Respuesta :
Answer:
The guarantee should be made of [tex]X = 1.555 + 14\sigma[/tex] years, in which [tex]\sigma[/tex] is the standard deviation given in part (a).
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions 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.
Assume that the average life of a refrigerator is 14 years
This means that [tex]\mu = 14[/tex]
The standard deviation given in part (a) before it breaks.
This will be the value of [tex]\sigma[/tex]
However, the company does not want to replace more than 6% of the refrigerators under guarantee. For how long should the guarantee be made?
We need to find the 100 - 6 = 94th percentile, which is X when Z has a pvalue of 0.94. So X when Z = 1.555. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.555 = \frac{X - 14}{\sigma}[/tex]
[tex]X = 1.555 + 14\sigma[/tex]
The guarantee should be made of [tex]X = 1.555 + 14\sigma[/tex] years, in which [tex]\sigma[/tex] is the standard deviation given in part (a).
