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Doing research for insurance rates, it is found that those aged 30 to 49 drive an average of 38.7 miles per day with a standard deviation of 6.7 miles. These distances are normally distributed. If a group of 60 drivers in that age group are randomly selected, what is the probability that the mean distance traveled each day is between 32.5 miles and 40.5 miles?

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JeanaShupp

Answer: 0.4302

Step-by-step explanation:

Given : Mean : [tex]\mu=\text{38.7 miles }[/tex]

Standard deviation : [tex]\sigma=\text{6.7 miles }[/tex]

Sample size : [tex]n=60[/tex]

Also, these distances are normally distributed.

Then , the formula to calculate the z-score is given by :-

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

For x=32.5

[tex]\\\\ z=\dfrac{32.5-38.7}{6.7}=-0.925373134\approx-0.93[/tex]

For x=40.5

[tex]\\\\ z=\dfrac{40.5-38.7}{6.7}=0.268656\approx0.27[/tex]

The p-value = [tex]P(-0.93<z<0.27)[/tex]

[tex]=P(0.27)-P(-0.93)=0.6064198- 0.1761855=0.4302343\approx0.4302[/tex]

Hence, the required probability :-0.4302

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