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From long experience, it is known that the time it takes to do an oil change and lubrication job on a vehicle has a normal distribution with a mean of 17.8 minutes and a standard deviation of 5.2 minutes. An auto service shop will give a free lube and oil change service to any customer who must wait beyond the guaranteed time to complete the work. If the shop does not want to give more than 1% of its customers a free lube and oil change service, how long should the guarantee be

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Answer: 29.916 minutes

Step-by-step explanation:

Given the following :

Mean of distribution = 17.8 minutes

Standard deviation of distribution = 5.2 minutes

Percentage of customers to give free lube and oil change service should not exceed 1% if the customer waits beyond the guaranteed time ;

Length of guarantee :

Since % of customers should not exceed 1%

Then probability of not being chosen = (1 - 0.01) = 0.99

Using the z table to find the Zscore of 0.99,

Zscore of 0.99 = 2.33

This is 2.33 standard deviations above the mean

(2.33 × standard deviation) + mean value

(2.33 × 5.2 minutes) + 17.8 minutes

12.116 + 17.8

= 29.916 minutes

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