0,x?0??1?x2,0?x?1?所以:
F?x???2??1x2?2x?1,1?x?2?2?1,x?2?9．解:每只器件寿命大于1500小时的概率
p?x?1500?????1500f(x)d(x)??10002d(x)?
1500x23??则任意取5只设其中寿命大与1500小时的器件为y只则y~B(5,2/3)
211232121p?y?2??1?p?y?0??p?y?1??1?C50()0?()5?C5()?()4?
33332435?32?3)??()??(1)??(?0.5)10．解: 22??(1)?(1??(0.5))?0.8413?0.6915?1?0.5328p?2?x?5???(10?3?4?3p??4?x?10???()??()?2?(3.5)?1?2?0.9998?1?0.9996
22p?x?2??p?x<-2??p?x?2???(?2?3?2?3)?1??()??(?2.5)?1??(?0.5)?1??(2.5)??(0.5)?0.697722 p?x?3??(2) C?3
1 2(3) p?x>d??0.9则p?x?d??0.1且d<3 即?(d?33?d)?0.1即1??()?0.1 223?d3?d?()?0.9则?1.29所以d?0.42
22211． 解设随机变量x表螺栓的长度x~N(10.05,0.06)
1?p?10.05?0.12?x?10.05?0.12?10.05?0.12?10.05??10.05?0.12?10.05?1???()??()?0.060.06??
?1???(2)??(?2)??2?2?(2)?2?2?0.9772?0.0456
12．解: x~N(?,?)
2 5
p?120 ?2?(要求2?(40?40?)?1?0.80 则?(40?)?0.90 则 40??1.29 则 ??40即??31 1.29 第三章 随机向量 1． 解：P{X?0,Y?0}?0;22C2C21P{X?0,Y?2}??;C7435112C3C2C26P{X?1,Y?1}??;C74352C32C23P{X?2,Y?0}??;C74352C32C23P{X?2,Y?2}??;4C73531C3C22P{X?3,Y?1}??;4C735P{X?0,Y?1}?0;P{X?1,Y?0}?0;121C3C2C26P{X?1,Y?2}??;C743511C32C2C212 P{X?2,Y?1}??;C743531C3C22P{X?3,Y?0}??;4C735P{X?3,Y?2}?02． 解：(1)F(??,??)???8k?1?????????f(x,y)dxdy??[?k(6?x?y)dy]dx?10224?k?118313 (6?x?y)dy]dx?02881.5411.5127(3)P{X?1.5}?F(1.5,??)??[?(6?x?y)dy]dx??(2?x)dx?028023224?x11212(4)P{X?Y?4}??[?(6?x?y)dy]dx??(x2?4x?6)dx?0288023(2)P{X?1,Y?3}??[?3． 6 x2???04.8y(2?x)dy?2.4(2?x)x解：fX(x)??f(x,y)dy?????0??14.8y(2?x)dx?2.4y(3?4y?y2)???fY(y)??f(x,y)dx???y???0???0?x?1其它0?y?1其它 4． 解：fX(x)?????????y?x???xedy?ef(x,y)dy????0?y?y???0edx?yef(x,y)dx????0yx?0其它y?0其它 fY(y)??5． ?????1解：(1)fX(x)???00?x?1其它0?x?1,y?0其它 2 y?1?2?e因为X,Y相互独立,所以f(x,y)?fX(x)fY(y)??2?0?(2)方程有实根则?=4X2?4Y?0即Y?X2P{Y?X2}??[?01x20yxx11??1?21edy]dx??(1?e2)dx?1?2??e2dx0022?2?1?2?[?(1)??(0)]?0.14456． 解: （1）F(??,??)??1?1dx?2cxydy?cx1214?1 故 c? 421?2124?x(1?x),?1?x?1（2）fX(x)??8 ?0,其它??75?y2,0?y?1 fY(y)??2?0,其它? 7．解：（1）由于X在（0，1）服从均匀分布 ?1,0?x?1 故f(x)?? 则1?y?e 0,其它?又y?e单调递增且可导，其反函数为：x?lny x 7 设Y?e的概率密度为：g(y) x?1'?1?(lny)??,1?y?e于是g(y)?? ??y??0,其它?0?（2）由于0?y，故 Y??2lnX的反函数为h(y)?ey??1??f[h(y)]?(h(y))?e2,y?0故 g(y)?? ??20??0,y?0??'1?y2 8．解法1： 由于X和Y是两个相互独立的随机变量， 由卷积公式fZ(z)??????fX(z?y)fY(y)dy可得 当z?0时， fZ(z)=0 当0?z?1时， fZ(z)??z0e?ydy?1?e?z 当1?z时，由0?x?1，知0?z?y?1，即：z?1?y?z fZ(z)??e?ydy?e1?z?e?z z?1z解法2：可有求密度函数的定义法计算得到。 9．解：（1） fX(x)?????????1?1?(x?y)?x(x?y)edy,x?0x?0??0?(x?1)ef(x,y)dx??? 2?2?0,x?0?0x?0???1?yy?0?(y?1)e同理 fY(y)???2 ?y?0?0由于f(x,y)?fX(x)fY(y)，故X和Y不相互独立的。 fz(z)???????z1?z?12?zz?0??0zedx?zef(x,z?x)dx??2??2 ??0??0z?0未完 8
