Question

It is estimated that the distance traveled by each volunteer driver at the Huntington VA Medical Center is normally distributed with a mean of 50 thousand miles and a standard deviation of 12 thousand miles each year. What percentage of drivers can be expected to travel either below 30 or above 60 thousand miles in a year?

Answer 1

Answer: 25.08%

Explanation:

Given: Distance traveled by each volunteer driver is normally distributed.

Mean (
`\mu`) = 50 thousand miles

standard deviation (
`\sigma`)= 12 thousand miles

Let X = distance traveled by each volunteer driver

Required probability :
`P(X<30)+P(X>60)`

`=P((X-\mu)/(\sigma)<(30-50)/(12))+P((X-\mu)/(\sigma)>(60-50)/(12))\\\\=P(z<-1.67)+P(z>0.83)\ \ \ [z=(X-\mu)/(\sigma)]\\\\= (1-P(z<1.67))+(1-P(z<0.83)\ \ \ [P(Z<-z)=1-P(Z<z)=P(Z>z)]\\\\= 2- P(z<1.67)-P(z<0.83)\\\\= 2- 0.9525-0.7967=0.2508\ \ [\text{By p-value table}]`

Hence, percentage of drivers can be expected to travel either below 30 or above 60 thousand miles in a year = 25.08%