Answer to Question #316663 in Management for abhi

Question

The top-selling Amar tire is rated 70,000 KMs, which means nothing. In fact, the distance

the tires can run until they wear out is a normally distributed random variable with a mean

of 82,000 KMs and a standard deviation of 6,400 KMs.

What is the probability that a tire wears out before 70,000 KMs?

What is the probability that a tire lasts more than 100,000 KMs?

Answer 1

X ~ $N(82000,6400^2)$

$P(X<70000)=P(N(82000,6400^2)<70000)=P(82000+6400Z<70000)=P(Z<-1.875)=0.0304$

$P(X>100000)=P(N(82000,6400^2)>100000)=P(82000+6400Z>100000)=P(Z>2.8125)=0.00246$