已知,一组数据x1,x2,…,xn,平均数为x(上面加一横)方差s^2=1/n[(x1-x)^2+(x2-x)^2+ … +(xn-x)^2判断s^2与1/n(x1^2+x2^2+ … +xn^2)-x的关系ps.上面单一个x的都是 平均数啊.

下面的推导过程中,也用x代替平均数．
(x1-x)^2+(x2-x)^2+ … +(xn-x)^2
= x1^2 - 2x1x + x^2 + x2^2 - 2x2x + x^2 + ...+ xn^2 - 2xn*x + x^2
= x1^2 + x2^2 +...+ xn^2 - 2(x1x + x2x + ...+ xn*x) + nx^2
= x1^2 + x2^2 +...+ xn^2 - 2x(x1 + x2 + ...+ xn) + nx^2
= x1^2 + x2^2 +...+ xn^2 - 2x * nx + nx^2
= x1^2 + x2^2 +...+ xn^2 - nx^2
将此结果代入s^2=1/n[(x1-x)^2+(x2-x)^2+ … +(xn-x)^2：
s^2=(1/n)[(x1-x)^2+(x2-x)^2+ … +(xn-x)^2
= (1/n)(x1^2 + x2^2 +...+ xn^2 - nx^2)
= (x1^2 + x2^2 +...+ xn^2 )/n - x^2