Question

A manufacturer of tires wants to advertise a mileage interval that ex-cludes no more than 10% of the mileage on tires he sells. All he knowsis that, for a large number of tires tested, the mean mileage was 25,000miles, and the standard deviation was 4000 miles. What interval wouldyou suggest?

Answer 1

This mileage interval is from 30120 miles and higher.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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:

All he knows is that, for a large number of tires tested, the mean mileage was 25,000 miles, and the standard deviation was 4000 miles. This means that
`\mu = 25000, \sigma = 4000`.

A manufacturer of tires wants to advertise a mileage interval that ex-cludes no more than 10% of the mileage on tires he sells. What interval wouldyou suggest?

The lower end of this interval is X when Z has a pvalue of 0.90. That is
`Z = 1.28`.

So

`Z = (X - \mu)/(\sigma)`

`1.28 = (X - 25000)/(4000)`

`X - 25000 = 4000*1.28`

`X = 30120`

This mileage interval is from 30120 miles and higher.