Answer to Question #333836 in Statistics and Probability for Gela

Question

There are five cards in a box containing the numbers 1, 3, 5, 7, and 9. A sample of size 3 is

to be drawn at a time. List all possible sample size of 3 from this population and compute the

mean and variance of the sampling distribution of the sample means.

1. Compute the population mean.

1. Compute the population variance.

1. Determine the number of possible samples.

1. List all possible samples and their corresponding means.

1. Construct the sampling distribution of the sample means.

1. Compute the mean of the sampling distribution of the sample means.

1. Compute the variance of the sampling distribution of the sample means.

1. Construct the histogram.

Answer 1

1. We have population values 1,3,5,7,9, population size N=5 and sample size n=3.

Mean of population $(\mu)$ = $\dfrac{1+3+5+7+9}{5}=5$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{16+4+0+4+16}{5}=8$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{8}=2\sqrt{2}$$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{5}C_3=10.$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,3,5 & 9/3 \\ \hdashline 2 & 1,3,7 & 11/3 \\ \hdashline 3 & 1,3,9 & 13/3 \\ \hdashline 4 & 1,5,7 & 13/3 \\ \hdashline 5 & 1,5,9 & 15/3 \\ \hdashline 6 & 1,7,9 & 17/3 \\ \hdashline 7 & 3,5,7 & 15/3 \\ \hdashline 8 & 3,5,9 & 17/3 \\ \hdashline 9 & 3,7,9 & 19/3 \\ \hdashline 10 & 5,7,9 & 21/3 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 9/3 & 1/10 & 9/30 & 81/90 \\ \hdashline 11/3 & 1/10& 11/30 & 121/90 \\ \hdashline 13/3 & 2/10 & 26/30 & 338/90 \\ \hdashline 15/3 & 2/10 & 30/30 & 450/90 \\ \hdashline 17/3 & 2/10 & 34/30 & 578/90 \\ \hdashline 19/3 & 1/10 & 19/30 & 361/90 \\ \hdashline 21/3 & 1/10 & 21/30 & 441/90 \\ \hdashline \end{array}$$

Mean of sampling distribution 

$$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=5=\mu$$

The variance of sampling distribution 

$$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$$ $$=\dfrac{12768}{90}-(5)^2=\dfrac{4}{3}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$$

$$\sigma_{\bar{X}}=\sqrt{\dfrac{4}{3}}\approx1.1547$$