The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 4178 miles, with a variance of 124,609. If he is correct, what is the probability that the mean of a sample of 40 cars would be less than 4265 miles? Round your answer to four decimal places.

P(X' < 4265) = 0.9418

Step-by-step explanation:

Due to the fact that the population is normal, the distribution of the sample mean of 40 cars is;

X' = (x1 + x2 + x3....... + x40)/40

We are given;

Normal mean;μ = 4178 miles

Variance = 124,609

Standard deviation; σ = √variance = √124609 = 353

Formula for standard error of mean = σ/√n = 353/√40 = 55.814

So from the formula;

z = (x - μ)/(σ/√n)

So for P(X' < 4265) we have;

z-value of P(X' < 4265) = (4265 - 4178)/55.814 ≈ 1.56

From the z-table attached, we have for P(z < 1.56) = 0.94179

Approximating to 4 decimal places gives 0.9418

answer: y = -2.5x - 15

              (-6,0)

step-by-step explanation: