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 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

Ansi Given that - Mean M) 1 It min, -) Slandard deviation (m) & Lrmin 19) p(x >26) = 1- P(x+29) X-4 1-P (1-9 e 20-17 2 = 1-P (2 2 2 2 1- (Z 2115) T-'0x4332 0,0663 = 6263/ - Therefore, the percentage of customers that receive The Seavice rate for Half price To 6.68%

by We want to Restrict it to no more 21. i, e; p (X>1) 20102 pix x) 1-01020198 Z (x - 19 (2-19 21054 = 2,054 (2) = (x-19) 4.108-X-19 X = 19 +41108 = 2.3, 108 min. - Therefore, Guaranteed Time limit should be made 23.108 min.