a.
We have population values "2,5,9" population size "N=3"
"\\mu=\\dfrac{2+5+9}{3}=\\dfrac{16}{3}""\\sigma^2=\\dfrac{1}{3}((2-16\/3)^2+(5-16\/3)^2+(9-16\/3)^2"
"=\\dfrac{74}{9}"
"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{\\dfrac{74}{9}}=\\dfrac{\\sqrt{74}}{3}"
We have population values "2,5,9" population size "N=3" and sample size "n=2." Thus, the number of possible samples which can be drawn without replacement is
"\\dbinom{N}{n}=\\dbinom{3}{2}=3""\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c}\n Sample & Sample & Sample \\ mean \\\\\n No. & values & (\\bar{X}) \\\\ \\hline\n 1 & 2,5 & 7\/2 \\\\\n \\hdashline\n 2 & 2,9 & 11\/2 \\\\\n \\hdashline\n 3 & 5,9 & 7 \\\\\n \\hline\n\\end{array}"The sampling distribution of the sample means.
"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c:c}\n& \\bar{X} & f & f(\\bar{X}) & \\bar{X}f(\\bar{X})& \\bar{X}^2f(\\bar{X}) \\\\ \\hline\n & 7\/2 & 1 & 1\/3 & 7\/6 & 49\/12 \\\\\n \\hdashline\n & 11\/2 & 1 & 1\/3 & 11\/6 & 121\/12 \\\\\n \\hdashline\n & 7 & 1 & 1\/3 & 7\/3 & 49\/3 \\\\\n \\hdashline\n Total & & 3 & 1 & 16\/3 & 183\/6 \\\\ \\hline\n\\end{array}"
"E(\\bar{X})=\\sum\\bar{X}f(\\bar{X})=\\dfrac{16}{3}"
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}}=\\dfrac{16}{3}=\\mu"
"Var(\\bar{X})=\\sum\\bar{X}^2f(\\bar{X})-(\\sum\\bar{X}f(\\bar{X}))^2"
"=\\dfrac{183}{6}-(\\dfrac{16}{3})^2=\\dfrac{37}{18}"
"\\sigma_{\\bar{X}}=\\sqrt{Var(\\bar{X})}=\\sqrt{\\dfrac{37}{18}}=\\dfrac{\\sqrt{74}}{6}"Verification:
"Var(\\bar{X})=\\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})=\\dfrac{\\dfrac{74}{9}}{2}(\\dfrac{3-2}{3-1})"
"=\\dfrac{37}{18}, True""\\bar{X}\\sim N(\\dfrac{16}{3}, \\dfrac{37}{18})"
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