设样本数据 x_ 1 , x_ 2 , , x_ 10 的…——2014 高考数学第 9 题答案解析

【答案】 $A$

【解析】

试题分析：由题得：$x_{1}+x_{2}+\cdots+x_{10}=10 \times 1=10 ;\left(x_{1}-1\right)^{2}+\left(x_{2}-1\right)^{2}+\cdots+\left(x_{10}-1\right)^{2}=10 \times 4=40 y_{1}, y_{2}, \cdots y_{10}$ 的均值和方差分别为：

均值 $\bar{y}=\frac{y_{1}+y_{2}+\cdots+y_{10}}{10}$

$$ =\frac{\left(x_{1}+a\right)+\left(x_{2}+a\right)+\cdots+\left(x_{10}+a\right)}{10}=\frac{\left(x_{1}+x_{2}+\cdots+x_{10}\right)+10 a}{10}=\frac{10+10 a}{10}=1+a $$

方差 $=\frac{\left(y_{1}-\bar{y}\right)^{2}+\left(y_{2}-\bar{y}\right)^{2}+\cdots+\left(y_{10}-\bar{y}\right)^{2}}{10}$

$=\frac{\left[\left(x_{1}+a\right)-(1+a)\right]^{2}+\left[\left(x_{2}+a\right)-(1+a)\right]^{2}+\cdots+\left[\left(x_{10}+a\right)-(1+a)\right]^{2}}{10}$

$$ =\frac{\left(x_{1}-1\right)^{2}+\left(x_{2}-1\right)^{2}+\cdots+\left(x_{10}-1\right)^{2}}{10}=\frac{40}{10}=4 $$

故选 $A$
考点：均值和方差．