Answer in Statistics and Probability for Alaric #251371
I.
“mean=mu=dfrac{8+10+12+14+16}{5}=12”
“variance=sigma^2=dfrac{1}{5}((8-12)^2+(10-12)^2”
“+(12-12)^2+(14-12)^2+(16-12)^2)=8”
“sigma=sqrt{sigma^2}=sqrt{8}=2sqrt{2}”
II. There are “dbinom{5}{2}=10” samples of size two which can be drawn without replacement:
“begin{matrix}n Sample & Sample mean \n (8,10) & 9 \n (8,12) & 10 \n (8,14) & 11 \n (8,16) & 12 \n (10,12) & 11 \n (10,14) & 12 \n (10,16) & 13\n (12,14) & 13 \n (12,16) & 14 \n (14,16) & 15 \n nend{matrix}”
III.
“begin{matrix}n bar{X} & P(bar{X}) \n 9 & 0.1 \n 10 & 0.1 \n 11 & 0.2 \n 12 & 0.2 \n 13 & 0.2 \n 14 & 0.1\n 15 & 0.1\nnend{matrix}”
IV.
“mu_{bar{X}}=9(0.1)+10(0.1)+11(0.2)+12(0.2)”
“+13(0.2)+14(0.1)+15(0.1)=12”
“sum_ibar{X}_i^2P(bar{X_i})=9^2(0.1)+10^2(0.1)+11^2(0.2)”
“+12^2(0.2)+13^2(0.2)+14^2(0.1)+15^2(0.1)=147”
“sigma_{bar{X}}^2=sum_ibar{X}_i^2P(bar{X_i})-mu_{bar{X}}^2=147-12^2=3”
“sigma_{bar{X}}=sqrt{sigma_{bar{X}}^2}=sqrt{3}”
“mu_{bar{X}}=12, sigma_{bar{X}}=sqrt{3}”
V.
The mean “mu_{bar{X}}” and standard deviation “sigma_{bar{X}}” of the sample mean “bar{X}” satisfy
“mu_{bar{X}}=12=mu,”
“sigma_{bar{X}}=sqrt{3}=dfrac{2sqrt{2}}{sqrt{2}}sqrt{dfrac{5-2}{5-1}}=dfrac{sigma}{sqrt{n}}sqrt{dfrac{N-n}{N-1}}”
