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

Answer:

This mileage interval is from 30120 miles and higher.

Step-by-step 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:

$Z = \frac{X - \mu}{\sigma}$

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

$Z = \frac{X - \mu}{\sigma}$

$1.28 = \frac{X - 25000}{4000}$

$X - 25000 = 4000*1.28$

$X = 30120$

This mileage interval is from 30120 miles and higher.