| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | B. μx+y = 76, σx+y = 13 Explanation: Since X and Y are independent, the mean of X + Y is the sum of the means of X and Y: μx+y = μx + μy = 32 + 44 = 76 The standard deviation of X + Y is the square root of the sum of the variances of X and Y: σx+y = √(σx^2 + σy^2) = √(5^2 + 12^2) = √(25 + 144) = √169 = 13 Therefore, the correct answer is B. μx+y = 76, σx+y = 13. |
B |
| deepseekr1 | To determine the mean and standard deviation of \( X + Y \): 1. **Mean (\(\mu_{X+Y}\))**: The mean of the sum of two independent random variables is the sum of their individual means: \[ \mu_{X+Y} = \mu_X + \mu_Y = 32 + 44 = 76 \] 2. **Standard Deviation (\(\sigma_{X+Y}\))**: For independent variables, variances add. First, compute the variances: \[ \text{Var}(X) = \sigma_X^2 = 5^2 = 25 \] \[ \text{Var}(Y) = \sigma_Y^2 = 12^2 = 144 \] Then, add the variances: \[ \text{Var}(X + Y) = 25 + 144 = 169 \] The standard deviation is the square root of the variance: \[ \sigma_{X+Y} = \sqrt{169} = 13 \] **Answer**: B. \(\mu_{X+Y} = 76\), \(\sigma_{X+Y} = 13\) |
B |
| gpto1 | To solve this problem, we'll use properties of expected value (mean) and standard deviation for independent random variables. **Mean (\( \mu \)) of \( X + Y \):** For any two random variables \( X \) and \( Y \), the expected value (mean) of their sum is: \[ \mu_{X+Y} = \mu_X + \mu_Y \] Given \( \mu_X = 32 \) and \( \mu_Y = 44 \), we have: \[ \mu_{X+Y} = 32 + 44 = 76 \] **Standard deviation (\( \sigma \)) of \( X + Y \):** For independent random variables, the variance of their sum is the sum of their variances: \[ \text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) \] Since variance is the square of the standard deviation: \[ \text{Var}(X) = \sigma_X^2 = 5^2 = 25 \] \[ \text{Var}(Y) = \sigma_Y^2 = 12^2 = 144 \] So: \[ \text{Var}(X+Y) = 25 + 144 = 169 \] \[ \sigma_{X+Y} = \sqrt{\text{Var}(X+Y)} = \sqrt{169} = 13 \] **Answer:** \[ \mu_{X+Y} = 76, \quad \sigma_{X+Y} = 13 \] **Therefore, the correct choice is B.** B |
B |
| GPT-4o | To find the mean and standard deviation of the sum of two independent random variables, we use the following properties: 1. The mean of the sum of two random variables is the sum of their means: \( μ_{X+Y} = μ_X + μ_Y \) 2. The variance of the sum of two independent random variables is the sum of their variances: \( σ^2_{X+Y} = σ^2_X + σ^2_Y \) The standard deviation is the square root of the variance. Given: - \( μ_X = 32 \) - \( σ_X = 5 \) - \( μ_Y = 44 \) - \( σ_Y = 12 \) First, calculate the mean of \( X + Y \): \[ μ_{X+Y} = μ_X + μ_Y = 32 + 44 = 76 \] Next, calculate the variance of \( X + Y \): \[ σ^2_{X+Y} = σ^2_X + σ^2_Y = 5^2 + 12^2 = 25 + 144 = 169 \] Then, the standard deviation of \( X + Y \) is: \[ σ_{X+Y} = \sqrt{σ^2_{X+Y}} = \sqrt{169} = 13 \] Thus, the mean and standard deviation of the random variable \( X + Y \) are: \[ μ_{X+Y} = 76 \] \[ σ_{X+Y} = 13 \] The correct answer is: \[ \text{B. } μ_{X+Y} = 76, σ_{X+Y} = 13 \] |
B |
| llama3.1-70B-Instruct | A |
B |
| llama3.1-405B-Instruct | B. μx+y = 76, σx+y = 13 Reasoning Skill: This question requires the application of the rules for the mean and standard deviation of the sum of independent random variables. The correct answer can be found by using the formulas μx+y = μx + μy and σx+y = √(σx^2 + σy^2). This type of question evaluates the ability to analyze and apply statistical concepts to solve problems, which is an essential skill in scientific evidence evaluation. Note: The other options are incorrect because: A. The standard deviation is incorrect, it should be 13, not 8.5. C. The standard deviation is incorrect, it should be 13, not 17. D. The mean is incorrect, it should be 76, not 38. |
B |
| DeepSeek-Coder-V2 | B. μx+y = 76, σx+y = 13 | B |
| Qwen2-72B-Instruct | B. μx+y = 76, σx+y = 13 | B |
| baichuan4 | B. μx+y = 76, σx+y = 13 | B |
| doubao | B. If X and Y are independent random variables, the mean of the sum is the sum of the means, so μx+y = μx + μy = 32 + 44 = 76. The variance of the sum of independent random variables is the sum of the variances, and the standard deviation is the square root of the variance. The variance of X is σx^2 = 5^2 = 25 and the variance of Y is σy^2 = 12^2 = 144. So the variance of X + Y is 25 + 144 = 169 and the standard deviation σx+y = √169 = 13. |
B |
