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)

The guarantee should be made of years, in which is the standard deviation given in part (a).

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.

Assume that the average life of a refrigerator is 14 years

This means that

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

This will be the value of

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

The guarantee should be made of years, in which is the standard deviation given in part (a).

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: