Question

Samples of 4 cards are drawn from a population of 6 cards numbered 1-6. Construct a sampling distribution of the sample means and answer the following questions:

1. How many samples of size 4 can be drawn from the population?

2. What are the possible means?

3. What is the probability of getting 4 as a mean?

4. What is the probability of getting 3.5 as a mean?

Answer 1

We have population values $1,2,3,4,5,6$ population size $N=6$ and sample size $n=4.$ Thus, the number of possible samples which can be drawn without replacement is

$$\dbinom{N}{n}=\dbinom{6}{4}=15$$ $$\def\arraystretch{1.5} \begin{array}{c:c:c} Sample & Sample & Sample \ mean \\ No. & values & (\bar{X}) \\ \hline 1 & 1,2,3,4 & 2.5 \\ \hdashline 2 & 1,2,3,5 & 2.75 \\ \hdashline 3 & 1,2,3,6 & 3 \\ \hdashline 4 & 1,2,4,5 & 3 \\ \hdashline 5 & 1,2,4,6 & 3.25 \\ \hline 6 & 1,2,5,6 & 3.5 \\ \hline 7 & 1,3,4,5 & 3.25 \\ \hline 8 & 1,3,4,6 & 3.5 \\ \hline 9 & 1,3,5,6 & 3.75 \\ \hline 10 & 1,4,5,6 & 4 \\ \hline 11 & 2,3,4,5 & 3.5 \\ \hline 12 & 2,3,4,6 & 3.75 \\ \hline 13 & 2,3,5,6 & 4 \\ \hline 14 & 2,4,5,6 & 4.25 \\ \hline 15 & 3,4,5,6 & 4.5 \\ \hline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f & f(\bar{X}) & \bar{X}f(\bar{X})& \bar{X}^2f(\bar{X}) \\ \hline 10/4 & 1& 1/15 & 9/20 & 81/80 \\ \hdashline 11/4 & 1 & 1/15 & 11/60 & 121/240 \\ \hdashline 12/4 & 2 & 2/15 & 24/60 & 288/240 \\ \hdashline 13/4 & 2 & 2/15 & 26/60 & 338/240 \\ \hdashline 14/4 & 3& 3/15 & 42/60 & 588/240 \\ \hdashline 15/4 & 2 & 2/15 & 30/60 & 450/240 \\ \hdashline 16/4 & 2 & 2/15 & 32/60 & 512/240 \\ \hdashline 17/4 & 1 & 1/15 & 17/60 & 289/240 \\ \hdashline 18/4 & 1 & 1/15 & 18/60 & 324/240 \\ \hdashline Total & 15 & 1 & 207/60 & 2910/240 \\ \hline \end{array}$$

2.

Possible means $2.5,2.75,3,3.5,3.75,4,4.25,4.5$

3.

$$P(\bar{X}=4)=\dfrac{2}{15}$$

4.

$$P(\bar{X}=3.5)=\dfrac{3}{15}$$