10.设总体 sim N(mu ,1) ,其中μ为未知参数,X1,X2,X3从为来自总体X的样本,下面4-|||-个关于μ的估计量中,最有效的一个是 ()-|||-(A) dfrac (1)(3)(X)_(1)+dfrac (1)(3)(X)_(2)+dfrac (1)(3)(X)_(3) . (B) dfrac (1)(4)(X)_(1)+dfrac (1)(2)(X)_(2)+dfrac (1)(4)(X)_(3)-|||-(C) dfrac (1)(6)(X)_(1)+dfrac (5)(6)(X)_(2) (D) dfrac (1)(5)(X)_(1)+dfrac (2)(5)(X)_(2)+dfrac (2)(5)(X)_(3)

本题考查了有效估计量的定义，属于基础题。
根据有效估计量的定义：设$\theta _{1}$和$\theta _{2}$都是$\theta $的无偏估计量，如果对一切$\theta $，估计量$\theta _{1}$的方差都不大于$\theta _{2}$的方差，则称$\theta _{1}$比$\theta _{2}$有效。由此可以出本题。
由于总体$X\sim N(\mu,1)$，其中$\mu$为未知参数，$X_{1}$，$X_{2}$，$X_{3}$是来自总体$X$的样本，
所以有$E(X_{i})=\mu$，$D(X_{i})=1$，$i=1,2,3$
选项$A$：$E(\frac{1}{3}X_{1}+\frac{1}{3}X_{2}+\frac{1}{3}X_{3})=\frac{1}{3}E(X_{1})+\frac{1}{3}E(X_{2})+\frac{1}{3}E(X_{3})=\mu$，
$D(\frac{1}{3}X_{1}+\frac{1}{3}X_{2}+\frac{1}{3}X_{3})=\frac{1}{9}D(X_{1})+\frac{1}{9}D(X_{2})+\frac{1}{9}D(X_{3})=\frac{1}{3}$；
选项$B$：$E(\frac{1}{4}X_{1}+\frac{1}{2}X_{2}+\frac{1}{4}X_{3})=\frac{1}{4}E(X_{1})+\frac{1}{2}E(X_{2})+\frac{1}{4}E(X_{3})=\mu$，
$D(\frac{1}{4}X_{1}+\frac{1}{2}X_{2}+\frac{1}{4}X_{3})=\frac{1}{16}D(X_{1})+\frac{1}{4}D(X_{2})+\frac{1}{16}D(X_{3})=\frac{3}{8}$；
选项$C$：$E(\frac{1}{6}X_{1}+\frac{5}{6}X_{2})=\frac{1}{6}E(X_{1})+\frac{5}{6}E(X_{2})=\mu$，
$D(\frac{1}{6}X_{1}+\frac{5}{6}X_{2})=\frac{1}{36}D(X_{1})+\frac{25}{36}D(X_{2})=\frac{13}{18}$；
选项$D$：$E(\frac{1}{5}X_{1}+\frac{2}{5}X_{2}+\frac{2}{5}X_{3})=\frac{1}{5}E(X_{1})+\frac{2}{5}E(X_{2})+\frac{2}{5}E(X_{3})=\mu$，
$D(\frac{1}{5}X_{1}+\frac{2}{5}X_{2}+\frac{2}{5}X_{3})=\frac{1}{25}D(X_{1})+\frac{4}{25}D(X_{2})+\frac{4}{25}D(X_{3})=\frac{9}{25}$；
由于$\frac{1}{3}<\frac{9}{25}<\frac{3}{8}<\frac{13}{18}$，所以最有效的估计量为选项$A$的估计量。
故本题答案为：$A$。