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

$P(21<X<30)$

$P(21<X<30)=P(\frac{21-27}{3}<Z<\frac{30-27}{3})$

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

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

From z tables, we have;

$0.84135-0.02286=0.8186$