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

Explanation:

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

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

The formula to calculate the z-score :-

`z=(x-\mu)/(\sigma)`

For x=36,000

`z=(36000-40000)/(5000)=-0.8`

For x=46,000

`z=(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