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Three randomly selected children are surveyed. The ages of the children are 2, 3, and 10. Assume that samples of size [tex]\( n=2 \)[/tex] are randomly selected with replacement from the population of 2, 3, and 10. Listed below are the nine different samples. Complete parts (a) through (d).

a. Find the value of the population variance [tex]\(\sigma^2\)[/tex].
[tex]\[
\sigma^2 = 12.667 \quad \text{(Round to three decimal places as needed.)}
\][/tex]

b. Find the variance of each of the nine samples, then summarize the sampling distribution of the variances in the format of a table representing the probability distribution of the distinct variance values.

Samples:
[tex]\[
\begin{array}{cc}
(2, 2) & (2, 3) & (2, 10) \\
(3, 2) & (3, 3) & (3, 10) \\
(10, 2) & (10, 3) & (10, 10) \\
\end{array}
\][/tex]

Complete the following tasks:
1. Calculate the variance for each sample.
2. Summarize the sampling distribution of the variances in a table.



Answer :

GinnyAnswer
Let's break down the solution in a detailed, step-by-step manner:

### Part (a): Calculate the Population Variance, [tex]\(\sigma^2\)[/tex]
1. Determine the Ages of the Children:
The ages of the children surveyed are 2, 3, and 10.

2. Calculate the Population Mean ([tex]\(\mu\)[/tex]):
[tex]\[ \mu = \frac{2 + 3 + 10}{3} = \frac{15}{3} = 5 \][/tex]

3. Calculate the Population Variance ([tex]\(\sigma^2\)[/tex]):
The variance is calculated using the formula:
[tex]\[ \sigma^2 = \frac{\sum_{i=1}^{N} (x_i - \mu)^2}{N} \][/tex]
Where [tex]\(x_i\)[/tex] are the individual ages and [tex]\(N\)[/tex] is the number of children.

Calculate each squared deviation from the mean:
[tex]\[ (2-5)^2 = 9, \quad (3-5)^2 = 4, \quad (10-5)^2 = 25 \][/tex]

Sum of squared deviations:
[tex]\[ 9 + 4 + 25 = 38 \][/tex]

Population variance:
[tex]\[ \sigma^2 = \frac{38}{3} \approx 12.667 \][/tex]

So, the value of the population variance [tex]\(\sigma^2\)[/tex] is approximately 12.667 (rounded to three decimal places).

### Part (b): Calculate the Variance for Each of the Nine Samples and Summarize the Sampling Distribution

1. List the Nine Different Samples of Size [tex]\(n=2\)[/tex]:
The samples are:
[tex]\[ (2, 2), (2, 3), (2, 10), (3, 2), (3, 3), (3, 10), (10, 2), (10, 3), (10, 10) \][/tex]

2. Calculate the Variance for Each Sample:
The variance of a sample [tex]\((x_1, x_2)\)[/tex] is calculated using the formula:
[tex]\[ \text{Variance} = \frac{(x_1 - \bar{x})^2 + (x_2 - \bar{x})^2}{2} \][/tex]
Where [tex]\(\bar{x}\)[/tex] is the mean of the sample.

- Sample (2, 2):
[tex]\[ \bar{x} = \frac{2+2}{2} = 2 \][/tex]
[tex]\[ \text{Variance} = \frac{(2-2)^2 + (2-2)^2}{2} = 0.0 \][/tex]

- Sample (2, 3):
[tex]\[ \bar{x} = \frac{2+3}{2} = 2.5 \][/tex]
[tex]\[ \text{Variance} = \frac{(2-2.5)^2 + (3-2.5)^2}{2} = 0.25 \][/tex]

- Sample (2, 10):
[tex]\[ \bar{x} = \frac{2+10}{2} = 6 \][/tex]
[tex]\[ \text{Variance} = \frac{(2-6)^2 + (10-6)^2}{2} = 16.0 \][/tex]

- Sample (3, 2):
[tex]\[ \bar{x} = \frac{3+2}{2} = 2.5 \][/tex]
[tex]\[ \text{Variance} = \frac{(3-2.5)^2 + (2-2.5)^2}{2} = 0.25 \][/tex]

- Sample (3, 3):
[tex]\[ \bar{x} = \frac{3+3}{2} = 3 \][/tex]
[tex]\[ \text{Variance} = \frac{(3-3)^2 + (3-3)^2}{2} = 0.0 \][/tex]

- Sample (3, 10):
[tex]\[ \bar{x} = \frac{3+10}{2} = 6.5 \][/tex]
[tex]\[ \text{Variance} = \frac{(3-6.5)^2 + (10-6.5)^2}{2} = 12.25 \][/tex]

- Sample (10, 2):
[tex]\[ \bar{x} = \frac{10+2}{2} = 6 \][/tex]
[tex]\[ \text{Variance} = \frac{(10-6)^2 + (2-6)^2}{2} = 16.0 \][/tex]

- Sample (10, 3):
[tex]\[ \bar{x} = \frac{10+3}{2} = 6.5 \][/tex]
[tex]\[ \text{Variance} = \frac{(10-6.5)^2 + (3-6.5)^2}{2} = 12.25 \][/tex]

- Sample (10, 10):
[tex]\[ \bar{x} = \frac{10+10}{2} = 10 \][/tex]
[tex]\[ \text{Variance} = \frac{(10-10)^2 + (10-10)^2}{2} = 0.0 \][/tex]

The variances of the nine samples are:
[tex]\[ [0.0, 0.25, 16.0, 0.25, 0.0, 12.25, 16.0, 12.25, 0.0] \][/tex]

3. Summarize the Sampling Distribution of the Variances:
Identify the unique variances and their probabilities:

- Unique Variances: [tex]\(0.0, 0.25, 12.25, 16.0\)[/tex]

- Count of Each Variance:
[tex]\[ 0.0 \text{ occurs } 3 \text{ times} \][/tex]
[tex]\[ 0.25 \text{ occurs } 2 \text{ times} \][/tex]
[tex]\[ 12.25 \text{ occurs } 2 \text{ times} \][/tex]
[tex]\[ 16.0 \text{ occurs } 2 \text{ times} \][/tex]

- Probability Distribution:
[tex]\[ P(0.0) = \frac{3}{9} \approx 0.333 \][/tex]
[tex]\[ P(0.25) = \frac{2}{9} \approx 0.222 \][/tex]
[tex]\[ P(12.25) = \frac{2}{9} \approx 0.222 \][/tex]
[tex]\[ P(16.0) = \frac{2}{9} \approx 0.222 \][/tex]

Summarized in a table, the probability distribution of the distinct variance values is:

| Variance | Probability |
|----------|-------------|
| 0.0 | 0.333 |
| 0.25 | 0.222 |
| 12.25 | 0.222 |
| 16.0 | 0.222 |