The toasters produced by a company have a normally distributed life span with a mean of 5.8 years and a standard deviation of 0.9 years, what warranty should be provided so that the company is replacing at most 5% of their toasters sold?

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Answer:

A warranty of 4.32 years should be provided.

Step-by-step explanation:

Problems of normally distributed samples are 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 question, we have that:

[tex]\mu = 5.8, \sigma = 0.9[/tex]

What warranty should be provided so that the company is replacing at most 5% of their toasters sold?

The warranty should be the 5th percentile, which is X when Z has a pvalue of 0.05. So it is X when Z = -1.645.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-1.645 = \frac{X - 5.8}{0.9}[/tex]

[tex]X - 5.8 = -1.645*0.9[/tex]

[tex]X = 4.32[/tex]

A warranty of 4.32 years should be provided.

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