Question

4. [20 Points] The time required for an automotive center to complete an oil change service an 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

a)

mean = 17, S.D = 2

Here, we need to calculate the z-score when X = 20

o = Z n - X

z = (20 - 17) / 2 = 3/2 = 1.5

Which gives a value of 0.9332 in the z-table.

This is a proportion of people who will get done in less than 20 minutes. For more than 20 minutes, we need to subtract this value with 1.

More than 20 minutes = 1 - 0.9332 = 0.0668

6.68% of the people will get it done in half-price.

b)

For that, we need to check the value of z-score corresponding with 98% of the people. which is 0.98.

It comes out to be 2.06.

The value will be 2.06 standard deviation above the mean.

Value = mean + 2.06*S.D = 17 + 2.06*2 = 17 + 4.12 = 21.12

The guaranteed time limit should be 21.12 minutes.