Answer to Question #349493 in Statistics and Probability for love

Question #349493

Samples of three cards are drawn at random

from a population of eight cards numbered

from 1 to 8.


  1. Construct a histogram of the sampling

distribution of the means.


1
Expert's answer
2022-06-10T13:51:12-0400

We have population values 1,2,3,4,5,6,7,8, population size N=8 and sample size n=3.

Mean of population "(\\mu)" = "\\dfrac{1+2+3+4+5+6+7+8}{8}=4.5"

b.Variance of population 


"\\sigma^2=\\dfrac{\\Sigma(x_i-\\bar{x})^2}{n}=\\dfrac{1}{8}(12.25+6.25+2.25+0.25""+0.25+2.25+6.25+12.25)=5.25"


"\\sigma=\\sqrt{\\sigma^2}=\\sqrt{5.25}\\approx2.29"


Select a random sample of size 2 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is "^{N}C_n=^{8}C_3=56."

"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c:c:c}\n no & Sample & Sample \\\\\n& & mean\\ (\\bar{x})\n\\\\ \\hline\n 1 & 1,2,3 & 6\/3 \\\\\n \\hdashline\n 2 & 1,2,4 & 7\/3 \\\\\n \\hdashline\n 3 & 1,2,5 & 8\/3 \\\\\n \\hdashline\n 4 & 1,2,6 & 9\/3 \\\\\n \\hdashline\n 5 & 1,2,7 & 10\/3 \\\\\n \\hdashline\n 6 & 1,2,8 & 11\/3 \\\\\n \\hdashline\n 7 & 1,3,4 & 8\/3 \\\\\n \\hdashline\n 8 & 1,3,5 & 9\/3 \\\\\n \\hdashline\n 9 & 1,3,6& 10\/3 \\\\\n \\hdashline\n 10 & 1,3,7 & 11\/3 \\\\\n \\hdashline\n 11 & 1,3,8 & 12\/3 \\\\\n \\hdashline\n 12 & 1,4,5 & 10\/3 \\\\\n \\hdashline\n 13 & 1,4,6 & 11\/3 \\\\\n \\hdashline\n 14 & 1,4,7 & 12\/3 \\\\\n \\hdashline\n 15 & 1,4,8 & 13\/3 \\\\\n \\hdashline\n 16 & 1,5,6 & 12\/3 \\\\\n \\hdashline\n 17 & 1,5,7 & 13\/3 \\\\\n \\hdashline\n 18 & 1,5,8 & 14\/3 \\\\\n \\hdashline\n 19 & 1,6,7 & 14\/3 \\\\\n \\hdashline\n 20 & 1,6,8 & 15\/3 \\\\\n \\hdashline\n 21 & 1,7,8 & 16\/3 \\\\\n \\hdashline\n 22 & 2,3,4 & 9\/3 \\\\\n \\hdashline\n 23 & 2,3,5 & 10\/3 \\\\\n \\hdashline\n 24 & 2,3,6 & 11\/3 \\\\\n \\hdashline\n 25 & 2,3,7 & 12\/3 \\\\\n \\hdashline\n 26 & 2,3,8 & 13\/3 \\\\\n \\hdashline\n 27 & 2,4,5 & 11\/3 \\\\\n \\hdashline\n 28 & 2,4,6 & 12\/3 \\\\\n \\hdashline\n 29 & 2,4,7 & 13\/3 \\\\\n \\hdashline\n 30 & 2,4,8 & 14\/3 \\\\\n \\hdashline\n 31 & 2,5,6 & 13\/3 \\\\\n \\hdashline\n 32 & 2,5,7 & 14\/3 \\\\\n \\hdashline\n 33 & 2,5,8 & 15\/3 \\\\\n \\hdashline\n 34 & 2,6,7 & 15\/3 \\\\\n \\hdashline\n 35 & 2,6,8 & 16\/3 \\\\\n \\hdashline\n 36 & 2,7,8 & 17\/3 \\\\\n \\hdashline\n 37 & 3,4,5 & 12\/3 \\\\\n \\hdashline\n 38 & 3,4,6 & 13\/3 \\\\\n \\hdashline\n 39 & 3,4,7 & 14\/3 \\\\\n \\hdashline\n 40 & 3,4,8 & 15\/3 \\\\\n \\hdashline\n 41 & 3,5,6 & 14\/3 \\\\\n \\hdashline\n 42 & 3,5,7 & 15\/3 \\\\\n \\hdashline\n 43 & 3,5,8 & 16\/3 \\\\\n \\hdashline\n 44 & 3,6,7 & 16\/3 \\\\\n \\hdashline\n 45 & 3,6,8 & 17\/3 \\\\\n \\hdashline\n 46 & 3,7,8 & 18\/3 \\\\\n \\hdashline\n 47 & 4,5,6 & 15\/3 \\\\\n \\hdashline\n 48 & 4,5,7 & 16\/3 \\\\\n \\hdashline\n 49 & 4,5,8 & 17\/3 \\\\\n \\hdashline\n 50 & 4,6,7& 17\/3 \\\\\n \\hdashline\n 51 & 4,6,8 & 18\/3 \\\\\n \\hdashline\n 52 & 4,7,8 & 19\/3 \\\\\n \\hdashline\n 53 & 5,6,7 & 18\/3 \\\\\n \\hdashline\n 54 & 5,6,8 & 19\/3 \\\\\n \\hdashline\n 55 & 5,7,8 & 20\/3 \\\\\n \\hdashline\n 56 & 6,7,8 & 21\/3 \\\\\n \\hdashline\n\\end{array}"





"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:}\n \\bar{X} & f(\\bar{X}) \n\\\\ \\hline\n 6\/3 & 1\/56 \\\\\n \\hdashline\n 7\/3 & 1\/56 \\\\\n \\hdashline\n 8\/3 & 2\/56 \\\\\n \\hdashline\n 9\/3 & 3\/56 \\\\\n \\hdashline\n 10\/3 & 4\/56 \\\\\n \\hdashline\n 11\/3 & 5\/56 \\\\\n \\hdashline\n 12\/3 & 6\/56 \\\\\n \\hdashline\n 13\/3 & 6\/56 \\\\\n \\hdashline\n 14\/3 & 6\/56 \\\\\n \\hdashline\n 15\/3 & 6\/56 \\\\\n \\hdashline\n 16\/3 & 5\/56 \\\\\n \\hdashline\n 17\/3 & 4\/56 \\\\\n \\hdashline\n 18\/3 & 3\/56 \\\\\n \\hdashline\n 19\/3 & 2\/56 \\\\\n \\hdashline\n 20\/3 & 1\/56 \\\\\n \\hdashline\n 21\/3 & 1\/56 \\\\\n \\hdashline\n\\end{array}"





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