Question

1) Consider the population distrībution of X as shown below: f(x) 0.2 0.2 0.6 7 Total Samples of size n-2 will be drawn from this population, by sampling with replacement.

Answer 1

In this Case

Number of possible samples are = 3² = 9

a) Now we make a List of samples with mean and variance

Mean = $(\sum X) / n$

Variance = $\sqrt{\frac{\sum (X - Xbar)^2}{n - 1 }}$

List of samples	Mean	Variance
(1,1)	1	0
(1,7)	4	18
(1,9)	5	32
(7,1)	4	18
(7,7)	7	0
(7,9)	8	2
(9,1)	5	32
(9,7)	8	2
(9,9)	9	0

b) Now construct probability distribution of mean

Mean	Probability
1	1/9
4	2/9
5	2/9
7	1/9
8	2/9
9	1/9

c) And probablity distribution of variance

Variance	Probability
0	3/9
2	2/9
18	2/9
32	2/9

2) E(Xbar) = ∑x* p(x) = 1 * 1/9 + 4*2/9 + 5*2/9 + 7*1/9 + 8*2/9 + 9*1/9 = 5.667

population mean = $\mu$ = (1 + 7 + 9) / 3 = 5.667

Hence E(xbar) = $\mu$

As well as

Population standard deviation($\sigma$) = $\sqrt\frac{\sum(X - \mu)^2}{N} = 3.399$

Standard deviation of mean, sd(xbar) = $\sqrt{\sum x^2 * p(x) ~-~ \mu^2}$ = 2.4

  $\frac{\sigma}{\sqrt{n}} = \frac{3.399}{\sqrt{2}} = 2.4$

Hence $SD(Xbar) = \frac{\sigma}{\sqrt{n}}$

Please let me know further clarifications regarding this work