C. assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?

The probability of such a tire lasting more than 60,000 miles is 0.0228, for the complete question provided in explanation.

Step-by-step explanation:

The lifetime of a certain type of car tire are normally distributed. The mean lifetime of a car tire is 50,000 miles with a standard deviation of 5,000 miles. Assume that the manufacturer's claim is true, what is the probability of such a tire lasting more than 60,000 miles?

This is the question of normal distribution:

First w calculate the value of Z corresponding to X = 60,000 miles

We, have; Mean = μ = 50,000 miles, and Standard Deviation = σ = 5,000 miles

Now, for Z, we know that:

Z = (x-μ)/σ

Z = (60,000 - 50,000)/5,000

Z = 2

Now, we have standard tables, for normal distribution in terms of values of Z. One is attached in this answer.

P(X > 60,000) = P(Z > 2) = 1 - P(Z = 2)

P(X > 60,000) = 1 - 0.9772

P(X > 60,000) = 0.0228

36/60

36/2=18

60/2=30

18/30

18/3=6

30/3=10

6/10

6/2=3

10/2=5

your final answer would be 3/5 of an

hope this !

i will pick c put check it over to be sure

step-by-step explanation: