Answer to Question #333456 in Statistics and Probability for nik

Question

Four rowers weighing 152, 156, 160, and 164 pounds make up a rowing team. Calculate the sample mean for each of the possible random samples with a size two replacement. They can be used to calculate the sample mean's probability distribution, mean, and standard deviation.

Answer 1

We have population values 152, 156, 160, and 164 population size N=4 and sample size n=2.

Mean of population $(\mu)$ = 

$\dfrac{152+156+160+164}{4}=158$

Variance of population 

$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{N}=20$

$\sigma=\sqrt{\sigma^2}=\sqrt{20}=2\sqrt{5}$

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{4}C_2=6.$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 152, 156 & 154 \\ \hdashline 2 & 152, 160 & 156 \\ \hdashline 3 & 152, 164 & 158 \\ \hdashline 4 & 156, 160 & 158 \\ \hdashline 5 & 156, 164 & 160 \\ \hdashline 6 & 160,164 & 162 \\ \hdashline \end{array}$

$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) &\bar{X}^2 f(\bar{X})\\ \hline 154 & 1/6 & 154/6 & 23716/6\\ \hdashline 156 & 1/6 & 156/6 & 24336/6\\ \hdashline 158 & 2/6 & 316/6 & 49928/6\\ \hdashline 160 & 1/6 & 160/6 &25600/6\\ \hdashline 162 & 1/6 & 162/6 & 26244/6\\ \hdashline \end{array}$

Mean of sampling distribution 

$\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=158=\mu$

The variance of sampling distribution 

$Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2$ $=\dfrac{149824}{6}-(158)^2=\dfrac{40}{6}= \dfrac{\sigma^2}{n}(\dfrac{N-n}{N-1})$

$\sigma_{\bar{X}}=\sqrt{\dfrac{20}{3}}\approx2.5820$