Get comprehensive answers to your questions with the help of IDNLearn.com's community. Ask your questions and receive detailed and reliable answers from our experienced and knowledgeable community members.
Sagot :
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 |
We appreciate your presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. Thank you for choosing IDNLearn.com for your queries. We’re here to provide accurate answers, so visit us again soon.
