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

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

Step-by-step explanation:

Normal Probability Distribution:

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

In a set with mean and standard deviation , the zscore of a measure X is given by:

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

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

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

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