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

Answer:

The guarantee should be made of $X = 1.555 + 14\sigma$ years, in which $\sigma$ 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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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

$Z = \frac{X - \mu}{\sigma}$

$1.555 = \frac{X - 14}{\sigma}$

$X = 1.555 + 14\sigma$

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