Question

3.b. 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?

Note: You may use Z-table for this.

Z-table link- Normal Table.xls (5 Marks)

Answer 1

We have a normal distribution, $\mu=82000, \sigma=6400.$

Let's convert it to the standard normal distribution, $z=\cfrac{x-\mu}{\sigma};$

$z_1=\cfrac{70000-82000}{64000}=-1.88, \\z_2=\cfrac{100000-82000}{6400}=2.81;$

$P(X<70000)=P(Z<-1.88)=0.3005;$

$P(X>100000)=\\ =P(Z>2.81)=1-P(Z<2.81)=$

$=1-0.99752=0.0024$ (from z-table).