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 tire will have a life of at least 47,500 miles?
a. 0668.
b. 4332.
c. 9332.
d. none of these.

The probability that a randomly selected tire will have a life of at least 47,500 miles

P( X≥47,500)= 0.0668

Step-by-step explanation:

Given 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

mean μ=40,000

standard deviation of 5,000 miles

σ = 5,000

by using normally distributed

here X = 47,500

z=1.5

Step(ii)

The probability that a randomly selected tire will have a life of at least 47,500 miles

P( X≥47,500) = P(z≥1.5)

                      = 0.5- A(z₁)

                     = 0.5 - A(1.5)

                     = 0.5-0.4332 ( see from normal table)

                     = 0.0668

Conclusion:-

The probability that a randomly selected tire will have a life of at least 47,500 miles

P( X≥47,500)= 0.0668

can you put the picture up?

step-by-step explanation:

this may not be at all, but i use the same website you're using for school, and if i'm stuck doing one of these i try to look for it on the section.