概率论与数理统计(龙永红)
更新时间:2024-01-21 17:17:01 阅读量: 教育文库 文档下载
第一章 1. (1) ?1?{(1,1),(1,2)(1,3)...(6,6)}
(2) ?2?{x|x1?x?x2} x1:当日最低价 x2:当日最高价 (3) ?3?{0,1,2,3,} (4) ?3?{1,2,3,?} 2. (1) (3) 3. ??{1,2,3,4,5,6} A?{1,3,5,} B?{1,2,3,4,} C?{2,4,} A?B?{1,2,3,4,5} A?B?{5} B?A?{2,4,} AB?{1,3} AC?? A?B?{1,2,3,4,6}
4. (5) ABC?ABC?ABC?ABC?ABC?ABC?ABC (8) ABC?ABC?ABC?ABC (10) AB?BC?AB (11) A?B?C
9. ①?P(A?B)?P(A?AB)?P(A)?P(AB)?0.25 又?P(A)?0.4 ?P(AB)?0.15
②?P(A?B)?P(A)?P(B)?P(AB) ?0.4?0.25?0.15 ?0.5
1
③?P(B?A)?P(B?AB) ?P(B)?P(AB) ?0.25?0.15 ?0.1
④P(AB)?P(A?B)?1?P(A?B) ?1?0.5 ?0.5
10. ?P(A?B?C)?1?P(ABC) 而 P(A)?1?P(A)?1?0.4?0.6 又P(A)?P(AB?AB) ?P(AB)?P(AB) ?P(AB)?P(A)?P(AB)?0.4 又 AB?ABC?ABC
?P(AB)?P(ABC)?P(ABC) ?P(ABC)?0.4?0.1?0.3 ?P(A?B?C)?0.7 11. A=“其中恰有K件” CKn?①?P(A)?1CkNN?N1Cn
N② B=“其中有次品” B?“一件次品也没有” ?P(B)?1?P(B)?Cn1?N?N1Cn
N③C=“其中至少有两件次品” C?“只有一件次品,或没有” ?P(C)?1?P(C)?1Cn?N?N1C1N1Cn?1N?N1Cn?NCnN.①: A=“男生比女生先到校”
2
12
P(A)?24!?6!24!1?24?24 30!P30C30 ②B=“李明比王先到学校” P(B)?
13. C=“至少两人生日同一天” C?“每个人生各不同” P(C)?1?P(C)?1?365?364??(365?n?1)
365n1214. ①A=“第2站停车”
A?“不停车”
?P(A)?1?P(A)?1?()25
②B=“第i和第J站至少有一站停车 B?“第i站到J站都不停” ?P(B)?1?P(B)
7?1?()25
989 ③Ai?“第i站有人下车(停车)” Aj?“第j站有人下车” P(Ai?Aj)?1?(Ai?Aj)?1?P(Ai?Aj)
?1?[P(Ai)?P(Aj)?P(AiAj)]
?1?P(Ai)?P(Aj)?P(AiAj)
87?1?()25?2?()25
99④D=“在第i站有3人下车”
3?()3?()22 (贝努里试验) P(D)?C25198915.(1)A=“前两个邮筒没有信” P(A)?2?21? 442(2)B=“第一个邮筒恰有一封信”
3
1C2?33P(B)?2?
8416. A=“前i次中恰好有取到k封信”
ki?kCa?Cb?i!(a?b?i)! P(A)?
(a?b)!ki?kCaCb ?i
Ca?b17. A3?“第三把钥匙可以开门” A2?“第二把钥匙可以开门”
① P(A3)?P(A1A2A3?A1A2A3?A1A2A3?A1A2A3)
?P(A1A2A3)?P(A1A2A3)?P(A1A2A3)?P(A1A2A3)
?432654463643??????????? 109810981098109824?120?144?
720288? 7204? 10 ② A3?“第三把钥匙才可以开门” P(A3)?6541201???? 10987206 ③ C=“最多试3把就可以开门”
464654????? 1010910985 ?
6 P(C)?18. 贝努里试验
A=“其中三次是正面”
11133P(A)?C10?()3?()7?C10?()10
22219.A=“恰有一红球,一白球,一黑球”
111C5?C3?C21 P(A)??34C101C2?3?2?2?24820. P(A)? ?13!13! 4
21. 几何概型
A=“等待时间不超过3分钟” X???到达汽车站的时间
??{xt?x?t?10} A?{xt?7?x?t?10}
?P(A)?S(A)3? S(?)1022. A=“需要等零出码头的概率”
x???第1条船到达时刻 y???第2条船到达时刻
??{(x,y)0?x?24 0?y?24? A?{(x,y)0?x?y?2 0?y?x?1?
