Question

If an automobile gets an average of 27 miles per gallon on a trip and the standard deviation is 3 miles per gallon, find the probability that on a randomly selected trip, the automobile will get between 21 and 30 miles per gallon. Assume the variable is normally distributed.

Answer 1

Let $X=$ distance per gallon on a trip in miles: $X\sim N(\mu, \sigma^2)$

Given $\mu=27$ miles per gallon, $\sigma=3$ miles per gallon.

Then

$$P(21<X<30)=P(X<30)-P(X\leq 21)$$

$$=P(Z<\dfrac{30-27}{3})-P(Z\leq \dfrac{21-27}{3})$$

$$=P(Z<1)-P(Z\leq -2)$$

$$\approx0.841345-0.022750\approx0.8186$$

The probability that on a randomly selected trip, the automobile will get between 21 and 30 miles per gallon is 0.8186.