Answer to Question #194337 in Statistics and Probability for Kit

Question #194337

Samples of three cards are drawn at random from a population of eight cards numbered from 1 to 8.

A. How many possible samples can be drawn?

B. Construct the sampling distribution of sample means.


1
Expert's answer
2021-05-18T14:09:44-0400

A. As order is of no account, this a combination problem


"\\dbinom{8}{3}=\\dfrac{8!}{3!(8-3)!}=\\dfrac{8(7)(6)}{1(2)(3)}=56"

56 possible samples can be drawn.


B.


"\\def\\arraystretch{1.5}\n \\begin{array}{c:c:c}\n Sample & \\bar{x} & Probability \\\\ \\hline\n 1,2,3 & 2 & 1\/56 \\\\\n 1,2,4 & 7\/3 & 1\/56 \\\\\n1,2,5 & 8\/3 & 1\/56 \\\\\n1,2,6 & 3 & 1\/56 \\\\\n1,2,7 & 10\/3 & 1\/56 \\\\\n1,2,8 & 11\/3 & 1\/56 \\\\\n1,3,4 & 8\/3 & 1\/56 \\\\\n1,3,5 & 3 & 1\/56 \\\\\n1,3,6 & 10\/3 & 1\/56 \\\\\n1,3,7 & 11\/3 & 1\/56 \\\\\n1,3,8 & 4 & 1\/56 \\\\\n1,4,5 & 10\/3 & 1\/56 \\\\\n1,4,6 & 11\/3 & 1\/56 \\\\\n1,4,7 & 4 & 1\/56 \\\\\n1,4,8 & 13\/3 & 1\/56 \\\\\n1,5,6 & 4 & 1\/56 \\\\\n1,5,7 & 13\/3 & 1\/56 \\\\\n1,5,8 & 14\/3 & 1\/56 \\\\\n1,6,7 & 14\/3 & 1\/56 \\\\\n1,6,8 & 5 & 1\/56 \\\\\n1,7,8 & 16\/3 & 1\/56 \\\\\n2,3,4 & 3 & 1\/56 \\\\\n2,3,5 & 10\/3 & 1\/56 \\\\\n2,3,6 & 11\/3 & 1\/56 \\\\\n2,3,7 & 4 & 1\/56 \\\\\n2,3,8 & 13\/3 & 1\/56 \\\\\n2,4,5 & 11\/3 & 1\/56 \\\\\n2,4,6 & 4 & 1\/56 \\\\\n2,4,7 & 13\/3 & 1\/56 \\\\\n2,4,8 & 14\/3 & 1\/56 \\\\\n2,5,6 & 13\/3 & 1\/56 \\\\\n2,5,7 & 14\/3 & 1\/56 \\\\\n2,5,8 & 5 & 1\/56 \\\\\n2,6,7 & 5 & 1\/56 \\\\\n2,6,8 & 16\/3 & 1\/56 \\\\\n2,7,8 & 17\/3 & 1\/56 \\\\\n3,4,5 & 4 & 1\/56 \\\\\n3,4,6 & 13\/3 & 1\/56 \\\\\n3,4,7 & 14\/3 & 1\/56 \\\\\n3,4,8 & 5 & 1\/56 \\\\\n3,5,6 & 14\/3 & 1\/56 \\\\\n3,5,7 & 5 & 1\/56 \\\\\n3,5,8 & 16\/3 & 1\/56 \\\\\n3,6,7 & 16\/3 & 1\/56 \\\\\n3,6,8 & 17\/3 & 1\/56 \\\\\n3,7,8 & 6 & 1\/56 \\\\\n4,5,6 & 5 & 1\/56 \\\\\n4,5,7 & 16\/3 & 1\/56 \\\\\n4,5,8 & 17\/3 & 1\/56 \\\\\n4,6,7 & 17\/3 & 1\/56 \\\\\n4,6,8 & 6 & 1\/56 \\\\\n4,7,8 & 19\/3 & 1\/56 \\\\\n5,6,7 & 6 & 1\/56 \\\\\n5,6,8 & 19\/3 & 1\/56 \\\\\n5,7,8 & 20\/3 & 1\/56 \\\\\n6,7,8 & 7 & 1\/56 \\\\\n\\end{array}"

"\\def\\arraystretch{1.5}\n \\begin{array}{c:c}\n \\bar{x} & p \\\\ \\hline\n 2 & 1\/56 \\\\\n 7\/3 & 1\/56 \\\\\n 8\/3 & 2\/56 \\\\ \n3 & 3\/56 \\\\\n 10\/3 & 4\/56 \\\\\n 11\/3 & 5\/56 \\\\\n 4 & 6\/56 \\\\\n 13\/3 & 6\/56 \\\\\n 14\/3 & 6\/56 \\\\\n 5 & 6\/56 \\\\\n16\/3 & 5\/56 \\\\\n 17\/3 & 4\/56 \\\\\n 6 & 3\/56 \\\\\n19\/3 & 2\/56 \\\\\n 20\/3 & 1\/56 \\\\\n 7 & 1\/56 \\\\\n\\end{array}"


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