1242?(222?232)S(A)2?P(A)?? S(?)24223. A=“第一次取出的是黑球”
B=“第二次取出的是黑球”
a?(a?1)P(AB)(a?b)?(a?b?1)a?1?? (1) P(BA)?
aP(A)a?b?1a?baa?1?P(AB)a?1a?ba?b?1 (2)P(AB)? ??aa?1baP(B)a?b?1???a?ba?b?1a?ba?b?1 (3)A=“取出两个球,有一个是黑球” B=“两个都是黑球” nA?a(a?b?1)?b?a?a?(a?2b?1) nB?a?(a?1)
P(BA)?nBa(a?1)a?1 ??nA[a(a?2b?1)]a?2b?1P(AB) P(A)5
24. (1)P(BA)? ?B?A
?AB=A ?P(BA)=P(AB)P(A)==1 P(A)P(A)P[A(B1?B2)]P(AB1?AB2) =P(A)P(A)(2)P(B1?B2A)= ?B1B2?? ?? ?P(AB1)?P(AB2)
P(A)P(AB1)P(AB2) ?P(A)P(A)?P(B1A)?P(B2A)
25. (1) ??{(男,男),(男,女)(女,男)(女,女)}
A=“已知一个是女孩,”=(男,女)(女,男){(女,女)} C=“两上都是女孩”= (女,女){} P(CA)=
(2)解略 P(A1A2)= Ai?“第i个是女孩” 26. A=“点数为4” P(A)?2?51? 6?63131227. A=“甲抽难签” B=“乙抽难签” C=“丙抽难签” ① P(A)?4 10 ② P(AB)?P(A)?P(BA) ?64? 10924? 904? 15 ③ P(ABC)?P(A)?PC(BA)?P(CAB)
432?? 109824 ?
720 ? 6
28. A=“试验成功,取到红球”
B0?“从第二个盒子中取到红球”
B1?“从第三个盒子中取到红球”
P(A)?P(AB0?AB1)
?P(AB0)?P(AB1)
?P(B0)?P(AB0)?P(B1)?P(AB1)
1738??? 210101059? 100?0.59 ?29. A=“废品” B1?“甲箱废品” B2?“乙箱废品” (1)P(A)?P(AB1?AB2)
?P(B1)?P(AB1)?P(B2)?P(AB2)
320?0.06??0.05 5050 ?0.056
3000?0.06?2400?0.05 (2)P(A)?
30?100?20?120180?120 ?
54001? 18 ?30. Bi?“第二次取球中有i个新球” i=0.1,2,3 Aj?“第一次取球中有j个新球” j=0,1,2,3 (1) P(B2)?P(B2A0?B2A1?B2A2?B2A3)
?P(A0)?P(B2A0)?P(A1)?P(B2A1)?P(A2)?P(B2A2) ?P(A3)?P(B2A3)
3?JC9JC3 P(Aj)? J?0,1,2,3 ① 3C121C92?JC3?J P(B2Aj)? J?0,1,2,3 ② 3C12 7
分别对应代入该式中,可得:
P(B2)?0.455
P(A1B2)P(A1)?P(B2A1) (2)P(A1B2)? ?P(B2)P(B2)将①,②代入该式,可得:
P(A1B2)?0.14
31、 A=“确实患有艾滋病” B=“检测结果呈阳性”
由题知:P(BA)?0.95 P(BA)?0.01 P(A)?0.001
P(A)?P(BA)P(AB)① P(AB)? ?P(B)P(A)?P(BA)?P(A)?P(BA)0.001?0.95
0.001?0.95?0.999?0.01 ?0.087
?② C=“高感染群体确实患有艾滋病” P(C)?0.01 P(CB)?P(C)?P(BC)P(BC) ?P(B)P(C)?P(BC)?P(C)?P(BC)0.01?0.95
0.01?0.95?0.99?0.01 ?0.49
?32. 解:不能说明“袭击者确为白人的概率”为0.8 设 A=“被袭击者正确识别袭击者种族”
A?“错误识别袭击者种族”
B=“袭击者为白人” B?“袭击者为非白人” 根据已知条件,有
P(A)?0.8 P(A)?0.2 P(B)?P(BA?BA) ?P(AB)?P(AB)
8
?P(A)?P(BA)?P(A)?P(BA) ?0.8?P(BA)?0.2?P(BA)
因 P(BA) 与 P(BA) 未给出,因而不能断定
P(B)?0.8
33. 解:P(A)?P(B)?P(C)? P(AB)?P(BC)?P(AC)? ?A,B,C两两独立, 又P(ABC)?11?P(A)P(B)P(C)? 481214 ?A,B,C不相互独立,只是两两独立。
34. ① P(A)?0 ?B?? 有 P(AB)?0?P(A)P(B) ?A,B独立 ②P(A)?1 ?B?? 有 P(A)?0 ?A与B独立 ?A,B独立
=P(B)-P(AB) ?P(AB)?P(B)?P(AB)?P(AB) =P(B)-P(A )P(B) =P(B)( 1-P(A))?P(A) =P(B)
35. P(A)>0 且 P(B)>0 且 A,B互不相容
则 A,B不可能相互独立
因为P(AB)?(?)?0 但因为 P(A)>0 P(B)>0
?P(AB)?0?P(A)P(B) ?不独立
B,C亦相互独立 36. A,B,C相互独立,证明 A,证:P(ABC)=P(A)P(B)P(C)且P(AB)=P(A)P(B)
P(BC)=P(B)P(C),P(AC)=P(A)P(C)
)?1?P(A?B)则P(AB)=P(A?B
?P(B)?P(AB) ?1?P(A)
?1?P(A)?P(B)?P(A)P(B) ?[1?P(A)][1?P(B)]
9
?P(A)P(B)
同理可证 P(AC)?P(A)P(C) P(BC)?P(B)P(C) 下证 P(ABC)?P(A?B?C)?1?P(A?B?C)
?1?P(A)?P(B)?P(C)?P(AB)?P(AC)?P(BC)?P(ABC)
?1?P(A)?P(B)?P(C)?P(A)P(B)?P(A)P(C) ?P(B)P(C)?P(A)P(B)P(C) ?[1?P(A)][1?P(B)][1?P(C)
?P(A)P(B)P(C)
?A,B,C相互独立
37. 证略,可用数学归纳法 38.
