(概率课后习题答案详解)
董永俊
P{X?n}?p(1?p)n?1,n?1,2,3......?E(X)??xipi??np(1?p)n?1?pin?1[1?(1?p)]2?1p
4.6参考课本230页参考答案
4.7解:设途中遇到红灯次数为X,则X~B(3,0.4)
E(X)?np?4?0.3?1.2
4.8解
??E(X)??f(x)xdx
??15002
3000??x2dx???101500150015002(x?3000)xdx ?500+1000 ?1500
4.9参见课本后面230页参考答案 4.10参见课本后面231页参考答案
4.11 解:设均值为?,方差为?2,则X~N(?,?2)根据题意有:
P(X?96)?1?P(X?96)
(概率课后习题答案详解)
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(概率课后习题答案详解)
董永俊
?1?P(X???72??96?)
?1??(t)
?2.3%
?(t)?0.997,解得t=2即?=12
所以成绩在60到84的概率为
P(60?X?84)?P(60-72X-?84-7212???12)
??(1-)?(-1 ) ?2?(1-)1 ?2?0.841-13 ?0.682 64.12E(X2)?0?0.4?12?0.3?22?0.2?32?0.1?2
E(5X2?4)?4?0.4?(5?12?4)?0.3?(5?22?4)?0.2?(5?32?4)?0.1?14
?x4.13解:
E(Y)?E(2X)???02xedx?2??xd(?e?x??x0)?2[?xe?x|?0??0edx]?2(?e?x)|?0?2E(Y)?E(e?2X)????x?0e?2xedx??0e?3xdx??13e?3x|?0?13
4.14解:3V?4?R3
(概率课后习题答案详解)
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(概率课后习题答案详解)
董永俊
设球的直径为X,则:
?1?f(x)??b?a?0?
a?x?其它b
4?(E(V)?E(X23)3)?E(?6X)=?3ba?6x31b?adx??6?1b?a?14x4|ba??24(b?a)(b?a)224.15参看课本后面231页答案 4.16 解: ffyx(x)??????f(x,y)dy??12y012yx2dy?4x
3(y)??????f(x,y)dy??12y?4x01014dx?1245y2?12y
3E(X)??f????x(x)?xdx?dx? y4E(Y)??f????y(x)?ydy??12y3?12dy?35
1E(XY)??????f(x,y)xydxdy???12x0?y?x?15y233dxdy??0?x012xy3dydx?12
0?y?x?1E(X)?22??f(x)?xdx?2?4x0101dx? y5E(Y)?2?????f(y)?y2dy??12y24?12dy?25
E(X?Y2)?E(X)?E(Y)?21615
4.17解
(概率课后习题答案详解)
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(概率课后习题答案详解)
董永俊
∵X与Y相互独立, ∴
E(XY)?E(X)E(Y)??10x2xdx??5ye5?ydy?(5?y23x3|10)??5yd(?e5?y)?23?(?ye5?y|?5???5e5?ydy)?23?[5?(?e)|]?5?23?(5?1)?4
4.18,4.19,4.20参看课本后面231,232页答案 4.21设X表示10颗骰子出现的点数之和,Xi颗骰子出现的点数,则X?10i(i?1,2,?10)表示第
?i?1Xi,且X161,X2,?X10是
独立同分布的,又E(X所以E(X)?E(?Xi?11010i)?1?16?2????6?16?216
)?i?i?1E(Xi)?10?216?35
4.22参看课本后面232页答案 4.23E(X2)?0?0.4?1?0.3?2?0.2?3?0.1?2222
D(X)?E(X)?[E(X)]?2?1?1
222E(Y)?0?0.3?1?0.5?2?0.2?3?0?1.3
2222D(Y)?E(Y)?[E(Y)]?1.3?0.9?0.49222
(概率课后习题答案详解)
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(概率课后习题答案详解)
董永俊
4.24
E(X2)??21421424130x24xdx??2x(?14x?1)dx?16x|0?[?116x?3x]|4?1?1123?143
D(X)?E(X2)?[E(X)]2?1423?4?3
?1?1?x?1?14.25
f?1?xy???14dyX(x)?
=??2
?1?x?1??0其它??0其它
Var(X)?E(X2)?[E(X)]2??1121?12xdx?[?12?12xdx]
?1?1x3|111123?1?2?2x2|1?1?3
?11?xyf)?????14dx?1?y?1?1
=?
?1?y?1Y(y?2
??0其它??0其它Var(Y)?E(Y2)?[E(Y)]2??11212?12ydy?[?1?12ydy]
?1?1y3|111123?1?2?2y2|1?1?3
4.26因为X~N(0,4),Y~U(0,4)所以有Var(X)=4 Var(Y)= 43
故:Var(X+Y)=Var(X)+Var(Y)=4+4163=
3
Var(2X-3Y)=4Var(X)+9Var(Y)=
4?4?9?43?28
4.27参看课本后面232页答案 4.28E(Z)?E(X1?X2???Xnn)?E(X1n)?E(X2n)???E(Xnn)
(概率课后习题答案详解)
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