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 19 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 3​% of its​ customers, how long should it make the guaranteed time​ limit?

​(a) The percent of customers that receive the service for​ half-price is ___ %?

Answer 1

Seavice on van Given't the time required to an automotive centre to complete an oil change Service on can automibile is given by mean (e) = 19 minutes standard deviation (o) = 2 minutes (a) Auto motive, that customers will take no longer than montere It us given by +(x>20) = 1-P(x220) we know that z=x-le - 1-P(< 20-10) =l-p( 22 ) El-p(720.5) using standard Normal table, we get -|-0.6915 = 0.3085 [ 30.8514

therefore, ** percentage I the service of recieve customers that recieve ] rate for hay price is 30.851 | 1. (b) In this case, . . If the automotive centre want to give the discount to more than 3.4. is, P(xxx) = 0.03 lie., P(x<x) = 0.03

we we know that z = X-le - p/x-he axle) = 0.03 Z-coitical value wat * = 0.03 cs 1.88 PLZ < -19) = 1.88 Ke-19 = 1-88 x = 2(1.88) +19 x= 3.76 +9 = 22.76 ~ 23 + x 23 minutes - 2

(a) The percentage of customers that receive the service for half -price is 30.85%

Answer 2

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