Question

The time required for an automotive center to complete an oil change service on an automobile approximately follows a normal distribution, with the mean of 17 minutes and a standard deviation of 2 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

The first question asks for the proportion of customers that wait more than 20 minutes, which is given by the probability:
\[P(X>20)=P\left(\frac{X-17}{2}>\frac{20-17}{2}\right)=P(Z>1.5)\]
For the second question, you want to find \(k\) such that
\[P(X>k)=P\left(\frac{X-17}{2}>\frac{k-17}{2}\right)=P\left(Z>\frac{k-17}{2}\right)=0.02\]