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

$12.50 is the difference between the amount of money Sonja started with and $30. So this is a subtraction problem that can be written as 30 = s – 12.50. Undo subtraction by adding 12.50 to both sides. The answer is $42.50 thats the sample response and answer

Step-by-step explanation: