Question

Suppose that the waiting time for a license plate renewal at a local office of a state motor vehicle department has been found to be normally distributed with a mean of 30 minutes and a standard deviation of 8 minutes. What is the probability that a randomly selected individual will have a waiting time between 15 and 45 minutes?

Answer 1

94.26% probability that a randomly selected individual will have a waiting time between 15 and 45 minutes.

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 30, \sigma = 8`

What is the probability that a randomly selected individual will have a waiting time between 15 and 45 minutes?

This is the pvalue of Z when
`X = 45` subtracted by the pvalue of Z when
`X = 15`.

So

X = 45

`Z = (X - \mu)/(\sigma)`

`Z = (45 - 30)/(8)`

`Z = 1.9`

`Z = 1.9` has a pvalue of 0.9713.

X = 15

`Z = (X - \mu)/(\sigma)`

`Z = (15 - 30)/(8)`

`Z = -1.9`

`Z = -1.9` has a pvalue of 0.0287.

So there is a 0.9713 - 0.0287 = 0.9426 = 94.26% probability that a randomly selected individual will have a waiting time between 15 and 45 minutes.