简单说：是为了保证估计的无偏性!
1.总体方差为σ2,均值为μ
S=[(X1-X)^2+(X2-X)^2.+(Xn-X)^2]/(n-1)
X表示样本均值=(X1+X2+...+Xn)/n
设A=(X1-X)^2+(X2-X)^2.+(Xn-X)^2
E(A)=E[(X1-X)^2+(X2-X)^2.+(Xn-X)^2]
=E[(X1)^2-2X*X1+X^2+(X2)^2-2X*X2+X^2+(X2-X)^2.+(Xn)^2-2X*Xn+X^2]
=E[(X1)^2+(X2)^2...+(Xn)^2+nX^2-2X*(X1+X2+...+Xn)]
=E[(X1)^2+(X2)^2...+(Xn)^2+nX^2-2X*(nX)]
=E[(X1)^2+(X2)^2...+(Xn)^2-nX^2]
而E(Xi)^2=D(Xi)+[E(Xi)]^2=σ2+μ2
E(X)^2=D(X)+[E(X)]^2=σ2/n+μ2
所以E(A)=E[(X1-X)^2+(X2-X)^2.+(Xn-X)^2]
=n(σ2+μ2)-n(σ2/n+μ2)
=(n-1)σ2
故为了保证样本方差的无偏性（即保证估计量的数学期望等于实际值,在此即要保证样本方差的期望等于总体方差）,应取：
S=[(X1-X)^2+(X2-X)^2.+(Xn-X)^2]/(n-1)
从而保证：E(S)=E(A)/(n-1）=(n-1)σ2/(n-1)=σ2