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 2.5 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 5% of its customers, how long should it make the guaranteed time limit?

a) 11.51%; b) 13 minutes

Step-by-step explanation:

We will use a z score to answer these questions.  The formula for a z score is

For part a, X is 20, the mean is 17 and the standard deviation is 2.5:

z = (20-17)/2.5 = 3/2.5 = 1.2

Using a z table, we see the area to the left of, or less than, this value is 0.8849; this means the area to the right of, or greater than, is 1-0.8849 = 0.1151.  This means 11.51% of people have a time greater than 20 minutes and get half off.

For part b, we start out in the z table finding values as close to 5%, or 0.05, as possible.  There are two possibilities, 0.0505 (z=-1.64) and 0.0495 (z=-1.65). Since they are equally distant from the desired value, we will use -1.645 for z.  Our values for the mean and standard deviation are the same:

-1.645 = (X-17)/2.5

Multiply both sides by 2.5:

2.5(-1.645) = ((X-17)/2.5)(2.5)

-4.1125 = X-17

Add 17 to each side:

-4.1125+17 = X-17+17

12.8875 = X

This means the garage should make the guaranteed time no more than 13 minutes.

eve and marks ratios are equivalent

step-by-step explanation: