The blenders produced by a company have a normally distributed life span with a mean of 8.2 years and a standard deviation of 1.3 years. What warranty should be provided so that the company is replacing at most 6% of their blenders sold?

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Answer:

A warranty of 6.185 years should be provided.

Step-by-step explanation:

When the distribution is normal, we use 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 = 8.2, \sigma = 1.3[/tex]

What warranty should be provided so that the company is replacing at most 6% of their blenders sold?

The warranty should be the 6th percentile, which is X when Z has a pvalue of 0.06. So X when Z = -1.55.

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

[tex]-1.55 = \frac{X - 8.2}{1.3}[/tex]

[tex]X - 8.2 = -1.55*1.3[/tex]

[tex]X = 6.185[/tex]

A warranty of 6.185 years should be provided.