| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To compute the covariance of two vectors x and y, you can use the following formula: Cov(x, y) = Σ((x[i] - mean(x)) * (y[i] - mean(y))) / (n - 1) where: - n is the number of elements in the vectors (which should be the same for both x and y) - mean(x) and mean(y) are the means of the vectors x and y, respectively Given x = (1, 2, 3, 4) and y = (2, 3, 4, 5): 1. Calculate the mean of x and y: mean(x) = (1 + 2 + 3 + 4) / 4 = 2.5 mean(y) = (2 + 3 + 4 + 5) / 4 = 3.5 2. Calculate the sum of the products of the differences: Σ((x[i] - mean(x)) * (y[i] - mean(y))) = ((1 - 2.5) * (2 - 3.5)) + ((2 - 2.5) * (3 - 3.5)) + ((3 - 2.5) * (4 - 3.5)) + ((4 - 2.5) * (5 - 3.5)) = (-1.5 * -1.5) + (-0.5 * -0.5) + (0.5 * 0.5) + (1.5 * 1.5) = 2.25 + 0.25 + 0.25 + 2.25 = 5 3. Calculate the covariance: Cov(x, y) = 5 / (4 - 1) = 5 / 3 ≈ 1.6667 So the covariance of x and y is approximately 1.6667. |
1.67 |
| deepseekr1 | To compute the covariance between the datasets \( x = (1, 2, 3, 4) \) and \( y = (2, 3, 4, 5) \): 1. **Calculate the means**: - Mean of \( x \): \( \bar{x} = \frac{1 + 2 + 3 + 4}{4} = 2.5 \) - Mean of \( y \): \( \bar{y} = \frac{2 + 3 + 4 + 5}{4} = 3.5 \) 2. **Compute deviations from the means**: - For each \( i \), calculate \( (x_i - \bar{x}) \) and \( (y_i - \bar{y}) \), then multiply these deviations: - \( (1 - 2.5)(2 - 3.5) = (-1.5)(-1.5) = 2.25 \) - \( (2 - 2.5)(3 - 3.5) = (-0.5)(-0.5) = 0.25 \) - \( (3 - 2.5)(4 - 3.5) = (0.5)(0.5) = 0.25 \) - \( (4 - 2.5)(5 - 3.5) = (1.5)(1.5) = 2.25 \) 3. **Sum the products of deviations**: - Sum = \( 2.25 + 0.25 + 0.25 + 2.25 = 5 \) 4. **Divide by the number of data points (population covariance)**: - Covariance = \( \frac{5}{4} = 1.25 \) Given that the data can be considered the entire population, the covariance is: \[ \boxed{\dfrac{5}{4}} \] |
1.67 |
| gpto1 | To compute the covariance between the two data sets \( x = (1, 2, 3, 4) \) and \( y = (2, 3, 4, 5) \), we'll use the formula for sample covariance: \[ \text{Cov}(X, Y) = \frac{1}{n - 1} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y}) \] where: - \( n \) is the number of observations. - \( \bar{x} \) and \( \bar{y} \) are the sample means of \( x \) and \( y \), respectively. **Step 1: Calculate the sample means.** For \( x \): \[ \bar{x} = \frac{1 + 2 + 3 + 4}{4} = \frac{10}{4} = 2.5 \] For \( y \): \[ \bar{y} = \frac{2 + 3 + 4 + 5}{4} = \frac{14}{4} = 3.5 \] **Step 2: Compute the deviations from the mean and their products.** \[ \begin{align*} \text{For } i=1:& \quad (x_1 - \bar{x}) = 1 - 2.5 = -1.5, \quad (y_1 - \bar{y}) = 2 - 3.5 = -1.5, \quad \text{Product} = (-1.5)(-1.5) = 2.25 \\ \text{For } i=2:& \quad (x_2 - \bar{x}) = 2 - 2.5 = -0.5, \quad (y_2 - \bar{y}) = 3 - 3.5 = -0.5, \quad \text{Product} = (-0.5)(-0.5) = 0.25 \\ \text{For } i=3:& \quad (x_3 - \bar{x}) = 3 - 2.5 = 0.5, \quad (y_3 - \bar{y}) = 4 - 3.5 = 0.5, \quad \text{Product} = (0.5)(0.5) = 0.25 \\ \text{For } i=4:& \quad (x_4 - \bar{x}) = 4 - 2.5 = 1.5, \quad (y_4 - \bar{y}) = 5 - 3.5 = 1.5, \quad \text{Product} = (1.5)(1.5) = 2.25 \\ \end{align*} \] **Step 3: Sum the products of deviations.** \[ \sum_{i=1}^{4} (x_i - \bar{x})(y_i - \bar{y}) = 2.25 + 0.25 + 0.25 + 2.25 = 5 \] **Step 4: Calculate the covariance.** Since \( n = 4 \): \[ \text{Cov}(X, Y) = \frac{5}{4 - 1} = \frac{5}{3} \approx 1.6667 \] **Answer:** The covariance of \( x \) and \( y \) is **\( \frac{5}{3} \) or approximately 1.6667**. |
