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

Answer:

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.

Step-by-step 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:

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

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.

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 $X = 5$.

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

$Z = \frac{5 - 8.2}{1.1}$

$Z = -2.91$

$Z = -2.91$ 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

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

$2.33 = \frac{X - 8.2}{1.1}$

$X - 8.2 = 2.33*1.1$

$X = 10.76$

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.