The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 4060 miles, with a standard deviation of 308. If he is correct, what is the probability that the me

We can use the central limit theorem to \approx imate the sampling distribution of the sample mean. The mean of the sampling distribution is equal to the population mean, which is given as 4060 miles. The standard deviation of the sampling distribution is calculated as {eq}frac{sigma}{sqrt{n}}=frac{308}{sqrt{41}}approx 48 {/eq} miles. Thus, the z-score for a sample mean with a deviation of {eq}133 {/eq} miles from the population mean is {eq}frac{133}{48}approx 2.77 {/eq}. To find the probability of a sample mean differing from the population mean by less than {eq}133 {/eq} miles, we need to find the area under the standard normal curve between z = -2.77 and z = 2.77. Using a standard normal distribution table (or a calculator), we find that this area is \approx imately 0.994, or 99.4%. Therefore, if the head of maintenance at XYZ Rent-A-Car is correct about the mean number of miles between services being 4060 miles with a standard deviation of 308 miles, there is a 99.4% probability that the mean of a sample of 41 cars would differ from the population mean by less than 133 miles.