Answer :
Given that:
- The tread life of a particular tire design is normally distributed.
- The Mean (in miles) is:
[tex]\mu=62000[/tex]- The Standard Deviation (in miles) is:
[tex]\sigma=5000[/tex]a. You need to find:
[tex]P(X\ge70000)[/tex]You need to find the corresponding z-score using this formula:
[tex]z=\frac{X-\mu}{\sigma}[/tex]In this case:
[tex]X=70000[/tex]Therefore, by substituting values and evaluating, you get:
[tex]z=\frac{70000-62000}{5000}=1.6[/tex]Then, you need to find:
[tex]P(z\ge1.6)[/tex]Using the Standard Normal Distribution Table, you get that:
[tex]P(z\ge1.6)\approx0.0548[/tex]Therefore:
[tex]P(X\ge70000)\approx0.0548[/tex]b. You need to find:
[tex]P(X\leq57000)[/tex]Find the z-score using:
[tex]X=57000[/tex]You get:
[tex]z=\frac{57000-62000}{5000}=-1[/tex]Then, you need to find the following using the Standard Normal Distribution Table:
[tex]P(X\leq-1)[/tex]You get:
[tex]P(X\leq-1)\approx0.1587[/tex]Therefore:
[tex]P(X\leq57000)\approx0.1587[/tex]c. You need to find:
[tex]P(57500Find the corresponding z-score for:[tex]X=57500[/tex]This is:
[tex]z=\frac{57500-62000}{5000}=-0.9[/tex]And find the z-score using:
[tex]X=65000[/tex]You get:
[tex]z=\frac{65000-62000}{5000}=0.6[/tex]You need to find:
[tex]P(-0.9Using the Standard Normal Distribution Table, you get that:[tex]P(z<0.6)\approx0.7257[/tex]And:
[tex]P(z<-0.9)\approx0.1841[/tex]Therefore:
[tex]P(-0.9Then:[tex]P(57500Hence, the answers are:a.
[tex]Probability\approx0.0548[/tex]b.
[tex]Probability\approx0.1587[/tex]c.
[tex]Probability\approx0.5416[/tex]