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

Solution-

Time to complete oil change service follows normal distribution with

Mean = $\mu$ = 19 minutes

Standard deviation = $\sigma$ = 9 minutes

(a)

Probability that the time taken is less than 20 minutes

= P( X ≤ 20)

20- = P2 <

20 – 19 = P2< 9

= P2 < 9

= P(Z < 0.111)

= 0.50 + 0.0438

= 0.5438

So, 54.38% times the time taken is less than 20 minutes.

So, 100 - 54.38 = 45.62 % times the time taken is more than 20 minutes.

Hence, 45.62 % of the customer receive servive for half price.

(b)

Let x minutes be the service time for which not more than 7% of customers will receive discount.

So, 100 -7 = 93 % of the time , the service time should be less then x minutes

So, P( X = x) ≤ 0.93 or

P2= < 0.93

2 - 19 P2= < 0.93 9

(

T - 19 < 1.48 9

x - 19 ≤ 1.48×9

x - 19 ≤ 13.32

x ≤ 13.32 + 19

x ≤ 32.32 minutes

Hence, minimum 32.32 minutes should be the guaranteed service time to the customer to have not more than 7% customer receive discount.