Question

Fast Auto Service provides oil and lube service for cars. It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is 2.4 minutes. The management wants to promote the business by guaranteeing a maximum waiting time for its customers. If a customer's car is not serviced within that period, the customer will receive a 50% discount on the charges. The company wants to limit this discount to at most 8% of the customers. What should the maximum guaranteed waiting time be

Answer 1

The maximum guaranteed waiting time should be of 18.37 minutes.

Explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

It is known that the mean time taken for oil and lube service at this garage is 15 minutes per car and the standard deviation is 2.4 minutes.

This means that
`\mu = 15, \sigma = 2.4`

The company wants to limit this discount to at most 8% of the customers. What should the maximum guaranteed waiting time be?

The 100 - 8 = 92th percentile, which is X when Z has a pvalue of 0.92. So X when Z = 1.405.

`Z = (X - \mu)/(\sigma)`

`1.405 = (X - 15)/(2.4)`

`X - 15 = 1.405*2.4`

`X = 18.37`

The maximum guaranteed waiting time should be of 18.37 minutes.