Answer to Question #173303 in Statistics and Probability for Denisse Bisuña

Question

The average number of driving miles before a certain kind of tire begins to show wear is 16,800 miles with a standard deviation of 3,300 miles. A car rental agency buys 36 of these tires for replacement purposes and puts each one on a different car:

a. What is the probability that the 36 tires will average less than 16,000 miles until they begin to show wear?

b. What is the probability that the 36 tires will average more than 18,000 miles until they begin to show wear?

Answer 1

We have that:

$\mu=16800$

$\sigma=3300$

$n=36$

a) $P(\bar X<16000)=P(Z<\frac{x-\mu}{\frac{\sigma}{\sqrt n}})=P(Z<\frac{16000-16800}{\frac{3300}{\sqrt {36}}})=P(Z<-1.45)=0.0735$

b) $P(\bar X>18000)=1-P(\bar X < 18000)=1-P(Z<\frac{18000-16800}{\frac{3300}{\sqrt {36}}})=1-P(Z<2.18)=1-0.9854=0.0146$

Answer:

a) 0.0735

b) 0.0146