A=“第一道工序出品”
B=“第二道工序出废品” C=“第三道工序出废品”
P(A?B?C)?1?P(ABC)
?1?P(A)?P(B)?P(C)
?1?0.9?0.95?0.8
?0.316
39. A=“雷达失灵” B=“计算机失灵” P(AB)?P(A)?P(B) (因为独立)
?0.9?0.7 ?0.63
40. B=“击落” A,B,C分别代表三收炮弹 Ai?i发炮弹击中敌机 i?1,2,3 P(A1)?P(ABC)?P(ABC)?P(ABC)
?0.3?0.7?0.7?0.7?0.7?0.3?0.7?0.7?0.3 ?0.441
P(A2)?P(ABC)?P(ABC)?P(ABC)
?0.3?0.3?0.7?0.3?0.3?0.7?0.3?0.3?0.7 ?0.189
P(A3)?P(ABC)?0.27
10
P(BA1)?0.2 P(BA2)?0.6 P(BA3)?1
?P(B)?0.441?0.2?0.189?0.6?0.027?1
?0.228 6 P(A2B)?P(A2B)
P(B) ? ?P(A2)?P(BA2)P(B)
0.189?0.6
0.2286 ?0.496
习题二(A)
1.解:X: 甲投掷一次后的赌本。
Y:乙???
x20 40 Y10 30 11p1p1 2222x?20?0,?1x~FX(x)??,20?x?40?2x?40?1,x?10?0, Y~F(y)??1,10?x?30?Y?2x?30?1,
2.解
(1)
11
?p(x?i)?1??a2i?1i?1100100i?1?a?2i?1?a?i?11001
i?2i?1i100(2)
?p(x?i)?1??2ai?1i?1100??112?2?ai?1??ai?i?1i?1??
?111?1??a?1?a23 3.解
X?5 ?2 0 2 1111p 510524.解
C17(1)X:有放回情形下的抽取次数。P(取到正品)=17?
C1010 P(取到次品)=
3 10X1 2 3 ? i? 3273i-17p737 ,? () ? ()? 10101010101010
(2)Y:无放回情形下。
Yp 1 2 3 4 7373273217 ? ? ? ??? 10109109810987
5.解
P(X??3)?1?p(X??3)?1?p(X??5)?1?14? 554 5P(X?3)?p(?3?X?3)?p(X??2)?P(X?0)?P(X?2)? 12
P(X?1?2)?p(X?1?2)?P(X?1??2)?p(X?1)?p(X??3)?p(X?2)?p(X??5)?710
6.解
(1)根据分布函数的性质
limx?1?2F(x)?F(1)?lim?1?A?1 ?Axx?122(2)P(0.5?X?0.8)?F(0.8)?F(0.5)?0.8?0.5=0.39 7.解:依据分布满足的性质进行判断: (1)???x???
