91Ó°ÊÓ

\(\mu =\lambda =8.7\)

\(\sigma=8.7\)

\(n=10\)

It is known that the mean of the sample means of distribution of a sample is same as the population mean.

\(\mu_{\bar{x}}=\mu=8.7\)

It is known that the standard deviation of the sample means is same as the population standard deviation divided by under root of the size of the sample

\(\sigma _{x}=\frac{\sigma}{\sqrt{n}}=\frac{8.7}{\sqrt{10}}\approx2.7512\)

The data values can be simulated from a normal distribution of

Mean \(=8.7\)

Standard Deviation \(=2.7512\)

using excel with following formula:

\(=-LN(1-RAND())*8.7\)

Each of the \(10\) sample values can be simulated using the above command, then, this can be repeated until \(1000\) samples each with \(10\) values are finished being calculated.

A snapshot of the result obtained is as per follows:

The mean can be calculated as the summation of all values divided by the quantity of values.

For example, the following calculation is done for the one of the columns:

\(\bar{x}=\frac{7.582461582+...+0.0767191717}{10}=8.529210815\)

This can be repeated for all the columns..

The answer is expected to be

Mean \(=8.7\)

Standard Deviation \(=42.7512\)

The reason here is that the \(1000\) sample means will have the similar distribution as the sampling distribution of sample mean.

The mean can be calculated as the summation of all values divided by the quantity of values.

\(\bar{x}=\frac{8.529210815+...+8.95736067}{1000}=8.70346197\)

\(s=\sqrt{\frac{(8.529210815-8.70346197)^{2}+...+(8.95736067-8.70346197)^{2}}{1000-1}}\approx 2.749376\)

The main difference between the results is parts (d) and (e) is because of sampling error. To be more precise, different samples would have different means and different standard deviations even though the values would be very close to the expected values.