The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 2135 miles, with a variance of 145,924. If he is correct, what is the probability that the mean of a sample of 40 cars would differ from the population mean by less than 29 miles

0.36878

Step-by-step explanation:

Given that:

Mean number of miles (m) = 2135 miles

Variance = 145924

Sample size (n) = 40

Standard deviation (s) = √variance = √145924 = 382

probability that the mean of a sample of 40 cars would differ from the population mean by less than 29 miles

P( 2135 - 29 < z < 2135 + 29)

Z = (x - m) / s /√n

Z = [(2106 - 2135) / 382 / √40] < z < [(2164 - 2135) / 382 / √40]

Z = (- 29 / 60.399503) < z < (29 / 60.399503)

Z = - 0.4801364 < z < 0.4801364

P(Z < - 0.48) = 0.31561

P(Z < - 0.48) = 0.68439

P(- 0.480 < z < 0.480) = 0.68439 - 0.31561 = 0.36878

= 0.36878