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Sagot :
Let's solve the problem of calculating the covariance for the dataset provided, given the numerical result.
### Step 1: Organize the Data
Given the contingency table, we can extract the following pairs of values for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
- [tex]\(x = [1, 1, 2, 2]\)[/tex]
- [tex]\(y = [4, 3, 1, 2]\)[/tex]
### Step 2: Calculate the Means
First, calculate the mean of [tex]\(x\)[/tex] and the mean of [tex]\(y\)[/tex].
#### Mean of [tex]\(x\)[/tex]
[tex]\[ \text{Mean of } x = \frac{1 + 1 + 2 + 2}{4} = \frac{6}{4} = 1.5 \][/tex]
#### Mean of [tex]\(y\)[/tex]
[tex]\[ \text{Mean of } y = \frac{4 + 3 + 1 + 2}{4} = \frac{10}{4} = 2.5 \][/tex]
### Step 3: Calculate the Covariance
Covariance is given by:
[tex]\[ \text{Cov}(x, y) = \frac{1}{n} \sum_{i=1}^{n} (x_i - \text{mean}_x) (y_i - \text{mean}_y) \][/tex]
For each pair [tex]\((x_i, y_i)\)[/tex]:
- [tex]\( (x_1, y_1) = (1, 4) \)[/tex]
- [tex]\( (x_2, y_2) = (1, 3) \)[/tex]
- [tex]\( (x_3, y_3) = (2, 1) \)[/tex]
- [tex]\( (x_4, y_4) = (2, 2) \)[/tex]
Calculate [tex]\( (x_i - \text{mean}_x) \)[/tex] and [tex]\( (y_i - \text{mean}_y) \)[/tex]:
1. For [tex]\((1, 4)\)[/tex]:
[tex]\[ (1 - 1.5) = -0.5; \quad (4 - 2.5) = 1.5; \quad \text{product} = -0.5 \times 1.5 = -0.75 \][/tex]
2. For [tex]\((1, 3)\)[/tex]:
[tex]\[ (1 - 1.5) = -0.5; \quad (3 - 2.5) = 0.5; \quad \text{product} = -0.5 \times 0.5 = -0.25 \][/tex]
3. For [tex]\((2, 1)\)[/tex]:
[tex]\[ (2 - 1.5) = 0.5; \quad (1 - 2.5) = -1.5; \quad \text{product} = 0.5 \times -1.5 = -0.75 \][/tex]
4. For [tex]\((2, 2)\)[/tex]:
[tex]\[ (2 - 1.5) = 0.5; \quad (2 - 2.5) = -0.5; \quad \text{product} = 0.5 \times -0.5 = -0.25 \][/tex]
Now, sum all the products:
[tex]\[ -0.75 + (-0.25) + (-0.75) + (-0.25) = -2.0 \][/tex]
Finally, divide by the number of pairs ([tex]\(n = 4\)[/tex]) to find the covariance:
[tex]\[ \text{Cov}(x, y) = \frac{-2.0}{4} = -0.5 \][/tex]
### Summary
The means of the data are:
- Mean of [tex]\(x\)[/tex] = 1.5
- Mean of [tex]\(y\)[/tex] = 2.5
The covariance between [tex]\(x\)[/tex] and [tex]\(y\)[/tex] is [tex]\( -0.5 \)[/tex].
### Step 1: Organize the Data
Given the contingency table, we can extract the following pairs of values for [tex]\(x\)[/tex] and [tex]\(y\)[/tex]:
- [tex]\(x = [1, 1, 2, 2]\)[/tex]
- [tex]\(y = [4, 3, 1, 2]\)[/tex]
### Step 2: Calculate the Means
First, calculate the mean of [tex]\(x\)[/tex] and the mean of [tex]\(y\)[/tex].
#### Mean of [tex]\(x\)[/tex]
[tex]\[ \text{Mean of } x = \frac{1 + 1 + 2 + 2}{4} = \frac{6}{4} = 1.5 \][/tex]
#### Mean of [tex]\(y\)[/tex]
[tex]\[ \text{Mean of } y = \frac{4 + 3 + 1 + 2}{4} = \frac{10}{4} = 2.5 \][/tex]
### Step 3: Calculate the Covariance
Covariance is given by:
[tex]\[ \text{Cov}(x, y) = \frac{1}{n} \sum_{i=1}^{n} (x_i - \text{mean}_x) (y_i - \text{mean}_y) \][/tex]
For each pair [tex]\((x_i, y_i)\)[/tex]:
- [tex]\( (x_1, y_1) = (1, 4) \)[/tex]
- [tex]\( (x_2, y_2) = (1, 3) \)[/tex]
- [tex]\( (x_3, y_3) = (2, 1) \)[/tex]
- [tex]\( (x_4, y_4) = (2, 2) \)[/tex]
Calculate [tex]\( (x_i - \text{mean}_x) \)[/tex] and [tex]\( (y_i - \text{mean}_y) \)[/tex]:
1. For [tex]\((1, 4)\)[/tex]:
[tex]\[ (1 - 1.5) = -0.5; \quad (4 - 2.5) = 1.5; \quad \text{product} = -0.5 \times 1.5 = -0.75 \][/tex]
2. For [tex]\((1, 3)\)[/tex]:
[tex]\[ (1 - 1.5) = -0.5; \quad (3 - 2.5) = 0.5; \quad \text{product} = -0.5 \times 0.5 = -0.25 \][/tex]
3. For [tex]\((2, 1)\)[/tex]:
[tex]\[ (2 - 1.5) = 0.5; \quad (1 - 2.5) = -1.5; \quad \text{product} = 0.5 \times -1.5 = -0.75 \][/tex]
4. For [tex]\((2, 2)\)[/tex]:
[tex]\[ (2 - 1.5) = 0.5; \quad (2 - 2.5) = -0.5; \quad \text{product} = 0.5 \times -0.5 = -0.25 \][/tex]
Now, sum all the products:
[tex]\[ -0.75 + (-0.25) + (-0.75) + (-0.25) = -2.0 \][/tex]
Finally, divide by the number of pairs ([tex]\(n = 4\)[/tex]) to find the covariance:
[tex]\[ \text{Cov}(x, y) = \frac{-2.0}{4} = -0.5 \][/tex]
### Summary
The means of the data are:
- Mean of [tex]\(x\)[/tex] = 1.5
- Mean of [tex]\(y\)[/tex] = 2.5
The covariance between [tex]\(x\)[/tex] and [tex]\(y\)[/tex] is [tex]\( -0.5 \)[/tex].
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