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: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 [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 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 [tex]\mu = 25000, \sigma = 4000[/tex].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 [tex]Z = 1.28[/tex].So[tex]Z = \frac{X - \mu}{\sigma}[/tex][tex]1.28 = \frac{X - 25000}{4000}[/tex][tex]X - 25000 = 4000*1.28[/tex][tex]X = 30120[/tex]This mileage interval is from 30120 miles and higher.