Question

The life expectancy of a particular brand of tire is normally distributed with a mean of 40,000 and a standard deviation of 5,000 miles. What is the probability that a randomly selected set of tires will have a life of 36,000 to 46,000 miles? Round your answer to 4 decimal places.

Answer 1

Answer:  0.6731

Step-by-step explanation:

Given : Mean : $\mu = 40,000\text{ miles}$

Standard deviation : $\sigma = 5,000\text{ miles}$

The formula to calculate the z-score :-

$z=\dfrac{x-\mu}{\sigma}$

For x=36,000

$z=\dfrac{36000-40000}{5000}=-0.8$

For x=46,000

$z=\dfrac{46000-40000}{5000}=1.2$

The P-value : $P(-0.8<z<1.2)=P(z<1.2)-P(z<-0.8)$

$=0.8849303-0.2118554=0.6730749\approx0.6731$

Hence, the probability that a randomly selected set of tires will have a life of 36,000 to 46,000 miles =0.6731