Answer to Question #110782 in Statistics and Probability for hitendra

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

There are 36 samples of size two which can be drawn with replacement:  

$(1,1),(1,2),(1,3),(1,4),(1,5),(1,6),$

$(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),$

$(3,1),(3,2),(3,3),(3,4),(3,5),(3,6),$

$(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),$

$(5,1),(5,2),(5,3),(5,4),(5,5),(5,6),$

$(6,1),(6,2),(6,3),(6,4),(6,5),(6,6).$

The corresponding sample means are 

$1.0,1.5,2.0,2.5,3.0,3.5,$

$1.5,2.0,2.5,3.0,3.5,4.0,$

$2.0,2.5,3.0,3.5,4.0,4.5,$

$2.5,3.0,3.5,4.0,4.5,5.0,$

$3.0,3.5,4.0,4.5,5.0,5.5,$

$3.5,4.0,4.5,5.0,5.5,6.0$

$$\begin{matrix} \bar{x} & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 & 3.5 & 4.0 & 4.5 & 5.0 & 5.5 & 6.0 \\ p(\bar{x}) & {1 \over36} & {1 \over18} & {1 \over12} & {1 \over9} & {5 \over36} & {1 \over6} & {5 \over36} & {1 \over9}& {1 \over12} & {1 \over18} & {1 \over36} \end{matrix}$$

$$\mu_{\bar{x}}=1( {1 \over36})+1.5( {1 \over18} )+2( {1 \over12} )+2.5( {1 \over9} )+3( {5 \over36} )+$$

$$+3.5( {1 \over6})+4( {5 \over36} )+4.5( {1 \over9} )+5( {1 \over12} )+5.5( {1 \over18} )+6({1\over36} )=$$

$$=3.5$$

$$\sigma_{\bar{x}}^2=(1-3.5)^2( {1 \over36})+(1.5-3.5)^2( {1 \over18} )+(2-3.5)^2( {1 \over12} )+$$

$$+(2.5-3.5)^2( {1 \over9} )+(3-3.5)^2( {1 \over9} )+(3.5-3.5)^2( {1 \over9} )+$$

$$+(4-3.5)^2( {1 \over9} )+(4.5-3.5)^2( {1 \over9} )+(5-3.5)^2( {1 \over9} )+$$

$$+(5.5-3.5)^2( {1 \over9} )+(6-3.5)^2( {1 \over9} )={35 \over 24}\approx1.458333$$

$$\sigma_{\bar{x}}=\sqrt{{35 \over 24}}\approx1.207615$$