Fast Auto Service guarantees that the maximum waiting time for its customers is
ID: 3129103 • Letter: F
Question
Fast Auto Service guarantees that the maximum waiting time for its customers is 20 minutes for oil/lube service on their cars. It also guarantees that any customer who has to wait longer than 20 minutes for this service will receive a 50% discount on the charges. It is estimated that the mean time taken for oil/lube service at this garage is 15 minutes per car and the standard deviation is 2.4 minutes. Suppose the time it takes for oil/lube service is normally distributed. What proportion of customers would you expect will receive a 50% discount on their charges? State your answer to the nearest thousandth.
Explanation / Answer
The mean time taken =15mins
Standard deviation=2.4
The time has to be greater than 20mins for 50% discount
Hence number of standard deviations greater than 15 to reach 20=(20-15)/2.4 =2.083
Hence those customers in the range above 2.083 standard deviations above the mean receive 50% discount.
Hence using the Normal distribution table
Z(2.083)=0.981374
Hence those recievinf 50% discount=1-Z(2.083)=0.01863 =0.019 (Rounded) or 1.9% of the customers receive 50% discount
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