Question

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 word counts of five essays are 504​, 639​, 542​, 601​, and 574. Use a sample size of 2.

the mean of the population is:

the standard deviation of the population is:

Answer 1

For the population,

   Mean = 572

   Standard deviation = 46.60043

For simple random sampling with replacement (SRSWR) we shall get total 5*5 = 25 samples.

Serial No	Samples of size 2	Mean	Standard deviation
1	504,504	504	0
2	504,639	571.5	67.5
3	504,542	523	19
4	504,601	552.5	48.5
5	504,574	539	35
6	639,504	571.5	67.5
7	639,639	639	0
8	639,542	590.5	48.5
9	639,601	620	19
10	639,574	606.5	32.5
11	542,504	523	19
12	542,639	590.5	48.5
13	542,542	542	0
14	542,601	571.5	29.5
15	542,574	558	16
16	601,504	552.5	48.5
17	601,,639	620	19
18	601,542	571.5	29.5
19	601,601	601	0
20	601,574	587.5	13.5
21	574,504	539	35
22	574,639	606.5	32.5
23	574,542	558	16
24	574,601	587.5	13.5
25	574,574	574	0