设总体 X 的数学期望为 mu，X_1, X_2, X_3 是取自于总体 X 的简单随机样本，则统计量(）是 mu 的无偏估计量。A. (1)/(2)X_1 + (1)/(3)X_2 + (1)/(4)X_3B. (1)/(2)X_1 + (1)/(3)X_2 + (1)/(5)X_3C. (1)/(2)X_1 + (1)/(3)X_2 + (1)/(6)X_3D. (1)/(2)X_1 + (1)/(3)X_2 + (1)/(7)X_3

步骤 1：理解无偏估计量的定义
无偏估计量是指统计量的期望值等于总体参数的值。对于本题，统计量的期望值应等于总体的数学期望 $\mu$。

步骤 2：计算各选项的期望值
- **选项 A**：$\frac{1}{2}X_1 + \frac{1}{3}X_2 + \frac{1}{4}X_3$
- 期望值：$E\left(\frac{1}{2}X_1 + \frac{1}{3}X_2 + \frac{1}{4}X_3\right) = \frac{1}{2}E(X_1) + \frac{1}{3}E(X_2) + \frac{1}{4}E(X_3) = \frac{1}{2}\mu + \frac{1}{3}\mu + \frac{1}{4}\mu = \left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}\right)\mu = \frac{13}{12}\mu \neq \mu$
- **选项 B**：$\frac{1}{2}X_1 + \frac{1}{3}X_2 + \frac{1}{5}X_3$
- 期望值：$E\left(\frac{1}{2}X_1 + \frac{1}{3}X_2 + \frac{1}{5}X_3\right) = \frac{1}{2}E(X_1) + \frac{1}{3}E(X_2) + \frac{1}{5}E(X_3) = \frac{1}{2}\mu + \frac{1}{3}\mu + \frac{1}{5}\mu = \left(\frac{1}{2} + \frac{1}{3} + \frac{1}{5}\right)\mu = \frac{31}{30}\mu \neq \mu$
- **选项 C**：$\frac{1}{2}X_1 + \frac{1}{3}X_2 + \frac{1}{6}X_3$
- 期望值：$E\left(\frac{1}{2}X_1 + \frac{1}{3}X_2 + \frac{1}{6}X_3\right) = \frac{1}{2}E(X_1) + \frac{1}{3}E(X_2) + \frac{1}{6}E(X_3) = \frac{1}{2}\mu + \frac{1}{3}\mu + \frac{1}{6}\mu = \left(\frac{1}{2} + \frac{1}{3} + \frac{1}{6}\right)\mu = 1\mu = \mu$
- **选项 D**：$\frac{1}{2}X_1 + \frac{1}{3}X_2 + \frac{1}{7}X_3$
- 期望值：$E\left(\frac{1}{2}X_1 + \frac{1}{3}X_2 + \frac{1}{7}X_3\right) = \frac{1}{2}E(X_1) + \frac{1}{3}E(X_2) + \frac{1}{7}E(X_3) = \frac{1}{2}\mu + \frac{1}{3}\mu + \frac{1}{7}\mu = \left(\frac{1}{2} + \frac{1}{3} + \frac{1}{7}\right)\mu = \frac{41}{42}\mu \neq \mu$

步骤 3：确定无偏估计量
只有选项 C 的系数和为1，满足无偏性条件，即统计量 $\frac{1}{2}X_1 + \frac{1}{3}X_2 + \frac{1}{6}X_3$ 是 $\mu$ 的无偏估计量。