Problem 47E

Life Spans of Tires A certain brand of automobile tire has a mean life span of 35,000 miles, with a standard deviation of 2250 miles. Assume the life spans of the tires have a bell-shaped distribution.

(a) The life spans of three randomly selected tires are 34,000 miles, 37,000 miles, and 30,000 miles. Find the z-score that corresponds to each life span. Determine whether any of these life spans are unusual.

(b) The life spans of three randomly selected tires are 30,500 miles, 37,250 miles, and 35,000 miles. Using the Empirical Rule, find the percentile that corresponds to each life span.

Solution :

Step 1 of 2:

Given a certain brand of automobile tire has a mean lifespan of 35,000 miles and a standard deviation of 2250 miles.

Here and .

Our goal is :

a). We need to find the z-score that corresponds to each life span.

b). We need to find the percentile that corresponds to each life span.

a). Given the life spans of 3 randomly selected tires :

34,000 miles, 37,000 miles, and 30,000 miles.

The z-score formula is

z =

The life span of 34,000 days is

z =

We know that and .

z =

z =

z = -0.4444

Therefore, the z-score is -0.4444.

The life span of 37,000 days is

z =

We know that and .

z =

z =

z = 0.8888

Therefore, the z-score is 0.8888.

The lifespan of 30,000 days is

z =

We know that and .

z =

z =

z = -2.2222

Therefore, the z-score is -2.2222.

Hence, the z-score is -2.22, has an unusual life span.