Answer :
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 |
### 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 |
