A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of 65 students in the high school and found a mean savings of 4700 dollars with a standard deviation of 1300 dollars. Use the normal distribution/empirical rule to estimate a 95\% confidence interval for the mean, rounding all values to the nearest whole number.

Step 1 ：A study of a local high school tried to determine the mean amount of money that each student had saved. The study surveyed a random sample of 65 students in the high school and found a mean savings of 4700 dollars with a standard deviation of 1300 dollars. We are asked to use the normal distribution/empirical rule to estimate a 95% confidence interval for the mean, rounding all values to the nearest whole number.

Step 2 ：To solve this problem, we need to use the formula for the confidence interval of the mean for a normal distribution. The formula is: \[\bar{x} \pm z \frac{s}{\sqrt{n}}\] where: \(\bar{x}\) is the sample mean, \(z\) is the z-score corresponding to the desired confidence level, \(s\) is the standard deviation of the sample, and \(n\) is the sample size.

Step 3 ：For a 95% confidence interval, the z-score is approximately 1.96. We can plug in the given values into this formula to find the confidence interval.

Step 4 ：Given that the mean is 4700, the standard deviation is 1300, the sample size is 65, and the z-score for a 95% confidence interval is 1.96, we can calculate the confidence interval.

Step 5 ：The 95% confidence interval for the mean amount of money that each student had saved is \(\boxed{(4384, 5016)}\) dollars.