1.67 |
| GPT-4o | To compute the covariance of the given data sets \( x = (1, 2, 3, 4) \) and \( y = (2, 3, 4, 5) \), follow the steps below: 1. **Calculate the means of \( x \) and \( y \):** \[ \bar{x} = \frac{1 + 2 + 3 + 4}{4} = \frac{10}{4} = 2.5 \] \[ \bar{y} = \frac{2 + 3 + 4 + 5}{4} = \frac{14}{4} = 3.5 \] 2. **Compute the products of the deviations of each pair \( (x_i, y_i) \):** \[ (x_1 - \bar{x})(y_1 - \bar{y}) = (1 - 2.5)(2 - 3.5) = (-1.5)(-1.5) = 2.25 \] \[ (x_2 - \bar{x})(y_2 - \bar{y}) = (2 - 2.5)(3 - 3.5) = (-0.5)(-0.5) = 0.25 \] \[ (x_3 - \bar{x})(y_3 - \bar{y}) = (3 - 2.5)(4 - 3.5) = (0.5)(0.5) = 0.25 \] \[ (x_4 - \bar{x})(y_4 - \bar{y}) = (4 - 2.5)(5 - 3.5) = (1.5)(1.5) = 2.25 \] 3. **Sum these products:** \[ \sum_{i=1}^{4} (x_i - \bar{x})(y_i - \bar{y}) = 2.25 + 0.25 + 0.25 + 2.25 = 5 \] 4. **Divide by the number of data points minus 1 to get the sample covariance:** \[ \text{Cov}(x, y) = \frac{\sum_{i=1}^{4} (x_i - \bar{x})(y_i - \bar{y})}{n-1} = \frac{5}{4-1} = \frac{5}{3} \approx 1.67 \] Thus, the covariance of \( x \) and \( y \) is approximately \( 1.67 \). |
1.67 |
| llama3.1-70B-Instruct | To compute the covariance of two datasets, x and y, we can use the following formula: cov(x, y) = Σ[(xi - x̄)(yi - ȳ)] / (n - 1) where: * xi and yi are the individual data points in x and y, respectively * x̄ and ȳ are the means of x and y, respectively * n is the number of data points * Σ denotes the sum of the products First, let's calculate the means of x and y: x̄ = (1 + 2 + 3 + 4) / 4 = 10 / 4 = 2.5 ȳ = (2 + 3 + 4 + 5) / 4 = 14 / 4 = 3.5 Next, let's calculate the deviations from the means for each data point: xi - x̄ = (1 - 2.5, 2 - 2.5, 3 - 2.5, 4 - 2.5) = (-1.5, -0.5, 0.5, 1.5) yi - ȳ = (2 - 3.5, 3 - 3.5, 4 - 3.5, 5 - 3.5) = (-1.5, -0.5, 0.5, 1.5) Now, let's compute the products of the deviations: (-1.5)(-1.5) = 2.25 (-0.5)(-0.5) = 0.25 (0.5)(0.5) = 0.25 (1.5)(1.5) = 2.25 Finally, let's sum the products and divide by (n - 1) = 3: cov(x, y) = (2.25 + 0.25 + 0.25 + 2.25) / 3 = 5 / 3 = 1.67 Therefore, the covariance of x and y is 1.67. |
1.67 |
| llama3.1-405B-Instruct | To compute the covariance between two vectors x and y, we can use the following formula: cov(x, y) = Σ[(xi - μx)(yi - μy)] / (n - 1) where xi and yi are the individual elements of the vectors x and y, μx and μy are the means of the vectors x and y, and n is the number of elements in the vectors. First, let's compute the means of the vectors x and y: μx = (1 + 2 + 3 + 4) / 4 = 10 / 4 = 2.5 μy = (2 + 3 + 4 + 5) / 4 = 14 / 4 = 3.5 Next, let's compute the deviations from the means for each element of the vectors: x - μx = (1 - 2.5, 2 - 2.5, 3 - 2.5, 4 - 2.5) = (-1.5, -0.5, 0.5, 1.5) y - μy = (2 - 3.5, 3 - 3.5, 4 - 3.5, 5 - 3.5) = (-1.5, -0.5, 0.5, 1.5) Now, let's compute the product of the deviations for each element: (-1.5)(-1.5) = 2.25 (-0.5)(-0.5) = 0.25 (0.5)(0.5) = 0.25 (1.5)(1.5) = 2.25 Finally, let's sum up the products and divide by the number of elements minus one: cov(x, y) = (2.25 + 0.25 + 0.25 + 2.25) / (4 - 1) = 5 / 3 = 1.67 Therefore, the covariance between the vectors x and y is 1.67. |
