Answer to Question #341216 in Statistics and Probability for Xchely

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 sample 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.

Mean of population $(\mu)$ = $\dfrac{1+2+3+4+5+6}{6}=3.5$

Variance of population 

$$\sigma^2=\dfrac{\Sigma(x_i-\bar{x})^2}{n}=\dfrac{1}{6}(6.25+2.25+0.25$$

$$+0.25+2.25+6.25)=\dfrac{17.5}{6}$$

$$\sigma=\sqrt{\sigma^2}=\sqrt{\dfrac{17.5}{6}}\approx1.707825$$

1. Select a random sample of size 4 without replacement. We have a sample distribution of sample mean.

The number of possible samples which can be drawn without replacement is $^{N}C_n=^{6}C_4=15.$

2.

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} no & Sample & Sample \\ & & mean\ (\bar{x}) \\ \hline 1 & 1,2,3,4 & 10/4 \\ \hdashline 2 & 1,2,3,5 & 11/4 \\ \hdashline 3 & 1,2,3,6 & 12/4 \\ \hdashline 4 & 1,2,4,5 & 12/4 \\ \hdashline 5 & 1,2,4,6 & 13/4 \\ \hdashline 6 & 1,2,5,6 & 14/4 \\ \hdashline 7 & 1,3,4,5 & 13/4 \\ \hdashline 8 & 1,3,4,6 & 14/4 \\ \hdashline 9 & 1,3,5,6 & 15/4 \\ \hdashline 10 & 1,4,5,6 & 16/4 \\ \hdashline 11 & 2,3,4,5 & 14/4 \\ \hdashline 12 & 2,3,4,6 & 15/4 \\ \hdashline 13 & 2,3,5,6 & 16/4 \\ \hdashline 14 & 2,4,5,6 & 17/4\\ \hdashline 15 & 3,4,5,6 & 18/4 \\ \hdashline \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c} \bar{X} & f(\bar{X}) &\bar{X} f(\bar{X}) & \bar{X}^2f(\bar{X}) \\ \hline 10/4 & 1/15 & 10/60 & 100/240 \\ \hdashline 11/4 & 1/15 & 11/60 & 121/240 \\ \hdashline 12/4 & 2/15 & 24/60 & 288/240 \\ \hdashline 13/4 & 2/15 & 26/60 & 338/240 \\ \hdashline 14/4 & 3/15 & 42/60 & 588/240 \\ \hdashline 15/4 & 2/15 & 30/60 & 450/240 \\ \hdashline 16/4 & 2/15 & 32/60 & 512/240 \\ \hdashline 17/4 & 1/15 & 17/60 & 289/240 \\ \hdashline 18/4 & 1/15 & 18/60 & 324/240 \\ \hdashline \end{array}$$

3.

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

4.

$$P(\bar{X}=3.5)=\dfrac{3}{15}=\dfrac{1}{5}$$