Answer: 25.08%
Step-by-step explanation:
Given: Distance traveled by each volunteer driver is normally distributed.
Mean ([tex]\mu[/tex]) = 50 thousand miles
standard deviation ([tex]\sigma[/tex])= 12 thousand miles
Let X = distance traveled by each volunteer driver
Required probability : [tex]P(X<30)+P(X>60)[/tex]
[tex]=P(\dfrac{X-\mu}{\sigma}<\dfrac{30-50}{12})+P(\dfrac{X-\mu}{\sigma}>\dfrac{60-50}{12})\\\\=P(z<-1.67)+P(z>0.83)\ \ \ [z=\dfrac{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}][/tex]
Hence, percentage of drivers can be expected to travel either below 30 or above 60 thousand miles in a year = 25.08%