For the following situation, find the mean and standard deviation of the population. List all samples (with replacement) of the given size from that population and find the mean of each. Find the mean and standard deviation of the sampling distribution and compare them with the mean and standard deviation of the population. The scores of three students in a study group on a test are 93,91,98. Use a sample size of 3
The mean of the population is
(Round to two decimal places as 'needed)

Step 1 ：First, we calculate the mean of the population. The mean is the sum of all the values divided by the number of values. In this case, the mean is \(\frac{93+91+98}{3} = 94\).

Step 2 ：Next, we calculate the standard deviation of the population. The standard deviation is the square root of the variance. The variance is the average of the squared differences from the mean. In this case, the variance is \(\frac{(93-94)^2+(91-94)^2+(98-94)^2}{3} = 9.33\). So, the standard deviation is \(\sqrt{9.33} = 3.06\).

Step 3 ：Now, we list all samples (with replacement) of size 3 from the population. The samples are (93, 93, 93), (93, 93, 91), (93, 93, 98), (93, 91, 93), (93, 91, 91), (93, 91, 98), (93, 98, 93), (93, 98, 91), (93, 98, 98), (91, 93, 93), (91, 93, 91), (91, 93, 98), (91, 91, 93), (91, 91, 91), (91, 91, 98), (91, 98, 93), (91, 98, 91), (91, 98, 98), (98, 93, 93), (98, 93, 91), (98, 93, 98), (98, 91, 93), (98, 91, 91), (98, 91, 98), (98, 98, 93), (98, 98, 91), (98, 98, 98).

Step 4 ：Next, we find the mean of each sample. The means are 93, 92.33, 94.67, 92.33, 91.67, 94, 94.67, 94, 96.33, 92.33, 91.67, 94, 91.67, 91, 93.33, 94, 93.33, 95.67, 94.67, 94, 96.33, 94, 93.33, 95.67, 96.33, 95.67, 98.

Step 5 ：Now, we calculate the mean of the sampling distribution. The mean of the sampling distribution is the average of the means of all the samples. In this case, the mean of the sampling distribution is \(\frac{93+92.33+94.67+92.33+91.67+94+94.67+94+96.33+92.33+91.67+94+91.67+91+93.33+94+93.33+95.67+94.67+94+96.33+94+93.33+95.67+96.33+95.67+98}{27} = 94\).

Step 6 ：Next, we calculate the standard deviation of the sampling distribution. The standard deviation of the sampling distribution is the square root of the variance of the sampling distribution. The variance of the sampling distribution is the average of the squared differences from the mean of the sampling distribution. In this case, the variance of the sampling distribution is \(\frac{(93-94)^2+(92.33-94)^2+(94.67-94)^2+(92.33-94)^2+(91.67-94)^2+(94-94)^2+(94.67-94)^2+(94-94)^2+(96.33-94)^2+(92.33-94)^2+(91.67-94)^2+(94-94)^2+(91.67-94)^2+(91-94)^2+(93.33-94)^2+(94-94)^2+(93.33-94)^2+(95.67-94)^2+(94.67-94)^2+(94-94)^2+(96.33-94)^2+(94-94)^2+(93.33-94)^2+(95.67-94)^2+(96.33-94)^2+(95.67-94)^2+(98-94)^2}{27} = 3.06\). So, the standard deviation of the sampling distribution is \(\sqrt{3.06} = 1.75\).

Step 7 ：Finally, we compare the mean and standard deviation of the sampling distribution with the mean and standard deviation of the population. The mean of the sampling distribution is the same as the mean of the population, which is 94. The standard deviation of the sampling distribution is smaller than the standard deviation of the population, which is 1.75 compared to 3.06. This is because the sampling distribution is a distribution of means, and the mean of a sample is less variable than individual data points.