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 15 minutes and a standard deviation of 3.5 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)
Let T be the service time for a customer.

Probability of customers receive the service for​ half-price = P(T > 20)

= P[Z > (20 - 15)/3.5]

= P[Z > 1.43]

= 0.0764

Percent of customers receive the service for​ half-price = 0.0764 * 100 = 7.64%

(B)

If the automotive center does not want to give the discount to more than 2% of its​ customers, we need to calculate 100 - 2 = 98% percentile of the service time T.

Z value for 98% percentile is 2.054

98% percentile of the service time T. = = 15 + 2.054 * 3.5 = 22.189

Thus, the automotive center should it make the guaranteed time​ limit of 22.189 minutes