Question

A popular oil change place decided to offer a guarantee for its customers based on the time it took to complete the oil change. Their research showed that these times were normally distributed with a mean time of 17.8 minutes with a standard deviation of 3.2 minutes. They wanted to place the guarantee at a time so that only 3% of customers would be expected to wait beyond the guaranteed time. Where should the company have placed that guarantee time

Answer 1

The company should have placed that guarantee time on 23.8 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 zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Their research showed that these times were normally distributed with a mean time of 17.8 minutes with a standard deviation of 3.2 minutes.

This means that
`\mu = 17.8, \sigma = 3.2`

They wanted to place the guarantee at a time so that only 3% of customers would be expected to wait beyond the guaranteed time.

So the 100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So X when Z = 1.88.

Where should the company have placed that guarantee time

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

`1.88 = (X - 17.8)/(3.2)`

`X - 17.8 = 1.88*3.2`

`X = 23.8`

The company should have placed that guarantee time on 23.8 minutes.