题目

5.设总体 sim N(1,4) ,X1,X2,···,Xn为来自总体X的一个样本则()-|||-A. dfrac (overline {X)-1}(2)sim N(0,1) B. dfrac (overline {X)-1}(4)sim N(0,1)-|||-C. dfrac (overline {X)-1}(2/sqrt {n)}sim N(0,1) D. dfrac (overline {X)-1}(sqrt {2)}sim N(0,1)

5. $\begin{gathered} \text{ 设总体 }X{\sim}N(1,4),X_1,X_2,\cdots,X_n\text{为来自总体 }X\text{ 的一个样本,则（）} \\ \mathrm{A.~}\frac{\overline{X}-1}{2}{\sim}N(0,1)\quad\mathrm{B.~}\frac{\overline{X}-1}{4}{\sim}N(0,1) \\ C.~\frac{\overline{X}-1}{2/\sqrt{n}}\sim N(0,1) \quad\text{D. }\frac{\overline{X}-1}{\sqrt{2}}\sim N(0,1) \end{gathered}$

题目解答

答案

解答如下：

$\begin{aligned} &\text{因为}E({\overline{X}})={}{\mu},D({\overline{X}})={\frac{\sigma^{2}}{n}}({\text{当X}\sim}\mathrm{N(\mu,~\alpha^{2})} \\ &\text{时,} \\ &\text{所以}E(\overline{X})=1,D(\overline{X})=\frac{2^{2}}{n}. \\ &\text{于是} \\ &E\Big(\frac{\overline{X}-1}{a}\Big)=\frac{1}{a}E(\overline{X}-1)=\frac{1}{a}\big[E(\overline{X})-1\big]=0, \\ &D\Big(\frac{\overline{X}-1}{a}\Big)=\frac{1}{a^{2}}D(\overline{X}-1)=\frac{1}{a^{2}}D(\overline{X})=\frac{1}{a^{2}}\cdot\frac{2^{2}}{n}, \\ &\text{当}a=\frac{2}{\sqrt{n}}\text{时},D\Big(\frac{\overline{X}-1}{a}\Big)=1\text{}\mathbb{}\text{}. \end{aligned}$

综上所述，选C