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
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
.
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
.
So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.28 = \frac{X - 25000}{4000}](https://tex.z-dn.net/?f=1.28%20%3D%20%5Cfrac%7BX%20-%2025000%7D%7B4000%7D)
![X - 25000 = 4000*1.28](https://tex.z-dn.net/?f=X%20-%2025000%20%3D%204000%2A1.28)
![X = 30120](https://tex.z-dn.net/?f=X%20%3D%2030120)
This mileage interval is from 30120 miles and higher.