设总体 X sim N(mu,4)，其中 mu 为未知参数，(X_(1),X_(2),X_(3)) 为样本，下面四个关于 mu 的无偏估计中，采用有效性这一标准来衡量，最好的一个是A. (1)/(6)X_(1)+(1)/(3)X_(2)+(1)/(2)X_(3)B. (1)/(5)X_(1)+(2)/(5)X_(2)+(2)/(5)X_(3)C. (2)/(7)X_(1)+(5)/(7)X_(2)D. (1)/(3)X_(1)+(1)/(3)X_(2)+(1)/(3)X_(3)

步骤 1：计算各选项的方差
- **选项 A**：$\frac{1}{6}X_{1}+\frac{1}{3}X_{2}+\frac{1}{2}X_{3}$
- 方差：$Var(\frac{1}{6}X_{1}+\frac{1}{3}X_{2}+\frac{1}{2}X_{3}) = (\frac{1}{6})^2 \times 4 + (\frac{1}{3})^2 \times 4 + (\frac{1}{2})^2 \times 4 = \frac{1}{9} \times 4 + \frac{4}{9} \times 4 + 1 \times 4 = \frac{4}{9} + \frac{16}{9} + 4 = \frac{4}{9} + \frac{16}{9} + \frac{36}{9} = \frac{56}{9}$
- **选项 B**：$\frac{1}{5}X_{1}+\frac{2}{5}X_{2}+\frac{2}{5}X_{3}$
- 方差：$Var(\frac{1}{5}X_{1}+\frac{2}{5}X_{2}+\frac{2}{5}X_{3}) = (\frac{1}{5})^2 \times 4 + (\frac{2}{5})^2 \times 4 + (\frac{2}{5})^2 \times 4 = \frac{4}{25} + \frac{16}{25} + \frac{16}{25} = \frac{36}{25}$
- **选项 C**：$\frac{2}{7}X_{1}+\frac{5}{7}X_{2}$
- 方差：$Var(\frac{2}{7}X_{1}+\frac{5}{7}X_{2}) = (\frac{2}{7})^2 \times 4 + (\frac{5}{7})^2 \times 4 = \frac{16}{49} + \frac{100}{49} = \frac{116}{49}$
- **选项 D**：$\frac{1}{3}X_{1}+\frac{1}{3}X_{2}+\frac{1}{3}X_{3}$
- 方差：$Var(\frac{1}{3}X_{1}+\frac{1}{3}X_{2}+\frac{1}{3}X_{3}) = 3 \times (\frac{1}{3})^2 \times 4 = 3 \times \frac{1}{9} \times 4 = \frac{4}{3}$

步骤 2：比较方差
- $\frac{56}{9} \approx 6.2222$
- $\frac{36}{25} = 1.44$
- $\frac{116}{49} \approx 2.3673$
- $\frac{4}{3} \approx 1.3333$

步骤 3：选择最小方差
最小方差为 $\frac{4}{3}$，对应选项 D。