Samples of size 3 are taken from the population 7, 8, 3, 4, 5, 6, 9, 2, 1 with replacement and without replacement. Find (1) The mean and standard deviation of the population. (u The mean of each of the sampling distribution of means. (u) The standard deviation of sampling distribution of means

We have population values $7, 8, 3, 4, 5, 6, 9, 2, 1$ population size $N=9$ and sample size $n=3.$

A. With replacement

Thus, the number of possible samples which can be drawn with replacement is $9^3=729.$

(1)

$\mu=\dfrac{7+8+3+4+5+6+9+2+1}{9}=5$

$\sigma^2=\dfrac{1}{9}((7-5)^2+(8-5)^2+(3-5)^2$

$+(4-5)^2+(5-5)^2+(6-5)^2$

$+(9-5)^2+(2-5)^2+(1-5)^2)=\dfrac{60}{9}=\dfrac{20}{3}$

$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{20}{3}}\approx2.5820$

(2) The mean of the sample means is

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

(3)The standard deviation of the sample means is

$\sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\dfrac{\sigma^2}{\sqrt{n}}=\sqrt{\dfrac{20}{3(3)}}\approx1.4907$

B. Without replacement

Thus, the number of possible samples which can be drawn without replacement is $\dbinom{9}{3}=84.$

(1)

$\mu=\dfrac{7+8+3+4+5+6+9+2+1}{9}=5$

$\sigma^2=\dfrac{1}{9}((7-5)^2+(8-5)^2+(3-5)^2$

$+(4-5)^2+(5-5)^2+(6-5)^2$

$+(9-5)^2+(2-5)^2+(1-5)^2)=\dfrac{60}{9}=\dfrac{20}{3}$

$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{20}{3}}\approx2.5820$

(2) The mean of the sample means is

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

(3)The standard deviation of the sample means is

$\sigma_{\bar{X}}=\sqrt{\sigma^2_{\bar{X}}}=\dfrac{\sigma^2}{\sqrt{n}}\cdot\sqrt{\dfrac{N-n}{N-1}}$

$=\sqrt{\dfrac{20}{3(3)}}\cdot\sqrt{\dfrac{9-3}{9-1}}\approx1.2910$