Answer to Question #251371 in Statistics and Probability for Alaric

Question #251371
Random variable of size n=2 are drawn from a population consisting the numbers 8,10, 12,14,and 16. Construct a sampling distribution of the sample mean to answer the questions.
1
Expert's answer
2021-10-15T09:28:53-0400

I.


mean=μ=8+10+12+14+165=12mean=\mu=\dfrac{8+10+12+14+16}{5}=12

variance=σ2=15((812)2+(1012)2variance=\sigma^2=\dfrac{1}{5}((8-12)^2+(10-12)^2

+(1212)2+(1412)2+(1612)2)=8+(12-12)^2+(14-12)^2+(16-12)^2)=8

σ=σ2=8=22\sigma=\sqrt{\sigma^2}=\sqrt{8}=2\sqrt{2}

II. There are (52)=10\dbinom{5}{2}=10 samples of size two which can be drawn without replacement: 


SampleSample mean(8,10)9(8,12)10(8,14)11(8,16)12(10,12)11(10,14)12(10,16)13(12,14)13(12,16)14(14,16)15\begin{matrix} Sample & Sample\ mean \\ (8,10) & 9 \\ (8,12) & 10 \\ (8,14) & 11 \\ (8,16) & 12 \\ (10,12) & 11 \\ (10,14) & 12 \\ (10,16) & 13\\ (12,14) & 13 \\ (12,16) & 14 \\ (14,16) & 15 \\ \end{matrix}

III.


XˉP(Xˉ)90.1100.1110.2120.2130.2140.1150.1\begin{matrix} \bar{X} & P(\bar{X}) \\ 9 & 0.1 \\ 10 & 0.1 \\ 11 & 0.2 \\ 12 & 0.2 \\ 13 & 0.2 \\ 14 & 0.1\\ 15 & 0.1\\ \end{matrix}

IV.


μXˉ=9(0.1)+10(0.1)+11(0.2)+12(0.2)\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+13(0.2)+14(0.1)+15(0.1)=12


iXˉi2P(Xiˉ)=92(0.1)+102(0.1)+112(0.2)\sum_i\bar{X}_i^2P(\bar{X_i})=9^2(0.1)+10^2(0.1)+11^2(0.2)

+122(0.2)+132(0.2)+142(0.1)+152(0.1)=147+12^2(0.2)+13^2(0.2)+14^2(0.1)+15^2(0.1)=147

σXˉ2=iXˉi2P(Xiˉ)μXˉ2=147122=3\sigma_{\bar{X}}^2=\sum_i\bar{X}_i^2P(\bar{X_i})-\mu_{\bar{X}}^2=147-12^2=3

σXˉ=σXˉ2=3\sigma_{\bar{X}}=\sqrt{\sigma_{\bar{X}}^2}=\sqrt{3}

μXˉ=12,σXˉ=3\mu_{\bar{X}}=12, \sigma_{\bar{X}}=\sqrt{3}

V.

The mean μXˉ\mu_{\bar{X}} and standard deviation σXˉ\sigma_{\bar{X}} of the sample mean Xˉ\bar{X} satisfy


μXˉ=12=μ,\mu_{\bar{X}}=12=\mu,

σXˉ=3=2225251=σnNnN1\sigma_{\bar{X}}=\sqrt{3}=\dfrac{2\sqrt{2}}{\sqrt{2}}\sqrt{\dfrac{5-2}{5-1}}=\dfrac{\sigma}{\sqrt{n}}\sqrt{\dfrac{N-n}{N-1}}


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