Question

Replacement times for TV sets are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years (Consumer Reports). Find the probability that a randomly selected TV will have a replacement time less than 5.0 years. If you want to provide a warranty so that only 1% of the TV sets will be replaced before the warranty expires, what is the time length of the warranty?

Answer 1

There is a 0.18% probability that a randomly selected TV will have a replacement time less than 5.0 years.

To provide a warranty so that only 1% of the TV sets will be replaced before the warranty expires, the length of the warranty is 10.76 years.

Explanation:

Problems of normally distributed samples 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.

In this problem, we have that:

Replacement times for TV sets are normally distributed with a mean of 8.2 years and a standard deviation of 1.1 years. This means that
`\mu = 8.2, \sigma = 1.1`.

Find the probability that a randomly selected TV will have a replacement time less than 5.0 years.

This is the pvalue of Z when
`X = 5`.

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

`Z = (5 - 8.2)/(1.1)`

`Z = -2.91`

`Z = -2.91` has a pvalue of 0.00181

This means that there is a 0.18% probability that a randomly selected TV will have a replacement time less than 5.0 years.

If you want to provide a warranty so that only 1% of the TV sets will be replaced before the warranty expires, what is the time length of the warranty?

This is the value of X when Z has a pvalue of 0.99. This is
`Z = 2.33`.

So

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

`2.33 = (X - 8.2)/(1.1)`

`X - 8.2 = 2.33*1.1`

`X = 10.76`

To provide a warranty so that only 1% of the TV sets will be replaced before the warranty expires, the length of the warranty is 10.76 years.