由题意得
X拔=1/n(x1+x2+...+xn),
S²=1/n[(x1-x拔)²+(x2-x拔)²+...+(Xn-X拔)²]
标准差=S
1、 kx1,kx2,kx3…kxn的平均数=1/n(kx1+kx2+kx3+…+kxn)
=k*(1/n)(x1+x2+...+xn)
=kx拔
方差=1/n[(kx1-kx拔)²+(kx2-kx拔)²+...+(kXn-kX拔)²]
=k²/n[(x1-x拔)²+(x2-x拔)²+...+(Xn-X拔)²]
=k²S²
标准差=kS
2、kx1+a,kx2+a,kx3+a…kxn+a的平均数
=1/n(kx1+a+kx2+a+kx3+a+…+kxn+a)
=1/n[k(x1+x2+...+xn)+na]
=kx拔+a
方差=1/n[(kx1+a-kx拔-a)²+(kx2+a-kx拔-a)²+...+(kXn+a-kX拔-a)²]
=k²/n[(x1-x拔)²+(x2-x拔)²+...+(Xn-X拔)²]
=k²S²
标准差=kS