题目

设X~N(-1,3^2),则随机变量（ ）~N(0,1).3^2

设X~N(-1,),则随机变量（ ）~N(0,1).

题目解答

答案

∵X~N(-1,)

∴E(X)=-1,D(X)=9;

∴该题的答案为A

解析

步骤 1：确定随机变量X的期望和方差
已知X~N(-1,),即X服从均值为-1，方差为9的正态分布。因此，E(X)=-1，D(X)=9。

步骤 2：计算每个选项的期望和方差
A. $\dfrac {x+1}{3}$
- 期望：$E(\dfrac {x+1}{3})=\dfrac {1}{3}E(X)+\dfrac {1}{3}=-\dfrac {1}{3}+\dfrac {1}{3}=0$
- 方差：$D(\dfrac {x+1}{3})=\dfrac {1}{9}D(X)=\dfrac {1}{9}\times 9=1$
B. $\dfrac {x-1}{9}$
- 期望：$E(\dfrac {x-1}{9})=\dfrac {1}{9}E(X)-\dfrac {1}{9}=-\dfrac {1}{9}-\dfrac {1}{9}=-\dfrac {2}{9}$
- 方差：$D(\dfrac {x-1}{9})=\dfrac {1}{81}D(X)=\dfrac {1}{81}\times 9=\dfrac {1}{9}$
C. $\dfrac {x-1}{3}$
- 期望：$E(\dfrac {x-1}{3})=\dfrac {1}{3}E(X)-\dfrac {1}{3}=-\dfrac {1}{3}-\dfrac {1}{3}=-\dfrac {2}{3}$
- 方差：$D(\dfrac {x-1}{3})=\dfrac {1}{9}D(X)=\dfrac {1}{9}\times 9=1$
D. $\dfrac {x+1}{9}$
- 期望：$E(\dfrac {x+1}{9})=\dfrac {1}{9}E(X)+\dfrac {1}{9}=-\dfrac {1}{9}+\dfrac {1}{9}=0$
- 方差：$D(\dfrac {x+1}{9})=\dfrac {1}{81}D(X)=\dfrac {1}{81}\times 9=\dfrac {1}{9}$

步骤 3：确定符合N(0,1)的随机变量
根据步骤2的计算结果，只有选项A的期望为0，方差为1，符合N(0,1)的分布。