1.67 |
| DeepSeek-Coder-V2 | To compute the covariance of two variables \( x \) and \( y \), we can use the following formula: \[ \text{Cov}(x, y) = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y}) \] where: - \( n \) is the number of observations, - \( x_i \) and \( y_i \) are the individual observations, - \( \bar{x} \) and \( \bar{y} \) are the means of \( x \) and \( y \), respectively. Given the data: - \( x = (1, 2, 3, 4) \) - \( y = (2, 3, 4, 5) \) First, we calculate the means of \( x \) and \( y \): \[ \bar{x} = \frac{1 + 2 + 3 + 4}{4} = \frac{10}{4} = 2.5 \] \[ \bar{y} = \frac{2 + 3 + 4 + 5}{4} = \frac{14}{4} = 3.5 \] Next, we compute the differences \( (x_i - \bar{x}) \) and \( (y_i - \bar{y}) \) for each observation: \[ \begin{align*} (x_1 - \bar{x}) &= 1 - 2.5 = -1.5 \\ (x_2 - \bar{x}) &= 2 - 2.5 = -0.5 \\ (x_3 - \bar{x}) &= 3 - 2.5 = 0.5 \\ (x_4 - \bar{x}) &= 4 - 2.5 = 1.5 \\ \end{align*} \] \[ \begin{align*} (y_1 - \bar{y}) &= 2 - 3.5 = -1.5 \\ (y_2 - \bar{y}) &= 3 - 3.5 = -0.5 \\ (y_3 - \bar{y}) &= 4 - 3.5 = 0.5 \\ (y_4 - \bar{y}) &= 5 - 3.5 = 1.5 \\ \end{align*} \] Now, we calculate the product \( (x_i - \bar{x})(y_i - \bar{y}) \) for each pair: \[ \begin{align*} (x_1 - \bar{x})(y_1 - \bar{y}) &= (-1.5)(-1.5) = 2.25 \\ (x_2 - \bar{x})(y_2 - \bar{y}) &= (-0.5)(-0.5) = 0.25 \\ (x_3 - \bar{x})(y_3 - \bar{y}) &= (0.5)(0.5) = 0.25 \\ (x_4 - \bar{x})(y_4 - \bar{y}) &= (1.5)(1.5) = 2.25 \\ \end{align*} \] Summing these products: \[ \sum_{i=1}^{4} (x_i - \bar{x})(y_i - \bar{y}) = 2.25 + 0.25 + 0.25 + 2.25 = 5 \] Finally, we compute the covariance: \[ \text{Cov}(x, y) = \frac{1}{4-1} \sum_{i=1}^{4} (x_i - \bar{x})(y_i - \bar{y}) = \frac{5}{3} \approx 1.6667 \] Thus, the covariance of \( x \) and \( y \) is: \[ \boxed{1.6667} \] |
1.67 |
| Qwen2-72B-Instruct | To compute the covariance of two vectors x and y, you can use the following formula: Cov(x, y) = Σ((x[i] - mean(x)) * (y[i] - mean(y))) / (n - 1) where: - n is the number of elements in the vectors (which should be the same for both x and y) - mean(x) and mean(y) are the means of the vectors x and y, respectively Given x = (1, 2, 3, 4) and y = (2, 3, 4, 5): 1. Calculate the mean of x and y: mean(x) = (1 + 2 + 3 + 4) / 4 = 2.5 mean(y) = (2 + 3 + 4 + 5) / 4 = 3.5 2. Calculate the sum of the products of the differences: Σ((x[i] - mean(x)) * (y[i] - mean(y))) = ((1 - 2.5) * (2 - 3.5)) + ((2 - 2.5) * (3 - 3.5)) + ((3 - 2.5) * (4 - 3.5)) + ((4 - 2.5) * (5 - 3.5)) = (-1.5 * -1.5) + (-0.5 * -0.5) + (0.5 * 0.5) + (1.5 * 1.5) = 2.25 + 0.25 + 0.25 + 2.25 = 5 3. Calculate the covariance: Cov(x, y) = 5 / (4 - 1) = 5 / 3 ≈ 1.6667 So the covariance of x and y is approximately 1.6667. |
