Question

Consider all possible samples of size 2 that can be drawn with replacement from the population 1, 5, and 8. Compute the following:

1. Population mean

2. Population variance

3. Population standard deviation

4. Illustrate the probability histogram of the sampling distribution of the means

Answer 1

We have population values 1,5,8 population size N=3 and sample size n=2.

1.Population mean

$$\mu=\dfrac{1+5+8}{3}=\dfrac{14}{3}$$

2. Population variance 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{\dfrac{121}{9}+\dfrac{1}{9}+\dfrac{100}{9}}{3}=\dfrac{74}{9}$$

3. Population standard deviation

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{74}{9}}=\dfrac{\sqrt{74}}{3}\approx2.867442$$

4. The number of possible samples which can be drawn with replacement is $N^n=3^2=9.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,1 & 2/2 \\ \hdashline 2 & 1,5 & 6/2 \\ \hdashline 3 & 1,8 & 9/2 \\ \hdashline 4 & 5,1 & 6/2 \\ \hdashline 5 & 5,5 & 10/2 \\ \hdashline 6 & 5,8 & 13/2 \\ \hdashline 7 & 8,1 & 9/2 \\ \hdashline 8 & 8,5 & 13/2 \\ \hdashline 9 & 8,8 & 16/2 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 2/2 & 1/9 & 2/18 & 4/36\\ \hdashline 6/2 & 2/9 & 12/18 & 72/36\\ \hdashline 9/2 & 2/9 & 18/18 & 162/36\\ \hdashline 10/2 & 1/9 & 10/18 & 100/36\\ \hdashline 13/2 & 2/9 & 26/18 & 338/36\\ \hdashline 16/2 & 1/9 & 16/18 & 256/36\\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=\dfrac{14}{3}=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{932}{36}-(\dfrac{14}{3})^2=\dfrac{37}{9}= \dfrac{\sigma^2}{n}$$

The standard error of the mean

$$\sigma_{\bar{X}}=\sqrt{\dfrac{37}{9}}\approx2.0276$$