设是取自正态总体的简单随机样本,下列统计量中是总体的数学期望的无偏估计量的是( ).(A) (B) (C) (D)

步骤 1：计算每个选项的期望值
对于每个选项，我们需要计算其期望值，以确定其是否为总体数学期望的无偏估计量。无偏估计量的期望值应该等于总体的数学期望μ。
步骤 2：计算选项A的期望值
${E}_{({\hat {1}}_{1})}=E(\dfrac {1}{3}{X}_{1}+\dfrac {1}{4}{X}_{2}-{X}_{3})=\dfrac {1}{3}E({X}_{1})+\dfrac {1}{4}E({X}_{2})-E({X}_{3})=\dfrac {1}{3}\mu +\dfrac {1}{4}\mu -\mu =-\dfrac {5}{12}\mu$
步骤 3：计算选项B的期望值
${E}_{({\lambda }_{2})}=E(\dfrac {1}{5}{X}_{1}+\dfrac {3}{10}{X}_{2}+\dfrac {1}{2}{X}_{3})=\dfrac {1}{5}E({X}_{1})+\dfrac {3}{10}E({X}_{2})+\dfrac {1}{2}E({X}_{3})=\dfrac {1}{5}\mu +\dfrac {3}{10}\mu +\dfrac {1}{2}\mu =\mu$
步骤 4：计算选项C的期望值
${E}_{({\lambda }_{3})}=E(\dfrac {1}{3}{X}_{1}+\dfrac {3}{4}{X}_{2}+{X}_{3})=\dfrac {1}{3}E({X}_{1})+\dfrac {3}{4}E({X}_{2})+E({X}_{3})=\dfrac {1}{3}\mu +\dfrac {3}{4}\mu +\mu =\dfrac {25}{12}\mu$
步骤 5：计算选项D的期望值
${E}_{({\lambda }_{4})}=E(\dfrac {1}{4}{X}_{1}+\dfrac {1}{4}{X}_{2}+\dfrac {1}{4}{X}_{3})=\dfrac {1}{4}E({X}_{1})+\dfrac {1}{4}E({X}_{2})+\dfrac {1}{4}E({X}_{3})=\dfrac {1}{4}\mu +\dfrac {1}{4}\mu +\dfrac {1}{4}\mu =\dfrac {3}{4}\mu$
步骤 6：确定无偏估计量
根据无偏估计量的定义，期望值等于总体数学期望μ的统计量是无偏估计量。因此，选项B是总体数学期望的无偏估计量。