题目

设总体 sim N(mu ,(sigma )^2), X1,X2,···,Xn是来自X的一个样本,则 () .-|||-A.N(μ,σ^2)-|||-B. (mu ,dfrac ({sigma )^2}(n))-|||-C.N(0,1)-|||-D.N(0,σ^2)

$\begin{align}&设总体X\sim N(\mu,\sigma^{2}),X_{1},X_{2},\cdots,X_{n}是来自X的一个样本,则\frac{1}{n}\sum_{i=1}^{n}X_{i}\sim(\quad).\\&A.N(\mu,\sigma^{2})\\&B.N\left(\mu,\frac{\sigma^{2}}{n}\right)\\&C.N(0,1)\\&D.N(0,\sigma^{2})\end{align}$

题目解答

答案

$\begin{align}&总体 X\sim N(\mu ,{\sigma }^{2}) , 则 E(X)=\mu , D(X)={\sigma }^{2} , \dfrac {1}{n}\sum _{i=1}^{n}{X}_{i} 是样本均值, \\&E(\dfrac {1}{n}\sum _{i=1}^{n}{X}_{i})=\dfrac {1}{n}\cdot n\mu =\mu , D(\dfrac {1}{n}\sum _{i=1}^{n}{X}_{i})=\dfrac {1}{{n}^{2}}\cdot n{\sigma }^{2}=\dfrac {{\sigma }^{2}}{n} ,\\&所以 \dfrac {1}{n}\sum _{i=1}^{n}{X}_{i}\sim N(\mu ,\dfrac {{\sigma }^{2}}{n}) .选B\end{align}$

解析

步骤 1：确定总体的均值和方差
总体 $X\sim N(\mu ,{\sigma }^{2})$ ,则 $E(X)=\mu $ $D(X)={\sigma }^{2}$

步骤 2：计算样本均值的期望
$\dfrac {1}{n}\sum _{i=1}^{n}{X}_{i}$ 是样本均值, $E(\dfrac {1}{n}\sum _{i=1}^{n}{X}_{i})=\dfrac {1}{n}\cdot n{u}_{n}=\mu $

步骤 3：计算样本均值的方差
$D(\dfrac {1}{n}\sum _{i=1}^{n}{X}_{i})=\dfrac {1}{{n}^{2}}\cdot n{\sigma }^{2}=\dfrac {{\sigma }^{2}}{n}$

步骤 4：确定样本均值的分布
由于总体 $X\sim N(\mu ,{\sigma }^{2})$ ,样本均值 $\dfrac {1}{n}\sum _{i=1}^{n}{X}_{i}$ 也服从正态分布,且均值为 $\mu $ ,方差为 $\dfrac {{\sigma }^{2}}{n}$ ,因此 $\dfrac {1}{n}\sum _{i=1}^{n}{X}_{i}\sim N(\mu ,\dfrac {{\sigma }^{2}}{n})$