Normal distribution
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The owners of Spiffy Lube want to offer their customers a 10-minute guarantee on their standard oil change service. If the oil change takes longer than 10 minutes to complete, the customer is given a coupon for a free oil change at the next visit. Based on past history, the owners believe that the time required to complete an oil change has a normal distribution with a mean of 8.6 minutes and a standard deviation of 1.2 minutes.
a. What percentage of customers will receive a free oil change coupon?
b. If management wants to limit the percentage of customers receiving a coupon to no more than 1 out of every 25 customers on average, what should they change the guaranteed time to?
c. Suppose management could improve the process by reducing the mean time required for an oil change (but keeping the standard deviation the same). How much change in the mean service time would be required to allow for a 10-minute guarantee that gives a coupon to no more than 1 out of every 25 customers on average?
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Solution Summary
This solution is an example of how to calculate probability and percentiles from a normal distribution.
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