Respuesta :
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 [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = 8.2, \sigma = 1.1[/tex].
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 [tex]X = 5[/tex].
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5 - 8.2}{1.1}[/tex]
[tex]Z = -2.91[/tex]
[tex]Z = -2.91[/tex] 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 [tex]Z = 2.33[/tex].
So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]2.33 = \frac{X - 8.2}{1.1}[/tex]
[tex]X - 8.2 = 2.33*1.1[/tex]
[tex]X = 10.76[/tex]
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.
