SOLUTION: 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 standa

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Question 1137717: 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 3 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 by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
the mean is 17 minutes and the standard deviation is 3 minutes.

the probability that it will take longer than 20 minutes is .1587.

that means that almost 16% of the customers would receive half price.

if the automotive service does not want to give the discount to more than 7% of its customers, then the guarenteed time limits has to be greater than 21.428 minutes.

these figures were derived using the following normal distribution calculator.

http://davidmlane.com/hyperstat/z_table.html

here's a display of the results.

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the first display calculates area from a value.

the second display calculates value from an area.

if you use the calculator with x-scores, the mean is 0 and the standard deviation is 1.

if you use the calculator with raw scores, as i did above, the mean is the mean of your distribution and the standard deviation is the standard deviation of your distribution.