Question

The tread life of a particular tire design is normally distributed with a mean of 62,000 miles and a standard deviation of 5,000 miles.a. What is the probability that the tread life of a random tire of this design will be at least 70,000 miles?

Answer 1

Given that:

- The tread life of a particular tire design is normally distributed.

- The Mean (in miles) is:

`\mu=62000`

- The Standard Deviation (in miles) is:

`\sigma=5000`

a. You need to find:

`P(X\ge70000)`

You need to find the corresponding z-score using this formula:

`z=(X-\mu)/(\sigma)`

In this case:

`X=70000`

Therefore, by substituting values and evaluating, you get:

`z=(70000-62000)/(5000)=1.6`

Then, you need to find:

`P(z\ge1.6)`

Using the Standard Normal Distribution Table, you get that:

`P(z\ge1.6)\approx0.0548`

Therefore:

`P(X\ge70000)\approx0.0548`

b. You need to find:

`P(X\leq57000)`

Find the z-score using:

`X=57000`

You get:

`z=(57000-62000)/(5000)=-1`

Then, you need to find the following using the Standard Normal Distribution Table:

`P(X\leq-1)`

You get:

`P(X\leq-1)\approx0.1587`

Therefore:

`P(X\leq57000)\approx0.1587`

c. You need to find:

`P(57500Find the corresponding z-score for:[tex]X=57500`

This is:

`z=(57500-62000)/(5000)=-0.9`

And find the z-score using:

`X=65000`

You get:

`z=(65000-62000)/(5000)=0.6`

You need to find:

`P(-0.9Using the Standard Normal Distribution Table, you get that:[tex]P(z<0.6)\approx0.7257`

And:

`P(z<-0.9)\approx0.1841`

Therefore:

`P(-0.9Then:[tex]P(57500Hence, <strong>the answers are:</strong><p><strong>a.</strong></p>[tex]Probability\approx0.0548`

b.

`Probability\approx0.1587`

c.

`Probability\approx0.5416`