Question

The probability distribution shown here describes a population of measurements that can assume values of 0, 2, 4, and 6, each of which occurs with the same relative frequency:

1. List all the different samples of n = 2 measurements that can be selected from this population. For example, (0, 6) is one possible pair of measurements; (2, 2) is another possible pair.
2. Calculate the mean of each different sample listed in part a.
3. If a sample of n = 2 measurements is randomly selected from the population, what is the probability that a specific sample will be selected.
4. Assume that a random sample of n = 2 measurements is selected from the population. List the different values of x found in part b and find the probability of each. Then give the sampling distribution of the sample mean x in tabular form.
5. Construct a probability histogram for the sampling distribution of$\overline{\mathbf{x}}$.

Short answer

a. The required answer is -

(0,2),(0,4),(0,6),(2,4),(2,6),(4,6),(0,0),(2,2),(4,4),(6,6),(2,0),(4,0),(6,0),(4,2),(6,2),(6,4).

b.

Samples	Mean
0,2	1
0,4	2
0,6	3
2,4	3
2,6	4
4,6	5
0.0	0
2,2	2
4,4	4
6,6	6
2,0	1
4,0	2
6,0	3
4,2	3
6,2	4
6,4	5

c.$\frac{\mathbf{1}}{\mathbf{16}}$

d.

Mean	Number of times
0	$\frac{\mathbf{1}}{\mathbf{16}}$
1	$\frac{\mathbf{1}}{\mathbf{8}}$
2	$\frac{\mathbf{3}}{\mathbf{16}}$
3	$\frac{\mathbf{1}}{\mathbf{4}}$
4	$\frac{\mathbf{3}}{\mathbf{16}}$
5	$\frac{\mathbf{1}}{\mathbf{18}}$
6	$\frac{\mathbf{1}}{\mathbf{16}}$

e.

Step-by-step solution

Formula forcalculating different samples

a.

The formula for calculating the total number of possible samples from the given values of x is shown below.

$$\begin{gathered} \mathrm{Numbers}\mathrm{of}\mathrm{samples}={(\mathrm{Totalnumber}\mathrm{of}\mathrm{values}\mathrm{of}\mathrm{x})}^{\mathrm{n}} \\ ={\left(4\right)}^2 \\ =16 \end{gathered}$$

Total number of samples

It has been found that only 16 different samples can be found, and accordingly, the possible samples are listed below.

(0,2),(0,4),(0,6),(2,4),(2,6),(4,6),(0,0),(2,2),(4,4),(6,6),(2,0),(4,0),(6,0),(4,2),(6,2),(6,4)

Total number of samples

b.

The table below shows the list of the means of different samples.

Samples	Mean
0,2	$\frac{0+2}{2}=1$
0,4	$\frac{0+4}{2}=2$
0,6	$\frac{0+6}{2}=3$
2,4	$\frac{2+4}{2}=3$
2,6	$\frac{2+6}{2}=4$
4,6	$\frac{4+6}{2}=5$
0.0	$\frac{0+0}{2}=0$
2,2	$\frac{2+2}{2}=2$
4,4	$\frac{4+4}{2}=4$
6,6	$\frac{6+6}{2}=6$
2,0	$\frac{2+0}{2}=1$
4,0	$\frac{4+0}{2}=2$
6,0	$\frac{6+0}{2}=3$
4,2	$\frac{4+2}{2}=3$
6,2	$\frac{6+2}{2}=4$
6,4	$\frac{6+4}{2}=5$

Formula for calculating different samples

c.

For each sample, the probability can be found by multiplying each of the probabilities of the respective values of x. Here, it can be seen that all the values have the same probabilities.

Total number of samples

The probability of each sample getting chosen is shown below.

Samples	Probability
0,2	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
0,4	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
0,6	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
2,4	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
2,6	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
4,6	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
0,0	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
2,2	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
4,4	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
6,6	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
2,0	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
4,0	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
6,0	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
4,2	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
6,2	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$
6,4	$\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

Therefore, it can be observed that for all the samples, the possibility is $\frac{\mathbf{1}}{\mathbf{16}}$.

Number of times the mean values are observed

d.

The number of times the mean values have been observed in Part b isshownbelow.

Mean	Number of times
0	1
1	2
2	3
3	4
4	3
5	2
6	1

Calculation of the probability

The probability of each mean ofgetting chosen is shown below

Mean	Number of times	Probability
0	1	$1\times \frac{1}{16}=\frac{1}{16}$
1	2	$2\times \frac{1}{16}=\frac{1}{16}$
2	3	$3\times \frac{1}{16}=\frac{1}{16}$
3	4	$4\times \frac{1}{16}=\frac{1}{16}$
4	3	$3\times \frac{1}{16}=\frac{1}{16}$
5	2	$2\times \frac{1}{16}=\frac{1}{16}$
6	1	$1\times \frac{1}{16}=\frac{1}{16}$

Therefore, it can be observed that for all the samples, the possibility for 0 and 6 is $\frac{\mathbf{1}}{\mathbf{16}}$,for 1 and 3, it is $\frac{\mathbf{1}}{\mathbf{8}}$, and for 2 and 4, it is.role="math" localid="1657972705521" $\frac{3}{\mathbf{16}}$

List of probabilities of x

The number of times the mean values have been observed in Part b isshownbelow.

Mean	Number of times
0	$\frac{\mathbf{1}}{\mathbf{16}}$
1	role="math" localid="1657972888816" $\frac{\mathbf{1}}{\mathbf{8}}$
2	$\frac{\mathbf{3}}{\mathbf{16}}$
3	$\frac{\mathbf{1}}{\mathbf{4}}$
4	$\frac{\mathbf{3}}{\mathbf{16}}$
5	$\frac{\mathbf{1}}{\mathbf{8}}$
6	$\frac{\mathbf{1}}{\mathbf{16}}$

Elucidation on the graph

The diagram shows the probabilities of the respective values of the mean in the form of a bar graph.The mean value 3 in the diagram shows the maximum probability, which is $\frac{\mathbf{1}}{\mathbf{4}}$.