Question #334058

A population consists of values (1, 4, 7). Consider all possible samples of size n= 3 that can be drawn without replacement from this population.

a. Find the mean of the population.

b. Find the standard deviation of the population.

c. Find the mean of the sampling distribution of means.

d. Find the standard deviation of the sample distribution of means,

e.Construct the probability histogram of x with replacement


1
Expert's answer
2022-04-27T14:39:40-0400

a. We have population values 1,4,7, population size N=3 and sample size n=3.

Mean of population (μ)(\mu) = 1+4+73=4\dfrac{1+4+7}{3}=4

b. Variance of population 


σ2=Σ(xixˉ)2n=13(9+0+9)=6\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{3}(9+0+9)=6


σ=σ2=62.4495\sigma=\sqrt{\sigma^2}=\sqrt{6}\approx2.4495

c. The number of possible samples which can be drawn with replacement is Nn=33=27N^n=3^3=27

noSampleSamplemean (xˉ)11,1,1121,1,4231,1,7341,4,1251,4,4361,4,7471,7,1381,7,4491,7,75104,1,12114,1,43124,1,74134,4,13144,4,44154,4,75164,7,14174,7,45184,7,76197,1,13207,1,44217,1,75227,4,14237,4,45247,4,76257,7,15267,7,46277,7,77\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 &1,1,1 & 1\\ \hdashline 2 & 1,1,4 & 2 \\ \hdashline 3 & 1,1,7 & 3\\ \hdashline 4 & 1,4,1 & 2\\ \hdashline 5 & 1,4,4 & 3 \\ \hdashline 6 & 1,4,7& 4 \\ \hdashline 7 & 1,7,1 & 3 \\ \hdashline 8 & 1,7,4 & 4 \\ \hdashline 9 & 1,7,7 & 5 \\ \hdashline 10 & 4,1,1 & 2\\ \hdashline 11 & 4,1,4 & 3 \\ \hdashline 12 & 4,1,7 & 4 \\ \hdashline 13 & 4,4,1 & 3 \\ \hdashline 14 & 4,4,4 & 4 \\ \hdashline 15 & 4,4,7 & 5 \\ \hdashline 16 & 4,7,1 & 4 \\ \hdashline 17 & 4,7,4 & 5 \\ \hdashline 18 & 4,7,7 & 6\\ \hdashline 19 & 7,1,1 & 3\\ \hdashline 20 & 7,1,4 & 4\\ \hdashline 21 & 7,1,7 & 5\\ \hdashline 22 & 7,4,1 & 4\\ \hdashline 23 & 7,4,4 & 5\\ \hdashline 24 & 7,4,7 & 6\\ \hdashline 25 & 7,7,1 & 5\\ \hdashline 26 & 7,7,4 & 6\\ \hdashline 27 & 7,7,7 & 7\\ \hdashline \end{array}




Xˉf(Xˉ)Xˉf(Xˉ)Xˉ2f(Xˉ)11/271/271/2723/276/2712/2736/2718/2754/2747/2728/27112/2756/2730/27150/2763/2718/27108/2771/277/2749/27\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 1 & 1/27 & 1/27 & 1/27 \\ \hdashline 2 & 3/27 & 6/27 & 12/27 \\ \hdashline 3 & 6/27 & 18/27 & 54/27 \\ \hdashline 4 & 7/27 & 28/27 & 112/27 \\ \hdashline 5 & 6/27 & 30/27 & 150/27 \\ \hdashline 6 & 3/27 & 18/27 & 108/27 \\ \hdashline 7 & 1/27 & 7/27 & 49/27 \\ \hdashline \end{array}


Mean of sampling distribution 

μXˉ=E(Xˉ)=Xˉif(Xˉi)=4=μ\mu_{\bar{X}}=E(\bar{X})=\sum\bar{X}_if(\bar{X}_i)=4=\mu



d. The variance of sampling distribution 

Var(Xˉ)=σXˉ2=Xˉi2f(Xˉi)[Xˉif(Xˉi)]2Var(\bar{X})=\sigma^2_{\bar{X}}=\sum\bar{X}_i^2f(\bar{X}_i)-\big[\sum\bar{X}_if(\bar{X}_i)\big]^2=48627(4)2=2=σ2n=\dfrac{486}{27}-(4)^2=2= \dfrac{\sigma^2}{n}

σXˉ=21.4142\sigma_{\bar{X}}=\sqrt{2}\approx1.4142

e.


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