Ch2-16 P=0.4k01 2 3 4 代入 Pk0.60.2400960.03840.0256 x01234 X x<0 F(x)06 0≤x<1 P(X≤x)= 0.6+0.24=0.84 1<x x<2 0.84+0.096=0.936,2≤x<3 0.936+0.0384=0.9744,3≤x<4 x≥4Ch2-16 • 0 • 1 • 2 • 3 • 4 x x ] ] 0.6 + 0.24 = 0.84, 1 x  2 0.6, 0  x 1 0, x  0 0.84 + 0.096 = 0.936, 2  x  3 0.936 + 0.0384 = 0.9744 , 3  x  4 1 x  4 F(x) ]• • ] k pk 0 1 2 3 4 0.6 0.24 0.096 0.0384 0.0256 代入 p = 0.4 P(X  x) =