《概率论与数理统计》第三版，科学出版社 - 课后习题答案

ffyx(x)??f(x,y)dy??12ydy?4x??0??12??x23

3(y)??f(x,y)dy??12ydx?12y?12y??y??2

E(X)??E(Y)????ff??x(x)?xdx??104xdx?344 54????y(x)?ydy??12y?12ydy?013 51E(XY)?f(x,y)xydxdy?0?y?x?10?y?x?1??12xydxdy??5300?x12xydydx?31 2E(X)??f(x)?xdx????2??22??2104xdx?42 35E(Y)??f(y)?ydy??12y?12ydy???012 5E(X2?Y2)?E(X)?E(Y)?2216 15 4.17解

∵X与Y相互独立， ∴

1?21?E(XY)?E(X)E(Y)??x2xdx?ye5?ydy?(x3|)?yd(?e5?y)05305???222??(?ye5?y|??e5?ydy)??[5?(?e5?y)|]??(5?1)?4555 3334.18，4.19，4.20参看课本后面231，232页答案

4.21设X表示10颗骰子出现的点数之和，Xi(i?1,2,?10)表示第

i颗骰子出现的点数，则X??Xi，且X1,X2,?X10是

i?110独立同分布的，又E(Xi)?1?1?2?1???6?1?216666

所以E(X)?E(?Xi)??E(Xi)?10?21?35

i?1i?1101064.22参看课本后面232页答案

4.23E(X2)?0?0.4?12?0.3?22?0.2?32?0.1?2

D(X)?E(X2)?[E(X)]2?2?12?1 E(Y2)?0?0.3?12?0.5?22?0.2?32?0?1.3 D(Y)?E(Y2)?[E(Y)]2?1.3?0.92?0.49

4.24

E(X2)??x202424111111114xdx??x2(?x?1)dx?x4|?[?x4?x3]|?1?? 224416016333D(X)?E(X2)?[E(X)]2??11?xydy?fX(x)????14

??022142?4? 33?1?x?1?1?=?2 ?其它?014.25

?1?x?1其它

1112Var(X)?E(X)?[E(X)]??xdx?[?xdx]2 ?12?121111111??x3|??x2|? 23?122?13

?1?y?1?1?11?xydx?? fY(y)????14=?2

??其它?0?0221

?1?y?1其它

1112Var(Y)?E(Y)?[E(Y)]??ydy?[?ydy]2 ?12?121111111??y3|??y2|? 23?122?134.26因为X~N(0,4),Y~U(0,4)所以有Var(X)=4 Var(Y)= 故：Var(X+Y)=Var(X)+Var(Y)=4+4=16

334 3Var(2X-3Y)=4Var(X)+9Var(Y)=

4?4?9?4?28 34.27参看课本后面232页答案 4.28E(Z)?E(?X1?X2???XnXXX)?E(1)?E(2)???E(n)

nnnn1111E(X1)?E(X2)???E(Xn)???n?? nnnnX1?X2???XnXXX)?D(1)?D(2)???D(n)

nnnnD(Z)?D(11112?2?2E(X1)?2E(X2)???2E(Xn)?2??n?nnnnn

后面4题不作详解

第五章 极限理

5.3

解：用Xi表示每包大米的重量，，则E(Xi)???10,D(Xi)??2?0.1

?Xi?1100i~N(n?,n?2)?N(100?10,100?0.1)

100100

Z??Xi?1100i?n?2n???Xi?1i?100?10??Xi?1i?100010~N(0,1)

100?0.1990?1000P(990??Xi?1010)?P(?10i?1100?Xi?1100i?100010?1010?1000) 101010?10001010?1000??()??(?)??(10)??(?10)?2?(10)?1?0.9986

10105.4解：因为Vi 服从区间[0,10]上的均匀分布，

0?10E(Vi)??5

22020

20102100D(Vi)??

1212100) 12?Vi?1i~N[?E(Vi),?D(Vi)]?N(20?5,20?i?1i?1Z??V??E(V)?V?20?5?V?100iiiii?1i?120202020?D(Vi)i?120?i?120?10012?i?110153~N(0,1)

P(V?105)?1?P(V?105)?1?P(?Vi?105)?1?P(i?120?V?100ii?12010153?105?100) 10153105?100?1??()?1??(0.387)?0.348

10153