The Central Limit Theorem (College Board AP® Statistics): Revision Note

The mean, $\mu$, and the standard deviation, $\frac{\sigma }{\sqrt{n}}$, are given in the exam under 'Sampling distributions for means', in the row called 'For one population'.

In your exam, you just need to state that the sample size is large enough. $n\ge 30$ is not a rule that you are expected to learn.

In Central Limit theorem questions, always show that you have checked the sample size satisfies $n\ge 30$!

The number of minutes that it takes to wait for a bus at a particular bus stop is distributed evenly between 0 minutes and 50 minutes. Any particular number of minutes that a person is required to wait between these two times is equally likely to occur. The standard deviation of waiting times is 14.4 minutes.

(a) If a person travels on the bus from this bus stop 40 times, estimate the probability that the mean time they have to wait exceeds 26 minutes.

(b) If, instead, the person travels on the bus from this bus stop 15 times, explain whether or not the method used in part (a) is still appropriate.

Answer:

(a)

This a probability question about the mean of a sample (a sample of 40 waiting times)

Waiting times are not normally distributed (they are evenly distributed)

This means the Central Limit theorem is required (check $n\ge 30$)

The sample size is $n=40$ which satisfies $n\ge 30$

The Central Limit theorem states that sample means follow an approximate normal distribution with mean $\mu$ and standard deviation $\frac{\sigma }{\sqrt{n}}$

The question gives $\sigma$ but not $\mu$, so work this out from the context

The mean waiting time between 0 and 50 minutes, if all times are equally likely, is 25 minutes

$\mu =25$

Use the standard deviation of 14.4 minutes and $n=40$ to find $\frac{\sigma }{\sqrt{n}}$

$\frac{\sigma }{\sqrt{n}}=\frac{14.4}{\sqrt{40}}=2.2768399...$

Find the probability the sample mean exceeds 26, $\mathrm{P}\left(\overline{X}>26\right)$

The z-score is

$\frac{26-25}{2.2768399...}=0.4392...$

Using 0.44 from the tables gives 0.6700 as the probability of being less than z

Subtract this from 1 to find the probability of exceeding z

1 - 0.6700

The probability that the mean waiting time exceeds 26 minutes is approximately 0.3300

(b)

The method used in part (a) involves the Central Limit theorem

The Central Limit theorem requires that $n\ge 30$

However the sample size here is $n=15$ which is less than 30

So the method in part (a) is not appropriate