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

Step-by-step 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 :-

$z=\dfrac{x-\mu}{\sigma}$

For x=32.5

$\\\\ z=\dfrac{32.5-38.7}{6.7}=-0.925373134\approx-0.93$

For x=40.5

$\\\\ z=\dfrac{40.5-38.7}{6.7}=0.268656\approx0.27$

The p-value = $P(-0.93$

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

Hence, the required probability :-0.4302