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) (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.
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.
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