI-Calculator: normalcdf [latex]\displaystyle{({1},{2},{3},\frac{{1.15}}{\sqrt{{75}}})}={0}[/latex]
| Remember that the smallest stress score is one. |
b. Find the 9