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 2.5 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 2​% of its​ customers, how long should it make the guaranteed time​ limit?

Answer 1

a: about 66%

b: 25 minutes (24 minutes, 8.25 seconds, but we round up)

Explanation:

Part a:

We will find the z-score if the situation.

µ = 19, σ = 2.5

x = 20.

The z-score is: z = (20 - 19)/2.5 = 0.40

P(z < 0.40) = 0.6554, so about 66% of customers would receive that discount

Part b:

We want to find the z-score that corresponds to 0.980, (top 2%). Looking of the chart we have to estimate since the exact value of 0.9800 isn't on the chart. z = 2.05 gives us 0.9798, and z = 2.06 gives us 0.9803. 0.9800 is just about in between them, so we average 2.05 and 2.06 giving us 2.055. We need to find the x-value that gives us a z-score of 2.055

2.055 = (x - 19)/2.5

5.1375 = x - 19 (multiply both sides by 2.5)

24.1375 = x (add 19 to both sides)

24.1375 minutes is 24 minutes 8.25 seconds. Rounding down to 24 minutes would make a little more than 2% get the discount, we need no more than 2%, so rounding up to 25 minutes would satisfy this requirement