Question

A company that produces an expensive stereo component is considering offering a warranty on the component. Suppose the population of lifetimes of the components is a normal distribution with a mean of 84 months and a standard deviation of 9 months. If the company wants no more than 2% of the components to wear out before they reach the warranty date, what number of months should be used for the warranty? (Enter your answer as a whole number.)

Answer 1

Warranty of 66 months.

Explanation:

Problems of normally distributed samples are 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.

In this problem, we have that:

`\mu = 84, \sigma = 9`

If the company wants no more than 2% of the components to wear out before they reach the warranty date, what number of months should be used for the warranty?

Only the lowest 2% will be replaced, so the warranty is the value of X when Z has a pvalue of 0.02. So it is X when Z = -2.055.

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

`-2.055 = (X - 84)/(9)`

`X - 84 = -2.055*9`

`X = 65.5`

Warranty of 66 months.