Answer to Question #187514 in Statistics and Probability for Joseph pardorla

Question #187514

random sample of size 4 from a finite population of 3,6,9


1
Expert's answer
2021-05-07T10:33:30-0400

a. the mean of the population



"\\mu=\\dfrac{2+3+7+8+10}{5}=6"

the standard deviation of the population



"\\sigma^2=\\dfrac{(2-6)^2+(3-6)^2+(7-6)^2+(8-6)^2+(10-6)^2}{5}"

"=9.2"

"\\sigma=\\sqrt{9.2}=3.033"


b. There are "\\dbinom{5}{4}" samples of size two which can be drawn without replacement: 



"\\dbinom{5}{4}=\\dfrac{5!}{4!(5-4)!}=5""\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c}\n Sample\\ No. & Sample\\ value & Sample\\ mean\\ \\bar{X} \\\\ \\hline\n 1 & (2,3,7,8) & 5 \\\\\n \\hdashline\n 2 & (2,3,7,10) & 5.5 \\\\\n \\hdashline\n 3 & (2,3,8,10) & 5.75 \\\\\n \\hdashline\n 4 & (2,7,8,10) & 6.75 \\\\\n \\hdashline\n 5 & (3,7,8,10) & 7 \n\\end{array}"

The sampling distribution of the sample mean "\\bar{X}" and its mean and standard deviation are:



"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n \\bar{X} & f & f(\\bar{X}) & \\bar{X}f(\\bar{X})& \\bar{X}^2f(\\bar{X}) \\\\ \\hline\n 5 & 1 & 0.2 & 1 & 5 \\\\\n \\hdashline\n 5.5 & 1 & 0.2 & 1.1 & 6.05 \\\\\n \\hdashline\n 5.75 & 1 & 0.2 & 1.15 & 6.6125 \\\\\n \\hdashline\n 6.75 & 1 & 0.2 & 1.35 & 9.1125 \\\\\n \\hdashline\n 7 & 1 & 0.2 & 1.4 & 9.8 \\\\\n \\hdashline\n Total & 5 & 1 & 6 & 36.575\n\\end{array}""E(\\bar{X})=\\sum\\bar{X}f(\\bar{X})=6""Var(\\bar{X})=\\sum\\bar{X}^2f(\\bar{X})-(E(\\bar{X}))^2=36.575-6^2""=0.575"

Verification:



"\\mu=6=E(\\bar{X})""Var(\\bar{X})=\\dfrac{\\sigma^2}{n}(\\dfrac{N-n}{N-1})=""=\\dfrac{9.2}{4}(\\dfrac{5-4}{5-1})=0.575"

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