Question

Consider all samples of size 4 from this population without replacement: 6 8 10 12 13 what is the mean and standard deviation of the sampling distribution

Answer 1

We have population values $6,8,10,12,13,$ population size $N=5$ and sample size $n=4.$ Thus, the number of possible samples which can be drawn without replacement is

$\binom{N}{n}=\binom{5}{4}=5$

The sampling distribution of the sample means.

$E(\bar{X})=\sum \bar{X} f(\bar{X})=9.8 \\\begin{array}{c}\mu=\frac{6+8+10+12+13}{5}=9.8\end{array}$

The mean of the sampling distribution of the sample means is equal to the

the mean of the population.

$E(\bar{X})=\mu_{\bar{X}}=9.8=\mu \\ \begin{array}{c}\operatorname{Var}(\bar{X})=\sum \bar{X}^{2} f(\bar{X})-\left(\sum \bar{X} f(\bar{X})\right)^{2} \\ =96.45-(9.8)^{2}=0.41 \\ \sigma_{\bar{X}}=\sqrt{\operatorname{Var}(\bar{X})}=\sqrt{0.41} \approx 0.6403\end{array}$