What the answer?

Question

The average number of miles (in thousands) that a car's tire will function before needing replacement is 67 and the standard deviation is 15. Suppose that 15 randomly selected tires are tested. Round all answers to 4 decimal places where possible and assume a normal distribution.

a. What is the distribution of X? X - N 67

, 15

1. What is the distribution of 5? I - N( 67
2. If a randomly selected individual tire is tested, find the probability that the number of miles (in thousands) before it will need replacement is between 69.1 and 72.4.
3. For the 15 tires tested, find the probability that the average miles (in thousands) before need of replacement is between 69.1 and 72.4.
4. For part d), is the assumption that the distribution is normal necessary? O Yes

Answer 1

Don't understand question #1, but here is the answer to #2.

Probability that the mileage figure for a single tire is between 69.1 and 72.4 thousand miles would be the area under the normal distribution curve corresponding to those values.

The z value provides the area under the normal distribution curve to the left of the given value and therefore its probability of occurrence.

z = (x - μ)/σ, where x is the value, μ is the mean and σ is the standard deviation.

z₁ = (69.1 - 67)/15 = 0.14 ; z₂ = (72.4 - 67)/15 = 0.36

P(69.1<x<72.4) = P(z<0.36) - P(z<0.14). In other words, you subtract the two areas under the curve.

z<0.36 = 0.6406; z<0.14 = 0.5557. 0.6406 - 0.5557 = 0.0849. Therefore, the probability that the mileage of the tire will fall between 69.1 and 72.4 thousand miles is 0.0849.

I'll leave question #3 to another tutor.