Question

Assume that the average life of a refrigerator is 14 years, with the standard deviation given in part (a) before it breaks. Suppose that a company guarantees refrigerators and will replace a refrigerator that breaks while under guarantee with a new one. However, the company does not want to replace more than 6% of the refrigerators under guarantee. For how long should the guarantee be made (rounded to the nearest tenth of a year)

Answer 1

The guarantee should be made of
`X = 1.555 + 14\sigma` years, in which
`\sigma` is the standard deviation given in part (a).

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.

Assume that the average life of a refrigerator is 14 years

This means that
`\mu = 14`

The standard deviation given in part (a) before it breaks.

This will be the value of
`\sigma`

However, the company does not want to replace more than 6% of the refrigerators under guarantee. For how long should the guarantee be made?

We need to find the 100 - 6 = 94th percentile, which is X when Z has a pvalue of 0.94. So X when Z = 1.555. So

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

`1.555 = (X - 14)/(\sigma)`

`X = 1.555 + 14\sigma`

The guarantee should be made of
`X = 1.555 + 14\sigma` years, in which
`\sigma` is the standard deviation given in part (a).