A trucking company determined that the distance traveled per truck per year is normally​ distributed, with a mean of thousand miles and a standard deviation of thousand miles. Complete parts​ (a) through​ (d) below.

Complete question is;

A trucking company determined that the distance traveled per truck per year in its entire fleet was normally distributed with a mean of 55 thousand miles and a standard deviation of 11 thousand miles.

a)A truck in the fleet traveled 42 thousand miles in the last vear. Calculate the z-score for this truck.

b) Interpret the z-score

c) The trucking company has a policy that any truck with mileage more than 2.5 standard deviations above the mean should be removed from the road and inspected. What is the probability that a randomly selected truck from the fleet will have to be inspected?

d) A randomly selected truck in the fleet traveled 85 thousand miles in the last year. Should this truck be removed from the road and inspected? Explain why or why not.

A) z = - 1.18

B) Interpretation of the z-score in A above shows that the distance traveled per truck per year in its entire fleet is 1.18 standard deviations lesser than the distance traveled per truck on average

C) P(z > 2.5) = 0.0062

D) it's value of z is greater than 2.5 minimum and so the truck will be removed from the road and inspected

Step-by-step explanation:

A) We are given;

Population mean; μ = 55000

Sample mean; x¯ = 42000

Standard deviation; σ = 11000

Z-score formula is;

z = (x¯ - μ)/σ

z = (42000 - 55000)/11000

z = - 1.18

B) Interpretation of the z-score in A above shows that the distance traveled per truck per year in its entire fleet is 1.18 standard deviations lesser than the distance traveled per truck on average.

C) since it has a policy that any truck with mileage more than 2.5 standard deviations above the mean should be removed from the road and inspected, then it means that probability that it will be inspected is;

P(z > 2.5) = 1 - P(z < 2.5)

From p-values from z-score table, we have;

P(z > 2.5) = 1 - 0.9938

P(z > 2.5) = 0.0062

D) A randomly selected truck in the fleet traveled 85 thousand miles in the last year. Thus; x¯ = 85000.

Thus;

z = (85000 - 55000)/11000

z = 2.73

This value of z is greater than 2.5 minimum and so the truck will be removed from the road and inspected