Question

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.

Answer 1

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.

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

$$\def\arraystretch{1.5} \begin{array}{c:c} \bar{x} & p \\ \hline 2 & 1/56 \\ 7/3 & 1/56 \\ 8/3 & 2/56 \\ 3 & 3/56 \\ 10/3 & 4/56 \\ 11/3 & 5/56 \\ 4 & 6/56 \\ 13/3 & 6/56 \\ 14/3 & 6/56 \\ 5 & 6/56 \\ 16/3 & 5/56 \\ 17/3 & 4/56 \\ 6 & 3/56 \\ 19/3 & 2/56 \\ 20/3 & 1/56 \\ 7 & 1/56 \\ \end{array}$$