已知一组数据x1,x2,x3.,x10的方差是2所以(x1-x)^2+(x2-x)^2+(x3-x)^2+...+(x10-x)^2=20(x1-3)^2+(x2-3)^2+(x3-3)^2+...+(x10-3)^2=120 两式相减法得到（x-3)(2x1-3-x+2x2-3-x...+2x10-3-x)=100(x-3)(2*10x-30-10x)=(...