若x~N(,a2),则z=x-∠~N(0,) 证Z的分布函数为 z(x)=P{z≤x}=P =P{X≤+σx}=F(A+ax) 对上式两边求导:fz(x)=fx(1+ax) (+Ox- e 2 2丌0 2丌 =0(x) 故z X-N(0,1证 Z的分布函数为 P Z x = 对上式两边求导: 2 2 2 ( ) 2 + − − = x e X P x − = + P X x F ( x) = X + FZ (x) = f Z (x) = f ( x) X + 2 2 2 1 x e − = =(x) ~ N (0,1) . X Z − 故 =