Question

Suppose a tire manufacturer wants to set a mileage guarantee on its new XB 70 tire. Tests revealed that the tire's mileage is normally distributed with a mean of 47,900 miles and a standard deviation of 2,050 miles. The manufacturer wants to set the guaranteed mileage so that no more than 5% of the tires will have to be replaced. What guaranteed mileage should the manufacturer announce

Answer 1

Answer:

The manufacturer should announce a guaranteed mileage of 44528 miles

Explanation:

Problems of normally distributed samples are 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:

$\mu = 47900, \sigma = 2050$

What guaranteed mileage should the manufacturer announce

Only until the 5th percentile will have to be replaced, which is the value of X when Z has a pvalue of 0.05. So it is X when Z = -1.645.

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

$-1.645 = \frac{X - 47900}{2050}$

$X - 47900 = -1.645*2050$

$X = 44528$

The manufacturer should announce a guaranteed mileage of 44528 miles

Answer 2

Answer:

x = -1.6449*2050 + 47900 = 44,528

Explanation:

no more than 5 percent of the tires will have to be replaced...

then z = -1.6449

x = -1.6449*2050 + 47900 = 44,528