Question

Consumer Reports indicated that the average life of a refrigerator before replacement is 14 years with a standard deviation of 2.5 years. Let X be the age at which a refrigerator is replaced. Assume that X has a distribution that is approximately normal. Suppose the company wants to set the warranty period so that it does not have to replace more than 2.5% of the refrigerators under guarantee.

Required:
a. State the random variable.
b. What is the probability that someone will keep a refrigerator fewer than 11 years before replacement?
c. What is the probability that someone will keep a refrigerator more than 18 years before replacement?
d. What is the probability that someone will replace a refrigerator between 8 and 15 years?

Answer 1

a) The random variable is the average life of a refrigerator.

b) 0.1151 = 11.51% probability that someone will keep a refrigerator fewer than 11 years before replacement.

c) 0.0548 = 5.48% probability that someone will keep a refrigerator more than 18 years before replacement

d) 0.6472 = 64.72% probability that someone will replace a refrigerator between 8 and 15 years

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 z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Consumer Reports indicated that the average life of a refrigerator before replacement is 14 years with a standard deviation of 2.5 years

This means that:

The random variable is the average life of a refrigerator.

`\mu = 14, \sigma = 2.5`

a. State the random variable.

The random variable is the average life of a refrigerator.

b. What is the probability that someone will keep a refrigerator fewer than 11 years before replacement?

This is the pvalue of Z when X = 11. So

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

`Z = (11 - 14)/(2.5)`

`Z = -1.2`

`Z = -1.2` has a pvalue of 0.1151

0.1151 = 11.51% probability that someone will keep a refrigerator fewer than 11 years before replacement.

c. What is the probability that someone will keep a refrigerator more than 18 years before replacement?

This is 1 subtracted by the pvalue of Z when X = 18. So

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

`Z = (18 - 14)/(2.5)`

`Z = 1.6`

`Z = 1.6` has a pvalue of 0.9452

1 - 0.9452 = 0.0548

0.0548 = 5.48% probability that someone will keep a refrigerator more than 18 years before replacement.

d. What is the probability that someone will replace a refrigerator between 8 and 15 years?

This is the pvalue of Z when X = 15 subtracted by the pvalue of Z when X = 8.

X = 15

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

`Z = (15 - 14)/(2.5)`

`Z = 0.4`

`Z = 0.4` has a pvalue of 0.6554

X = 8

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

`Z = (8 - 14)/(2.5)`

`Z = -2.4`

`Z = -2.4` has a pvalue of 0.0082

0.6554 - 0.0082 = 0.6472

0.6472 = 64.72% probability that someone will replace a refrigerator between 8 and 15 years