The mean per capita income is 24,863 dollars per annum with a standard deviation of 743 dollars per annum. What is the probability that the sample mean would differ from the true mean by more than 88 dollars if a sample of 422 persons is randomly selected? Round your answer to four decimal places. Answer How to enter your answer (opens in new window)

Step 1 ：Given that the population mean is \(24863\) dollars per annum and the population standard deviation is \(743\) dollars per annum. A sample of \(422\) persons is randomly selected.

Step 2 ：We are asked to find the probability that the sample mean differs from the true mean by more than \(88\) dollars.

Step 3 ：According to the Central Limit Theorem, the sampling distribution of the mean is normally distributed with mean equal to the population mean and standard deviation equal to the population standard deviation divided by the square root of the sample size.

Step 4 ：First, we calculate the standard error, which is the population standard deviation divided by the square root of the sample size. The standard error is approximately \(36.16867409488479\).

Step 5 ：Next, we calculate the z-score, which is the difference between the sample mean and the population mean divided by the standard error. The z-score is approximately \(2.433044677533411\).

Step 6 ：Finally, we use the standard normal distribution (Z-distribution) to find the probability that the sample mean differs from the true mean by more than \(88\) dollars. The probability is \(0.015\).

Step 7 ：So, the probability that the sample mean would differ from the true mean by more than \(88\) dollars if a sample of \(422\) persons is randomly selected is \(\boxed{0.015}\).