Question

The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal​ distribution, with a mean of 17 minutes and a standard deviation of 4 minutes. ​(a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take​ longer, the customer will receive the service for​ half-price. What percent of customers receive the service for​ half-price? ​(b) If the automotive center does not want to give the discount to more than 2​% of its​ customers, how long should it make the guaranteed time​ limit?

Answer 1

a) 22.66% of customers receive the service for​ half-price.

b) The guaranteed time​ limit should be of 25.2 minutes.

Explanation:

When the distribution is normal, we use 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 question, we have that:

`\mu = 17, \sigma = 4`

​(a) The automotive center guarantees customers that the service will take no longer than 20 minutes. If it does take​ longer, the customer will receive the service for​ half-price. What percent of customers receive the service for​ half-price?

Longer than 20 minutes is 1 subtracted by the pvalue of Z when X = 20. So

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

`Z = (20 - 17)/(4)`

`Z = 0.75`

`Z = 0.75` has a pvalue of 0.7734

1 - 0.7734 = 0.2266

22.66% of customers receive the service for​ half-price.

(b) If the automotive center does not want to give the discount to more than 2​% of its​ customers, how long should it make the guaranteed time​ limit?

The time limit should be the 100 - 2 = 98th percentile, which is X when Z has a pvalue of 0.98. So X when Z = 2.054.

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

`2.054 = (X - 17)/(4)`

`X - 17 = 4*2.054`

`X = 25.2`

The guaranteed time​ limit should be of 25.2 minutes.