7.[单选题]设X_(1),X_(2),X_(3)是来自总体X、容量是3的一样本，则总体均值的无偏估计是A. (1)/(6)X_(1)+(5)/(6)X_(2)-(2)/(6)X_(3)B. (1)/(6)X_(1)+(1)/(3)X_(2)+(1)/(6)X_(3)C. (1)/(3)X_(1)+(1)/(3)X_(2)+(1)/(3)X_(3)D. (1)/(5)X_(1)+(4)/(5)X_(2)-(3)/(5)X_(3)

本题考查总体均值均值无偏估计的知识点。解题思路是根据无偏估计的定义，若一个估计量$\hat{\theta}$是总体参数$\theta$的无偏估计，则$E(\hat{\hat{\theta}\}) = \theta$。对于本题，我们需要判断哪个选项作为总体均值$\mu$的估计量时，其数学期望等于$\mu$。

设总体$X$的均值为$\mu$，即$E(X_{1},X_{2},X_{3})$是来自总体$X$的样本，则$E(X_{1}) = E(X_{2}) = E(X_{3})=\mu$。

• 选项A：
  设$\hat{\mu}_{A}=\frac{1}{6}X_{1}+\frac{5}{6}X_{2}-\frac{2}{6}X_{3}$，根据数学期望的线性性质$E(aX + bY)=aE(X)+bE(Y)$（$a,b$为常数)），可得：
  $E(\hat{\mu}_{A}) = E(\frac{1}{6}X_{1}+\frac{5}{6}X_{2}-\frac{2}{6}X_{3})=\frac{1}{6}E(X_{1})+\frac{5}{6}E(X_{2})-\frac{2}{6}E(X_{3})}$
  将$E(X_{1}) = E(X_{2}) = E(X_{3})=\mu$代入上式得：
  $E(\hat{\mu}_{A})=\frac{1}{6}\mu+\frac{5}{6}\mu-\frac{2}{6}\mu=\frac{1 + 5 - 2}{6}\mu=\frac{4}{6}\mu\neq\mu$
  所以选项A不是总体均值的无偏估计。

• 选项B：
  设$\hat{\mu}_{B}=\frac{1}{6}X_{1}+\frac{1}{3}X_{2}+\frac{1}{6}X_{3}$，同理可得：
  $E(\hat{\mu}_{B}) = E(\frac{1}{6}X_{1}+\frac{1}{3}X_{2}+\frac{1}{6}X_{3})=\frac{1}{6}E(X_{1})+\frac{1}{3}E(X_{2})+\frac{1}{6}E(X_{3})$
  将$E(X_{1}) = E(X_{2}) = E(X_{3})=\mu$代入上式得：
  $E(\hat{\mu}_{B})=\frac{1}{6}\mu+\frac{1}{3}\mu+\frac{1}{6}\mu=\frac{1 + 2+1}{6}\mu=\frac{4}{6}\mu\neq\mu$
  所以选项B不是总体均值的无偏估计。

• 选项C：
  设$\hat{\mu}_{C}=\frac{1}{3}X_{1}+\frac{1}{3}X_{2}+\frac{1}{3}X_{3}$，同理可得：
  $E(\hat{\mu}_{C}) = E(\frac{1}{3}X_{1}+\frac{1}{3}X_{2}+\frac{1}{3}X_{3})=\frac{1}{3}E(X_{1})+\frac{1}{3}E(X_{2})+\frac{1}{3}E(X_{3})$
  将$E(X_{1}) = E(X_{2}) = E(X_{3})=\mu$代入上式得：
  $E(\hat{\mu}_{C})=\frac{3}\mu+\frac{1}{3}\mu+\frac{1}{3}\mu=\frac{1 + 1+1}{3}\mu=\mu$
  所以选项C是总体均值的无偏估计。

• 选项D：
  设$\hat{\mu}_{D}=\frac{1}{5}X_{1}+\frac{4}{5}X_{2}-\frac{3}{5}X_{3}$为0.95的置信区间为$(\bar{X}-z_{\alpha/2}\frac{\sigma}{\sqrt{n}},\bar{X}+z_{\alpha/2}\frac{\sigma}{\sqrt{n}})$，其中$z_{\alpha/2}$是标准正态分布的上$\frac{\alpha}{2}$分位数。
  已知$1 - \alpha = 0.95$，则$\alpha = 0.05$，$\frac{\alpha}{2}=0.025$，查\ to get along with others.
  $z_{\alpha/2}=z_{0.025}=1.96$，$\bar{X}=10$，$\sigma = 2$，$n = 16$。
  将这些值代入置信区间为：
  $(10 - 1.96\times\frac{2}{\sqrt{16}},10 + 1.96\times\frac{2}{\sqrt{16}})$
  $=(10 - 1.96\times\frac{2}{可以}{4},10 + 1.96\times\frac{2}{4})$
  $=(10 - 0.98,10 + 0.98)$
  $=(9.02,10.98)$