6.4 Using the central limit theorem  (Page 28/8)

It is important for you to understand when to use the CLT . If you are being asked to find the probability of the mean, use the CLT for the mean. If youare 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.

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 $\overline{x}$ of the sample tends to get closer and closer to $\mu$ . From the Central Limit Theorem, we know that as $n$ gets larger and larger, the sample means follow a normal distribution. The larger n gets, the smaller thestandard deviation gets. (Remember that the standard deviation for $\overline{X}$ is $\frac{\sigma }{\sqrt{n}}$ .) This means that the sample mean $\overline{x}$ must be close to the population mean $\mu$ . We can say that $\mu$ 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 done on a college campus among the students. The stress scores follow a uniform distribution with the lowest stress score equal to 1 and the highest equal to 5. Using a sample of 75 students, find:

1. The probability that the mean stress score for the 75 students is less than 2.
2. The 90th percentile for the mean stress score for the 75 students.
3. The probability that the total of the 75 stress scores is less than 200.
4. The 90th percentile for the total stress score for the 75 students.

Let $X$ = one stress score.

Problems 1. and 2. ask you to find a probability or a percentile for a mean . Problems 3 and 4 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 $b=5$ (See Continuous Random Variables for the uniform).

${\mu }_X=\frac{a+b}{2}=\frac{1+5}{2}=3$

${\sigma }_X=\sqrt{\frac{(b-a{)}^2}{12}}=\sqrt{\frac{(5-1{)}^2}{12}}=1.15$

For problems 1. and 2., let $\overline{X}$ = the mean stress score for the 75 students. Then,

$\overline{X}$ ~ $N(3,\frac{1.15}{\sqrt{75}})$ where $\mathrm{n\; =\; 75}$ .

For problems c and d, let $\mathrm{\Sigma X}$ = the sum of the 75 stress scores. Then, $\mathrm{\Sigma X}$ ~ $N\left[\right(\mathrm{75})\cdot (3),\sqrt{75}\cdot 1.15]$