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The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a population mean of 60,000 miles and a population standard deviation of 4,000 miles. Suppose we select a sample of 40 tires and use a simulator to determine the tread life. What is the likelihood of finding that the sample mean is between 59,050 and 60,950?

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JeanaShupp

Answer: 0.8664

Step-by-step explanation:

Given : Mean : [tex]\mu =  60,000\text{ miles}[/tex]

Standard deviation : [tex]\sigma = 4,000\text{ miles}[/tex]

Sample size : [tex]n=40[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}[/tex]

For x=  59,050

[tex]z=\dfrac{59050-60000}{\dfrac{4000}{\sqrt{40}}}\approx-1.50[/tex]

For x= 60,950

[tex]z=\dfrac{60950-60000}{\dfrac{4000}{\sqrt{40}}}\approx1.50[/tex]

The P-value : [tex]P(-1.5<z<1.5)=P(z<1.5)-P(z<-1.5)[/tex]

[tex]=0.9331927-0.0668072=0.8663855\approx0.8664[/tex]

Hence, the likelihood of finding that the sample mean is between 59,050 and 60,950=0.8664

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