1.67 |
| baichuan4 | To compute the covariance of two sets of numbers, we use the formula: \[ \text{Cov}(X, Y) = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x}) (y_i - \bar{y}) \] where: - \(n\) is the number of observations, - \(x_i\) is the \(i\)-th observation of the \(X\) variable, - \(\bar{x}\) is the mean of the \(X\) variable, - \(y_i\) is the \(i\)-th observation of the \(Y\) variable, - \(\bar{y}\) is the mean of the \(Y\) variable. Given \(x = (1, 2, 3, 4)\) and \(y = (2, 3, 4, 5)\), first we need to calculate the means: \[ \bar{x} = \frac{1+2+3+4}{4} = 2.5 \] \[ \bar{y} = \frac{2+3+4+5}{4} = 3.5 \] Now we can calculate the covariance: \[ \text{Cov}(X, Y) = \frac{1}{4-1} [(1-2.5)(2-3.5) + (2-2.5)(3-3.5) + (3-2.5)(4-3.5) + (4-2.5)(5-3.5)] \] \[ \text{Cov}(X, Y) = \frac{1}{3} [(-1.5)(-1.5) + (-0.5)(-0.5) + (0.5)(0.5) + (1.5)(1.5)] \] \[ \text{Cov}(X, Y) = \frac{1}{3} [2.25 + 0.25 + 0.25 + 2.25] \] \[ \text{Cov}(X, Y) = \frac{1}{3} [5] \] \[ \text{Cov}(X, Y) = \frac{5}{3} \] So the covariance between \(x\) and \(y\) is \(\frac{5}{3}\). |
1.67 |
| doubao | First, we need to find the mean of x and y. Mean of x = (1 + 2 + 3 + 4) / 4 = 10 / 4 = 2.5. Mean of y = (2 + 3 + 4 + 5) / 4 = 14 / 4 = 3.5. Then, we calculate the deviations of each element of x and y from their respective means. For x: (-1.5, -0.5, 0.5, 1.5). For y: (-1.5, -0.5, 0.5, 1.5). Next, we multiply the corresponding deviations together and sum them up: ((-1.5) * (-1.5) + (-0.5) * (-0.5) + 0.5 * 0.5 + 1.5 * 1.5) = (2.25 + 0.25 + 0.25 + 2.25) = 5. Finally, we divide by the number of elements (4) to get the covariance: 5 / 4 = 1.25. |
1.67 |
| minimax | To compute the covariance of two variables x and y, you can use the following formula: Cov(x, y) = Σ[(xi - mean(x)) * (yi - mean(y))] / (n - 1) where xi and yi are the individual data points, mean(x) and mean(y) are the means of the x and y data sets, and n is the number of data points. First, we need to calculate the means of x and y: mean(x) = (1 + 2 + 3 + 4) / 4 = 10 / 4 = 2.5 mean(y) = (2 + 3 + 4 + 5) / 4 = 14 / 4 = 3.5 Now, we can calculate the covariance: Cov(x, y) = [(1 - 2.5) * (2 - 3.5)] + [(2 - 2.5) * (3 - 3.5)] + [(3 - 2.5) * (4 - 3.5)] + [(4 - 2.5) * (5 - 3.5)] / (4 - 1) Simplify the terms inside the brackets: = [(-1.5) * (-1.5)] + [(-0.5) * (-0.5)] + [(0.5) * (0.5)] + [(1.5) * (1.5)] / 3 Calculate the products: = [2.25] + [0.25] + [0.25] + [2.25] / 3 Sum the products: = 5 / 3 Finally, divide by (n - 1), which is 3 in this case: Cov(x, y) = 5 / 3 ≈ 1.6667 So, the covariance of x and y is approximately 1.6667. |
1.67 |
