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
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
.
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
.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{5 - 8.2}{1.1}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B5%20-%208.2%7D%7B1.1%7D)
![Z = -2.91](https://tex.z-dn.net/?f=Z%20%3D%20-2.91)
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
.
So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![2.33 = \frac{X - 8.2}{1.1}](https://tex.z-dn.net/?f=2.33%20%3D%20%5Cfrac%7BX%20-%208.2%7D%7B1.1%7D)
![X - 8.2 = 2.33*1.1](https://tex.z-dn.net/?f=X%20-%208.2%20%3D%202.33%2A1.1)
![X = 10.76](https://tex.z-dn.net/?f=X%20%3D%2010.76)
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.