题目
设是来自正态总体的样本，则（ ）是无偏估计．A. B. C. D.
设
是来自正态总体
的样本，则（ ）是
无偏估计．
B.
C.
D.
题目解答
答案
C. $\dfrac {1}{5}{x}_{1}+\dfrac {1}{5}{x}_{2}+\dfrac {3}{5}{x}_{3}$
解析
步骤 1：理解无偏估计的概念
无偏估计是指估计量的期望值等于被估计参数的真实值。对于正态总体的样本，样本均值是总体均值的无偏估计。
步骤 2：计算每个选项的期望值
A. $\dfrac {1}{5}{x}_{1}+\dfrac {1}{5}{x}_{2}+\dfrac {1}{5}{x}_{3}$
期望值为：$\dfrac {1}{5}E(x_{1})+\dfrac {1}{5}E(x_{2})+\dfrac {1}{5}E(x_{3}) = \dfrac {1}{5}\mu + \dfrac {1}{5}\mu + \dfrac {1}{5}\mu = \dfrac {3}{5}\mu$，不是无偏估计。
B. ${x}_{1}+{x}_{2}+{x}_{3}$
期望值为：$E(x_{1})+E(x_{2})+E(x_{3}) = \mu + \mu + \mu = 3\mu$，不是无偏估计。
C. $\dfrac {1}{5}{x}_{1}+\dfrac {1}{5}{x}_{2}+\dfrac {3}{5}{x}_{3}$
期望值为：$\dfrac {1}{5}E(x_{1})+\dfrac {1}{5}E(x_{2})+\dfrac {3}{5}E(x_{3}) = \dfrac {1}{5}\mu + \dfrac {1}{5}\mu + \dfrac {3}{5}\mu = \mu$，是无偏估计。
D. $\dfrac {2}{5}{x}_{1}+\dfrac {2}{5}{x}_{2}+\dfrac {2}{5}{x}_{3}$
期望值为：$\dfrac {2}{5}E(x_{1})+\dfrac {2}{5}E(x_{2})+\dfrac {2}{5}E(x_{3}) = \dfrac {2}{5}\mu + \dfrac {2}{5}\mu + \dfrac {2}{5}\mu = \dfrac {6}{5}\mu$，不是无偏估计。
无偏估计是指估计量的期望值等于被估计参数的真实值。对于正态总体的样本，样本均值是总体均值的无偏估计。
步骤 2：计算每个选项的期望值
A. $\dfrac {1}{5}{x}_{1}+\dfrac {1}{5}{x}_{2}+\dfrac {1}{5}{x}_{3}$
期望值为：$\dfrac {1}{5}E(x_{1})+\dfrac {1}{5}E(x_{2})+\dfrac {1}{5}E(x_{3}) = \dfrac {1}{5}\mu + \dfrac {1}{5}\mu + \dfrac {1}{5}\mu = \dfrac {3}{5}\mu$，不是无偏估计。
B. ${x}_{1}+{x}_{2}+{x}_{3}$
期望值为：$E(x_{1})+E(x_{2})+E(x_{3}) = \mu + \mu + \mu = 3\mu$，不是无偏估计。
C. $\dfrac {1}{5}{x}_{1}+\dfrac {1}{5}{x}_{2}+\dfrac {3}{5}{x}_{3}$
期望值为：$\dfrac {1}{5}E(x_{1})+\dfrac {1}{5}E(x_{2})+\dfrac {3}{5}E(x_{3}) = \dfrac {1}{5}\mu + \dfrac {1}{5}\mu + \dfrac {3}{5}\mu = \mu$，是无偏估计。
D. $\dfrac {2}{5}{x}_{1}+\dfrac {2}{5}{x}_{2}+\dfrac {2}{5}{x}_{3}$
期望值为：$\dfrac {2}{5}E(x_{1})+\dfrac {2}{5}E(x_{2})+\dfrac {2}{5}E(x_{3}) = \dfrac {2}{5}\mu + \dfrac {2}{5}\mu + \dfrac {2}{5}\mu = \dfrac {6}{5}\mu$，不是无偏估计。