Answer to Question #108663 in Statistics and Probability for Suraj Singh

Question

Consider a population of 6 units with values 1, 2, 3, 4, 5 and 6. Write down all possible samples of size 2 from this population. Obtain the sampling distribution of the sample mean. Compute mean and variance of the sampling distribution obtained.

Answer 1

We randomly choose two unts, without replacement. The order doesn't matter.

$$\begin{matrix} Sample & sample\ mean, \ \bar{x} \\ 1,2 & 1.5 \\ 1,3 & 2\\ 1,4 & 2.5 \\ 1,5 & 3 \\ 1,6 & 3.5 \\ 2,3 & 2.5 \\ 2,4 & 3 \\ 2,5 & 3.5 \\ 2,6 & 4\\ 3,4 & 3.5 \\ 3,5 & 4 \\ 3,6 & 4.5 \\ 4,5 & 4.5 \\ 4,6 & 5 \\ 5,6 & 5.5 \end{matrix}$$

$$\begin{matrix} \bar{x} & f & probability \\ 1.5 & 1 & 1/15 \\ 2.0 & 1 & 1/15\\ 2.5 & 2 & 2/15\\ 3.0 & 2 & 2/15 \\ 3.5 & 3 & 1/5\\ 4.0 & 2 & 2/15\\ 4.5 & 2 & 2/15\\ 5.0 & 1 & 1/15\\ 5.5 & 1 & 1/15\\ \end{matrix}$$

$$\mu_{\bar{x}}=1.5({1\over15})+2.0({1\over15})+2.5({2\over15})+3.0({2\over15})+3.5({1\over5})+$$

$$+4.0({2\over15})+4.5({2\over15})+5.0({1\over15})+5.5({1\over5})=3.5$$

$$(1.5)^2({1\over15})+(2.0)^2({1\over15})+(2.5)^2({2\over15})+(3.0)^2({2\over15})+(3.5)^2({1\over5})+$$

$$+(4.0)^2({2\over15})+(4.5)^2({2\over15})+(5.0)^2({1\over15})+(5.5)^2({1\over5})=$$

$$={161\over 12}$$

$$Var(\bar{x})=\sigma_{\bar{x}}^2={161\over 12}-(3.5)^2={7\over 6}\approx1.166667$$