Question

A population consists of the five measurements 1, 3, 4, 6, and 8. Suppose samples of size 3 are drawn from this population. 

Construct a table, solve for the mean, variance and standard deviation.

Answer 1

We have population values $1,3,4,6,8,$ population size N= 5  and sample size n = 3.  Thus, the number of possible samples which can be drawn without replacement is

( NCn) = (5C3) = 10

Sample no sample values sample means

1 1,3,4 8/3

2 1,3,6 10/3

3 1,3,8 4

4 1,4,6 11/3

5 1,4,8 13/3

6 1,6,8 5

7 3,4,6 13/3

8 3,4,8 5

9 3,6,8 17/3

10 4,6,8 6

we obtain the mean, variance and standard deviation as below

The sampling distribution of the sample means.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 8/3 & 1& 1/10 & 8/30 & 64/90 \\ \hdashline 10/3 & 1& 1/10 & 10/30 & 100/90 \\ \hdashline 11/3 & 1& 1/10 & 11/30 & 121/90 \\ \hdashline 4 & 1& 1/10 & 12/30 & 144/90 \\ \hdashline 13/3 & 2& 2/10 & 26/30 & 338/90 \\ \hdashline 5 & 2& 2/10 & 30/30 & 450/90 \\ \hdashline 17/3 & 1& 1/10 & 17/30 & 289/90 \\ \hdashline 6 & 1& 1/10 & 18/30 & 324/90 \\ \hdashline Total & 10 & 1 & 132/30 & 1830/90 \\ \hline \end{array}$$ $$mean=(132/30)=4.4 ,\\variance=(1830/90)-(4.4*4.4)=0.97333, \\std =(0.97333)^{1/2}=0.98657$$