| kimi | To compute the covariance of two variables, \( x \) and \( y \), you can follow these steps: 1. Calculate the mean of \( x \) and \( y \). 2. Subtract each element of \( x \) and \( y \) from their respective means and multiply the results for each pair. 3. Sum up all the products from step 2. 4. Divide the sum by the number of pairs minus 1 (if you're calculating the sample covariance) or by the number of pairs (if you're calculating the population covariance). Given the vectors \( x = (1, 2, 3, 4) \) and \( y = (2, 3, 4, 5) \), let's calculate the covariance. Step 1: Calculate the means \[ \text{Mean of } x = \frac{1+2+3+4}{4} = 2.5 \] \[ \text{Mean of } y = \frac{2+3+4+5}{4} = 3.5 \] Step 2: Subtract the means and multiply \[ (1-2.5)(2-3.5) = -1.5 \times -1.5 = 2.25 \] \[ (2-2.5)(3-3.5) = -0.5 \times -0.5 = 0.25 \] \[ (3-2.5)(4-3.5) = 0.5 \times 0.5 = 0.25 \] \[ (4-2.5)(5-3.5) = 1.5 \times 1.5 = 2.25 \] Step 3: Sum up all the products \[ \text{Sum} = 2.25 + 0.25 + 0.25 + 2.25 = 5 \] Step 4: Divide by the number of pairs minus 1 (sample covariance) \[ \text{Covariance} = \frac{5}{4-1} = \frac{5}{3} \approx 1.67 \] So, the sample covariance of \( x \) and \( y \) is approximately 1.67. |
1.67 |
| wenxinyiyan | 首先，我们需要计算x和y的均值。 x的均值 = (1 + 2 + 3 + 4) / 4 = 2.5 y的均值 = (2 + 3 + 4 + 5) / 4 = 3.5 接下来，我们计算协方差。协方差的公式为： Cov(x, y) = Σ[(xi - x的均值) * (yi - y的均值)] / (n - 1) 其中，xi和yi是x和y的各个元素，n是元素的数量。 将给定的值代入公式，我们得到： Cov(x, y) = [(1 - 2.5) * (2 - 3.5) + (2 - 2.5) * (3 - 3.5) + (3 - 2.5) * (4 - 3.5) + (4 - 2.5) * (5 - 3.5)] / (4 - 1) = [(-1.5) * (-1.5) + (-0.5) * (-0.5) + (0.5) * (0.5) + (1.5) * (1.5)] / 3 = [2.25 + 0.25 + 0.25 + 2.25] / 3 = 5 / 3 = 1.6667 所以，x和y的协方差是1.6667。 |
1.67 |
| zhipu | To compute the covariance between two sets of data, x and y, you can use the following formula for the sample covariance: \[ \text{Cov}(x, y) = \frac{1}{n-1} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y}) \] Where: - \( n \) is the number of pairs of scores, - \( x_i \) and \( y_i \) are the individual sample points, - \( \bar{x} \) is the mean of x, - \( \bar{y} \) is the mean of y. Let's calculate the covariance step by step for the given data: 1. Calculate the means (averages) of x and y: \[ \bar{x} = \frac{1 + 2 + 3 + 4}{4} = \frac{10}{4} = 2.5 \] \[ \bar{y} = \frac{2 + 3 + 4 + 5}{4} = \frac{14}{4} = 3.5 \] 2. Compute the product of the deviations of each pair of scores from their respective means: For each pair \( (x_i, y_i) \): \[ (x_i - \bar{x})(y_i - \bar{y}) \] Calculating these products for each pair: \[ (1 - 2.5)(2 - 3.5) = (-1.5)(-1.5) = 2.25 \] \[ (2 - 2.5)(3 - 3.5) = (-0.5)(-0.5) = 0.25 \] \[ (3 - 2.5)(4 - 3.5) = (0.5)(0.5) = 0.25 \] \[ (4 - 2.5)(5 - 3.5) = (1.5)(1.5) = 2.25 \] 3. Sum these products: \[ \text{Sum} = 2.25 + 0.25 + 0.25 + 2.25 = 5 \] 4. Divide by \( n - 1 \) (where \( n \) is the number of pairs, which is 4 in this case): \[ \text{Cov}(x, y) = \frac{1}{4-1} \times 5 = \frac{1}{3} \times 5 = \frac{5}{3} \approx 1.67 \] So, the covariance of x and y is approximately 1.67. This positive value indicates that x and y tend to increase together. Note: The above formula is for the sample covariance, which is typically used when the data represents a sample of a larger population. If x and y represent the entire population, the denominator would be \( n \) instead of \( n - 1 \). However, in most practical scenarios and particularly in statistical estimation, the sample covariance formula is used. |
1.67 |
