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 2.5 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 5​% of its​ customers, how long should it make the guaranteed time​ limit?

Answer:a) 11.51%; b) 13 minutesStep-by-step explanation:We will use a z score to answer these questions.  The formula for a z score is[tex]z=\frac{X-\mu}{\sigma}[/tex]For part a, X is 20, the mean is 17 and the standard deviation is 2.5:z = (20-17)/2.5 = 3/2.5 = 1.2Using a z table, we see the area to the left of, or less than, this value is 0.8849; this means the area to the right of, or greater than, is 1-0.8849 = 0.1151.  This means 11.51% of people have a time greater than 20 minutes and get half off.For part b, we start out in the z table finding values as close to 5%, or 0.05, as possible.  There are two possibilities, 0.0505 (z=-1.64) and 0.0495 (z=-1.65). Since they are equally distant from the desired value, we will use -1.645 for z.  Our values for the mean and standard deviation are the same:-1.645 = (X-17)/2.5Multiply both sides by 2.5:2.5(-1.645) = ((X-17)/2.5)(2.5)-4.1125 = X-17Add 17 to each side:-4.1125+17 = X-17+1712.8875 = XThis means the garage should make the guaranteed time no more than 13 minutes.