Question

Albert Abbasi, VP of Operations at Ingleside International Bank, is evaluating the service level provided to walk-in customers. Accordingly, he plans a sample of waiting times for walk-in customers. If the population of waiting times has a mean of 15 minutes and a standard deviation of 4 minutes, the probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is ________.

Answer 1

The probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is 0.9973.

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

`X \sim N(\mu=15,\sigma=4)`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

On this case
`\bar X \sim N(15,(4)/(√(64)))`

Solution to the problem

We are interested on this probability

`P(13.5<\bar X<16.5)`

If we apply the Z score formula to our probability we got this:

`P(13.5<\bar X<16.5)=P((13.5-\mu)/((\sigma)/(√(n)))<(X-\mu)/((\sigma)/(√(n)))<(16.5-\mu)/((\sigma)/(√(n))))`

`=P((13.5-15)/((4)/(√(64)))<Z<(16.5-15)/((4)/(√(64))))=P(-3<Z<3)`

And we can find this probability on this way:

`P(-3<Z<3)=P(Z<3)-P(Z<-3)`

And in order to find these probabilities we can find tables for the normal standard distribution, excel or a calculator.

`P(-3<Z<3)=P(Z<3)-P(Z<-3)=0.99865-0.00135=0.9973`

The probability that Albert's sample of 64 will have a mean between 13.5 and 16.5 minutes is 0.9973.