题目
4.某公司10位员工的月工资(单位:元)为x1,x2 ,····-|||-x10,其均值和方差分别为x和s^2,若从下月起每位员-|||-工的月工资增加100元,则这10位员工下月工资的均-|||-值和方差分别为 ()-|||-A. overline (x)cdot (s)^2+(100)^2 B. overline (x)+100cdot (s)^2+(100)^2-|||-C.x,s^2 D. overline (x)+100cdot (s)^2
题目解答
答案
解析
步骤 1：计算新数据的均值
原数据的均值为 $\overline {x}=\dfrac {1}{10}({x}_{1}+{x}_{2}+\cdots +{x}_{10})$。每位员工的月工资增加100元后，新数据的均值为 $\overline {x'}=\dfrac {1}{10}({x}_{1}+100+{x}_{2}+100+\cdots +{x}_{10}+100)=\dfrac {1}{10}({x}_{1}+{x}_{2}+\cdots +{x}_{10})+100=\overline {x}+100$。
步骤 2：计算新数据的方差
原数据的方差为 ${s}^{2}=\dfrac {1}{10}[ {({x}_{1}-\overline {x})}^{2}+{({x}_{2}-\overline {x})}^{2}+\cdots +{({x}_{10}-\overline {x})}^{2}]$。每位员工的月工资增加100元后，新数据的方差为 ${s'}^{2}=\dfrac {1}{10}{[ ({x}_{1}+100-\overline {x}-100)}^{2}+{({x}_{2}+100-\overline {x}-100)}^{2}+\cdots +{({x}_{10}+100-\overline {x}-100)}^{2}] =\dfrac {1}{10}[ {({x}_{1}-\overline {x})}^{2}+{({x}_{2}-\overline {x})}^{2}+\cdots +{({x}_{10}-\overline {x})}^{2}] ={{s}_{1}}^{2}$。
原数据的均值为 $\overline {x}=\dfrac {1}{10}({x}_{1}+{x}_{2}+\cdots +{x}_{10})$。每位员工的月工资增加100元后，新数据的均值为 $\overline {x'}=\dfrac {1}{10}({x}_{1}+100+{x}_{2}+100+\cdots +{x}_{10}+100)=\dfrac {1}{10}({x}_{1}+{x}_{2}+\cdots +{x}_{10})+100=\overline {x}+100$。
步骤 2：计算新数据的方差
原数据的方差为 ${s}^{2}=\dfrac {1}{10}[ {({x}_{1}-\overline {x})}^{2}+{({x}_{2}-\overline {x})}^{2}+\cdots +{({x}_{10}-\overline {x})}^{2}]$。每位员工的月工资增加100元后，新数据的方差为 ${s'}^{2}=\dfrac {1}{10}{[ ({x}_{1}+100-\overline {x}-100)}^{2}+{({x}_{2}+100-\overline {x}-100)}^{2}+\cdots +{({x}_{10}+100-\overline {x}-100)}^{2}] =\dfrac {1}{10}[ {({x}_{1}-\overline {x})}^{2}+{({x}_{2}-\overline {x})}^{2}+\cdots +{({x}_{10}-\overline {x})}^{2}] ={{s}_{1}}^{2}$。