Question

4. [20 Points] 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 19 minutes and a standard variance of 9 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 7% of its customers, how long should it make the guaranteed time limit?

Answer 1

Sl": Griven that XNN (4, 62) where le = 1g (a) Also given that, if the service time will take longer than 20 minutes then the customer will receive Service for half-price. PLX720) = ? ie. Pldt > 20-19 P(Z > 0.11) 2015-P10-ZZŁ011) 0.5-0.0438 0.4562 ras Z-0 Z-0.1) half price 0.4562 X/00 7 Hence, percentage of customeo's receive the service for 45 62 % Ay b) Now, center does not want to give_discount to more. than 7% of its customer. die PLX 270 = 7%. then find x'. e P(x - 72-19) P(2 7 2') = 0.07, where z'=2549 0:5- Plo<z«Z') = 0.07 POCZ LZE 0.5-0.07 PloczLz')= 0.43 z'= 1.476 {from normal tables 2-19 = 1.476 2=0 Z=2' ta 9 > X = (9 X 1.476) +19 32: 284 minuta