注意： 这是第一稿（存在一些错误） 第四章概率论习题__偶数.doc
2方案一：平均年薪为3万
方案二：记年薪为X，则p(X?1.2)?0.2，p(X?4.2)?0.8
EX?1.2?0.2?4.2?0.8?3.6?3
故应采用方案二 4p?X?2??11315，p?X?3??，p?X?4??，p?X?5??，p?X?6??，28142872831p?X?7??，p?X?2??，
1441131531EX?2??3??4??5??6??7??8??6。
281428728144x6不会
2?2x?2x，0?x??? edy?2e?00x??1?2x（1）EX??xfX?x?dx??2xedx?
0021（2）E?3X?1??3EX?1?
2???x2?2x1（3）E?X,Y????xyf?x,y?dxdy???xyedydx?。
????00x48fX?x??f?x,y?dy??x10XU?8,9?，Y98U?8,9?
1?小时? 3EX?Y???98x?yf?x,y?dxdy?即先到的人等待的平均时间为20分钟。
??e??t,t?0,12f(t)??
t?0.?0,ET??t?edt??8?edt?088??t???t1?e?8??。
211C14C1C1414（1）a?1时，p??n?0??2，p??n?1?? 2C15C1511C1C24E?n?214??
C151532112C15CaC15?aCa?a，p??2?2??2 a?2时，p??2?0??2，p??2?1??2C15C15C15112CaC15?aCa2a E?2?1??2??22C15C1515由E?2?4得，a?10。 345563C10C5C10C54C10C5（2）p??9?4??，p??9?5??，p??9?6??， 999C15C15C157819C10C52C10C5C10C50，p??9?8??，p??9?4?? p??9?7??999C15C15C154556378190C10C5C10C54C10C5C10C52C10C5C10C5E?9?4?9?5?9?6?9?7?9?8?9?9?9?6。
C15C15C15C15C15C1516记Y为进入购物中心的人数，X为购买冷饮的人数，则
kkpX?X?k???pY?Y?m?Cmp?1?p?m?k??m?k
?m?k???me??m!kkCmp?1?p?m?k
?m?k??m?k?!k!?ke??pkk!k?m?0?me??pk?1?p?m?k
??m!??m?1?p?m
??p? ?e??pk!k[??1?p?]m???1?p? e?m!m?0??p? ?e??p
k!故购买冷饮的顾客人数服从参数为?p的泊松分布，易知期望为?p。
2112C15C15?a?4?Ca4?Ca4?26??a?18 D??2??. 0??1??2??????2?22?C153C3C363?????1515?22220 EX?????xf?x?dx?0，
1?2?xDX?EX??xedx?2
2??1??xEX??xedx?1
2??2DX?EX?EX2??21?2?x??xedx?1?1 2??1 2
22 p?X?0??p?X?1??p?Y?0??p?Y?1??（1）
p?X?Y?1??p?X?0,Y?1??p?X?1,Y?1??p?X?1,Y?0? ?p?X?0?p?Y?1??p?X?1?p?Y?1??p?X?1?p?Y?0? ?（2）EX???1?3 401?Y??0???1?p?X?0?p?Y?0??0???1?p?X?0?p?Y?1??
01 1???1?p?X?1?p?Y?0??1???1?p?X?1?p?Y?1??0
???? ?E?X???1??
DX???1?Y??EX???1?Y2?[EX???1?]2
Y?Y222[0??1]pX?0pY?0?[0??1]p?X?0?p?Y?1?? ???????? ?01 [1???1?]p?X?1?p?Y?0??[1???1?]p?X?1?p?Y?1? ?02121 2?2e?2x,x?0,?4e?4y,24fX?x??? fY?y???0,x?0.??0,?1?e?2x,x?0,?1?e?4y, FY?y???FX?x???x?0.?0,?0,（1）Z?min?X,Y?
y?0,y?0.
y?0, y?0.?1?e?6z,z?0,FZ?z??p?Z?z??1????1?FX?z???1?FY?z??????0,z?0. ?故Z服从参数为??6的指数分布，故EZ?
（2）Z?max?X,Y?
DZ11?1。 ，DZ?。故Cv?Z??EZ636?2z?4z???1?e??1?e?,z?0,FZ?z??p?Z?z??FX?z?FY?z???
0,z?0.???7EZ??zdFZ?z??，
012?49332， DZ?EZ2??EZ???z2dFZ?z???0144144故Cv?Z??DZ33?。 EZ7（3)Z?X?Y，
EZ?EX?EY?Cv?Z??35，DZ?DX?DY?， 416DZ5? EZ3?1fx,ydy??????2，x?1
1112E(X)?0，DX?EX2??EX???x2dx?，
2?13?1同理fY?y???f?x,y?dx?，y?1
??21E(Y)?0，DY?
31111cov?X,Y??E(XY)?E(X)E(Y)???xy??1?xy?dxdy?
?1?14926(1)fX?x???X,Y?cov?X,Y?1?，故X和Y正相关。
DXDY3又f(x,y)?fX?x?fY?y?，故X和Y不独立。 （2）
cov?X,Y22??E(XY22)?E(X)E(Y)?E(XY)?D(X)D(Y)??22221?1?1?1x2y2?11?1?xy?dxdy??049故??0，即X和Y不相关。 又FX2,Y2?x,y??pX?x,Y?y
22?? ?p?x?X? ??x,?y?Y?y
????yyx?xf?t,v?dtdv?xy 所以fX2,Y2?x,y??111???m(x)?n(y)，故X2和Y2相互独立。 4xy4xy28（1）不会写
?1n?1n11（2）cov?X,Xi??cov??xj,Xi???cov?Xj,Xi??D?Xi??
nn?nj?1?nj?1n0?kkn0?k?k?（3）cov?Sk,Tk??cov??Xi,?Xj? ???cov?Xi,Xj?
j?n0?1i?1j?n0?1?i?1?
??kn0n0?ki?1j?n0?1?cov?Xi,Xj??i?n0?1j?n0?1??kkcov?Xi,Xj??i?n0?1j?k?1??cov?X,X?
ijkn0?k?i?n0?1?DXki?k?n0，
DSk??DXi?k，
i?1
DTk?n0?kj?n0?1?DXj?k，
k?n0。 k??cov?Sk,?DSkDTk?30 （1）p?X?0,Y?0??pY?0X?0p?X?0????2， 51p?X?0,Y?1??p?Y?1X?0?p?X?0??，
51p?X?1,Y?0??p?Y?0X?1?p?X?1??，
51p?X?1,Y?1??p?Y?1X?1?p?X?1??，
53232p?X?0??，p?X?1??，p?Y?0??，p?Y?1??，
5555p?X?0,Y?0??p?X?0?p?Y?0?，故X和Y不独立。
（2）cov?X,Y??E(XY)?E(X)E(Y)
?0?p?X?0,Y?0??p?X?0,Y?1??p?X?1,Y?0? ?1?p?X?1,Y?1??p?X?1?p?Y?1??故X和Y正相关。
32 （1）由??0知，X和Y不相关，等价于X和Y相互独立。
??1 25
浙江大学《概率论、数理统计与随机过程》课后习题答案张帼奋主编第四章概率论习题_偶数
