Question

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?

Answer 1

Answer: 0.4302

Explanation:

Given : Mean :
`\mu=\text{38.7 miles }`

Standard deviation :
`\sigma=\text{6.7 miles }`

Sample size :
`n=60`

Also, these distances are normally distributed.

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

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

For x=32.5

`\\\\ z=(32.5-38.7)/(6.7)=-0.925373134\approx-0.93`

For x=40.5

`\\\\ z=(40.5-38.7)/(6.7)=0.268656\approx0.27`

The p-value =
`P(-0.93<z<0.27)`

`=P(0.27)-P(-0.93)=0.6064198- 0.1761855=0.4302343\approx0.4302`

Hence, the required probability :-0.4302