| gpto1 | To compute the covariance between the two data sets \( x = (1, 2, 3, 4) \) and \( y = (2, 3, 4, 5) \), we'll use the formula for sample covariance: \[ \text{Cov}(X, Y) = \frac{1}{n - 1} \sum_{i=1}^{n} (x_i - \bar{x})(y_i - \bar{y}) \] where: - \( n \) is the number of observations. - \( \bar{x} \) and \( \bar{y} \) are the sample means of \( x \) and \( y \), respectively. **Step 1: Calculate the sample means.** For \( x \): \[ \bar{x} = \frac{1 + 2 + 3 + 4}{4} = \frac{10}{4} = 2.5 \] For \( y \): \[ \bar{y} = \frac{2 + 3 + 4 + 5}{4} = \frac{14}{4} = 3.5 \] **Step 2: Compute the deviations from the mean and their products.** \[ \begin{align*} \text{For } i=1:& \quad (x_1 - \bar{x}) = 1 - 2.5 = -1.5, \quad (y_1 - \bar{y}) = 2 - 3.5 = -1.5, \quad \text{Product} = (-1.5)(-1.5) = 2.25 \\ \text{For } i=2:& \quad (x_2 - \bar{x}) = 2 - 2.5 = -0.5, \quad (y_2 - \bar{y}) = 3 - 3.5 = -0.5, \quad \text{Product} = (-0.5)(-0.5) = 0.25 \\ \text{For } i=3:& \quad (x_3 - \bar{x}) = 3 - 2.5 = 0.5, \quad (y_3 - \bar{y}) = 4 - 3.5 = 0.5, \quad \text{Product} = (0.5)(0.5) = 0.25 \\ \text{For } i=4:& \quad (x_4 - \bar{x}) = 4 - 2.5 = 1.5, \quad (y_4 - \bar{y}) = 5 - 3.5 = 1.5, \quad \text{Product} = (1.5)(1.5) = 2.25 \\ \end{align*} \] **Step 3: Sum the products of deviations.** \[ \sum_{i=1}^{4} (x_i - \bar{x})(y_i - \bar{y}) = 2.25 + 0.25 + 0.25 + 2.25 = 5 \] **Step 4: Calculate the covariance.** Since \( n = 4 \): \[ \text{Cov}(X, Y) = \frac{5}{4 - 1} = \frac{5}{3} \approx 1.6667 \] **Answer:** The covariance of \( x \) and \( y \) is **\( \frac{5}{3} \) or approximately 1.6667**. |
1.67 |
| deepseekv3 | To compute the covariance of the two datasets \( x = (1, 2, 3, 4) \) and \( y = (2, 3, 4, 5) \), follow these steps: --- ### Step 1: Compute the mean of \( x \) and \( y \) The mean of \( x \) is: \[ \bar{x} = \frac{1 + 2 + 3 + 4}{4} = \frac{10}{4} = 2.5 \] The mean of \( y \) is: \[ \bar{y} = \frac{2 + 3 + 4 + 5}{4} = \frac{14}{4} = 3.5 \] --- ### Step 2: Compute the deviations from the mean For each data point, subtract the mean: | \( x_i \) | \( y_i \) | \( x_i - \bar{x} \) | \( y_i - \bar{y} \) | |-----------|-----------|----------------------|----------------------| | 1 | 2 | \(-1.5\) | \(-1.5\) | | 2 | 3 | \(-0.5\) | \(-0.5\) | | 3 | 4 | \(0.5\) | \(0.5\) | | 4 | 5 | \(1.5\) | \(1.5\) | --- ### Step 3: Multiply the deviations for each pair Multiply \( (x_i - \bar{x}) \) and \( (y_i - \bar{y}) \) for each pair: | \( x_i - \bar{x} \) | \( y_i - \bar{y} \) | Product | |----------------------|----------------------|-----------------------| | \(-1.5\) | \(-1.5\) | \(2.25\) | | \(-0.5\) | \(-0.5\) | \(0.25\) | | \(0.5\) | \(0.5\) | \(0.25\) | | \(1.5\) | \(1.5\) | \(2.25\) | --- ### Step 4: Sum the products \[ \sum (x_i - \bar{x})(y_i - \bar{y}) = 2.25 + 0.25 + 0.25 + 2.25 = 5.0 \] --- ### Step 5: Divide by the number of data points The covariance is: \[ \text{Cov}(x, y) = \frac{\sum (x_i - \bar{x})(y_i - \bar{y})}{n} = \frac{5.0}{4} = 1.25 \] --- ### Final Answer: The covariance of \( x \) and \( y \) is: \[ \text{Cov}(x, y) = 1.25 \] |
1.67 |