第2个回答 2013-06-23
答案:16和4
过程:
设平均数是X,方差是Y吧,方便一点
x1,x2,x3,...,xn
Xa=1/n(x1+x2+x3+...+xn)
Ya=1/n[(x1-Xa)^2+(x2-Xa)^2+...+(xn-Xa)^2]
x1+c,x2+c,x3+c,...,xn+c
Xb=1/n[(x1+c)+(x2+c)+...+(xn+c)]
``=1/n(x1+x2+x3+...+xn)+c
``=Xa+c
Yb=1/n[(x1+c-Xb)^2+(x2+c-Xb)^2+...+(xn+c-Xb)^2]
``=1/n[(x1-Xa)^2+(x2-Xa)^2+...+(xn-Xa)^2]
``=Ya
dx1,dx2,dx3,...,dxn
Xc=1/n(dx1+dx2+...+dxn)
``=d/n(x1+x2+...+xn)
``=dXa
Yc=1/n[(dx1-Xc)^2+(dx2-Xc)^2+...+(dxn-Xc)^2]
``=1/n[(dx1-dXa)^2+(dx2-dXa)^2+...+(dxn-dXa)^2]
``=1/n[d^2(x1-Xa)^2+d^2(x2-Xa)^2+...+d^2(xn-Xa)^2]
``=d^2/n[(x1-Xa)^2+(x2-Xa)^2+...+(xn-Xa)^2]
``=d^2Ya