Example:
Draw all possible samples of size 2 without replacement from a population consisting of 3, 6, 9, 12, 15. Form the sample distribution of sample means and verify the results.
(i) (ii)
Solution:
We have population values 3, 6, 9, 12, 15, population size and sample size Thus, the number of possible samples which can be drawn without replacement is
Sample No. 
Sample Values 
Sample Mean 
Sample No. 
Sample Values 
Sample Mean 
1 
3, 6 
4.5 
6 
6, 12 
9.0 
2 
3, 9 
6.0 
7 
6, 15 
10.5 
3 
3, 12 
7.5 
8 
9, 12 
10.5 
4 
3, 15 
9.0 
9 
9, 15 
12.0 
5 
6, 9 
7.5 
10 
12, 15 
13.5 
The sampling distribution of the sample mean and its mean and standard deviation are:





4.5 
1 
1/10 
4.5/10 
20.25/10 
6.0 
1 
1/10 
6.0/10 
36.00/10 
7.5 
2 
2/10 
15.0/10 
112.50/10 
9.0 
2 
2/10 
18.0/10 
162.00/10 
10.5 
2 
2/10 
21.0/10 
220.50/10 
12.0 
1 
1/10 
12.0/10 
144.00/10 
13.5 
1 
1/10 
13.5/10 
182.25/10 
Total 
10 
1 
90/10 
877.5/10 
The mean and variance of the population are:

3 
6 
9 
12 
15 


9 
36 
81 
144 
225 

and
Verification:
(i) (ii)
Example:
If random samples of size three and drawn without replacement from the population consisting of four numbers 4, 5, 5, 7. Find sample mean for each sample and make sampling distribution of . Calculate the mean and standard deviation of this sampling distribution. Compare your calculations with population parameters.
Solution:
We have population values 4, 5, 5, 7, population size and sample size . Thus, the number of possible samples which can be drawn without replacement is
Sample No. 
Sample Values 
Sample Mean 
1 
4, 5, 5 
14/3 
2 
4, 5, 7 
16/3 
3 
4, 5, 7 
16/3 
4 
5, 5, 7 
17/3 
The sampling distribution of the sample mean and its mean and standard deviation are:





14/3 
1 
1/4 
14/12 
196/36 
16/3 
2 
2/4 
32/12 
512/36 
17/3 
1 
1/4 
17/12 
289/36 
Total 
4 
1 
63/12 
997/36 
The mean and standard deviation of the population are:

4 
5 
5 
7 


16 
25 
25 
49 

and
Hence and