| minimax | The mean of the sum of two independent random variables is the sum of their means. Therefore, μx+y = μx + μy = 32 + 44 = 76. The standard deviation of the sum of two independent random variables is the square root of the sum of their variances (since they are independent, the covariance term drops out). Therefore, σx+y = √(σx^2 + σy^2) = √(5^2 + 12^2) = √(25 + 144) = √169 = 13. The correct answer is B. μx+y = 76, σx+y = 13. |
B |
| kimi | B. μx+y = 76, σx+y = 13 Explanation: For independent random variables X and Y, the mean of the sum of X and Y is the sum of their means: μx+y = μx + μy = 32 + 44 = 76 The standard deviation of the sum of independent random variables is the square root of the sum of their variances (since variance is the square of standard deviation): σx+y = √(σx^2 + σy^2) = √(5^2 + 12^2) = √(25 + 144) = √169 = 13 |
B |
| wenxinyiyan | B Explanation: 1. Mean of X + Y: * When two random variables X and Y are independent (or even if they are not independent), the mean of their sum is the sum of their means. * Formula: μx+y = μx + μy * Given: μx = 32 and μy = 44 * Calculation: μx+y = 32 + 44 = 76 2. Standard Deviation of X + Y: * When two random variables X and Y are independent, the variance of their sum is the sum of their variances. The standard deviation is then the square root of this sum. * Formula for variance: Var(X+Y) = Var(X) + Var(Y) * Formula for standard deviation: σx+y = sqrt[Var(X) + Var(Y)] * Given: σx = 5 and σy = 12 * Calculation of variances: Var(X) = 5^2 = 25, Var(Y) = 12^2 = 144 * Calculation of standard deviation: σx+y = sqrt[25 + 144] = sqrt[169] = 13 Combining the results for mean and standard deviation, we get μx+y = 76 and σx+y = 13, which corresponds to option B. |
B |
| zhipu | To determine the mean and standard deviation of the random variable \(X + Y\), we can use the properties of means and variances for independent random variables. ### Mean of \(X + Y\) The mean of the sum of two independent random variables is the sum of their means: \[ \mu_{X+Y} = \mu_X + \mu_Y \] Given: \[ \mu_X = 32 \] \[ \mu_Y = 44 \] So, \[ \mu_{X+Y} = 32 + 44 = 76 \] ### Standard Deviation of \(X + Y\) The variance of the sum of two independent random variables is the sum of their variances. The standard deviation is the square root of the variance. \[ \sigma_{X+Y}^2 = \sigma_X^2 + \sigma_Y^2 \] Given: \[ \sigma_X = 5 \] \[ \sigma_Y = 12 \] So, \[ \sigma_X^2 = 5^2 = 25 \] \[ \sigma_Y^2 = 12^2 = 144 \] \[ \sigma_{X+Y}^2 = 25 + 144 = 169 \] \[ \sigma_{X+Y} = \sqrt{169} = 13 \] Thus, the mean and standard deviation of the random variable \(X + Y\) are: \[ \mu_{X+Y} = 76 \] \[ \sigma_{X+Y} = 13 \] The correct answer is: B. μx+y = 76, σx+y = 13 Answer: B |
B |
| gpto1 | To solve this problem, we'll use properties of expected value (mean) and standard deviation for independent random variables. **Mean (\( \mu \)) of \( X + Y \):** For any two random variables \( X \) and \( Y \), the expected value (mean) of their sum is: \[ \mu_{X+Y} = \mu_X + \mu_Y \] Given \( \mu_X = 32 \) and \( \mu_Y = 44 \), we have: \[ \mu_{X+Y} = 32 + 44 = 76 \] **Standard deviation (\( \sigma \)) of \( X + Y \):** For independent random variables, the variance of their sum is the sum of their variances: \[ \text{Var}(X+Y) = \text{Var}(X) + \text{Var}(Y) \] Since variance is the square of the standard deviation: \[ \text{Var}(X) = \sigma_X^2 = 5^2 = 25 \] \[ \text{Var}(Y) = \sigma_Y^2 = 12^2 = 144 \] So: \[ \text{Var}(X+Y) = 25 + 144 = 169 \] \[ \sigma_{X+Y} = \sqrt{\text{Var}(X+Y)} = \sqrt{169} = 13 \] **Answer:** \[ \mu_{X+Y} = 76, \quad \sigma_{X+Y} = 13 \] **Therefore, the correct choice is B.** B |
B |
| deepseekv3 | To solve this problem, we need to calculate the mean and standard deviation of the random variable \( X + Y \). 1. **Mean of \( X + Y \):** Since \( X \) and \( Y \) are independent, the mean of \( X + Y \) is the sum of the individual means: \[ \mu_{X+Y} = \mu_X + \mu_Y = 32 + 44 = 76 \] 2. **Standard Deviation of \( X + Y \):** For independent random variables, the variance of \( X + Y \) is the sum of the individual variances. The standard deviation is the square root of the variance: \[ \sigma_{X+Y} = \sqrt{\sigma_X^2 + \sigma_Y^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 \] Therefore, the mean and standard deviation of \( X + Y \) are \( \mu_{X+Y} = 76 \) and \( \sigma_{X+Y} = 13 \). **Answer: B** |
B |