单调性:x1?x2?F(x1)?F(x2).在0?x???时不满足。 (2)0?x???,不满足单调性。
?11?,???x?0(3)???x?0,满足单调性,定义F(x)??1?x2是可以做分布函数的.所以,F(x)?能21?x?0?x???0,做分布函数。
8 解
(1) F(x)在x=0,x=1处连续,所以X是连续型。
?2x,0?x?1 f(x)?F'(x)??0,其他?(2) F(x)在x=0处连续,但在X=1处间断,所以X不是连续型。
9解: (1) ⅰ)求a,由
?????f(x)dx?1??ae?????x0???xdx?1??0?2a?edx?1??2a?e?xd(?x)?1 ?a?12x1?x???2edx, x1?x1?x??1xF(x)?edx??e?x?e , 当x<0, ???22201?xx11edx??e?xdx?1?e?x 当x≥0, F(x)????2022?1xx?0?2e,所以F(x)?? ,
1?x?1?e,x?0?2ⅱ)F(x)?P(X?x)?ⅲ)
13
P(?1?X?22)?F()?F(?1) 221?21?11?2?1?e?e?1?(e?e?1)
22211p(X?1)?F(??)?F(1)?1?(1?e?1)?e?1
222221P(?X?2)?1?e?22(2) ⅰ)求a:
21?21?2?(1?e)?(e?e2)
2222?????f(x)dx?1
??0??0dx??xdx??(2?x)dx??0dx?1
01a1a??112 ?x21?(2x?x)?1 0221 ?a?4a?4?0 ?a?2 ⅱ)F(x)?P(X?x)?X<0,F(X)=0.
0≤X<1,F(x)??xdx??xdx???0xxa2
?x??f(x)dx
12x 21131?2x?x2??2x?x2?1 22221≤x<2, ,F(x)?X≥2,F(x)=1.
?10xdx??(2?x)dx?1x0,x?0??12x,0?x?1??2所以:F(x)?? , 12?2x?x?1,1?x?22??1,x?2?ⅲ)
P(?1?X?2121)??()2?0? , 2224P(2119?x?2)?2?2??(2)2?1??22? , 2244121?(1)? , 22P(X>1) ?1? 10.
?????f(x)dx?1,因f(x)关于x=u对称?f(u?x)?f(u?x)
14
?f(2u?x)?f(u?(u?x))?f(u?(u?x))?f(x) ① 下面证明
?u?x??f(y)dy??u?x??u?xf(z)dz,②
u?x令z+y =2u?y=2u-z
?=
u?x??f(y)dy????f(2u?z)d(2u?z)?????u?x??f(2u?z)dz
???u?xf(2u?z)dz??f(z)dz????u?xf(z)dz (由①式有f(2u-z)=f(z))
????又
?u?x??f(z)dz??f(z)dz?1,由于②式
??f(z)dz??f(y)dy?1
????u?xu?x?F(u?x)?F(u?x)?1
11.解
(1)第2题(2):
?1i1EX??i?2a??i?2()?2?i?2()i
33i?1i?1i?1i??(2)第3题:由分布律得: EX??5?11111?(?2)??0??2??? 51052512.解:ER=1%×0.1+2%×0.1+┅+6%×0.1=3.7%,
若投资额为10万元,则预期收入为 10×(1+3.7%)=10.37(万元)
22-42-4-4
DR=ER-(ER)=15.7×10-(3.7)×10=2.01×10 2222222
ER=(1%)×0.1+(2%)×0.1+(3%)×0.2+(4%)×0.3+(5%)×0.2+(6%)×0.1 -5-5-5-5-5-5
=10+4×10+18×10+48×10+50×10+36×10
-4
=15.7×10
13.解:题意不清晰,条件不足,未给出分期期类.
解一.设现在拥用Y,收益率k%, 假设现在至1100时仅一期,则 K%=
1100?Y11001100 ??1?Y?KYY1?10011001100FY(x)?p(Y?x)?p(?x)?p(k?(?1)?100
kx1?1001110022000 dx?1?20(?1)?21?1100?(?1)?1005xxx5220001100dFY(x)??2,0?(?1)?100?5fY(x)???x xdx?其他?0,?220001100?,?x?1100??x2 1.05?其他?0,
15
EY??11001.0511001100y?fy(y)dy?22000lny1100?22000?ln1.05?1073.4元
1.05解二,由于0≤x≤5题意是否为五期呢?由贴现公式
1100?Y1100?Y? Y1?5K001100?x)?p(k?20(?1)) P(Y≤X)= P(1?5K%x5K%=
??14400?1100?1 dx?1?20(?1)??5?1100??20(?1)5x5x??x51100?4400?44001100?20?20(?1)?5?2,?x?1100fY(x)??x=?xx1.25??其他其他?0?0,EY??110011001.25xfY(x)dx?4400?lnx11001100/1。25?4400?ln1.25?981.2
14.证明:E(X-EX)2
??(x?EX)f(x)dx??(x2?2xEX?(EX)2)f(x)dx
??????2?? ??????x2f(x)dx?2EX?xf(x)dx?(EX)2?f(x)dx
???????? ?EX2?2EX?EX?(EX)2 ?EX2?(EX)2 15.证明:(2.31)
D(ax)?E?ax?E(ax)?
2 ?E?a(x?Ex)??Ea2(x?Ex)2?a2E(x?Ex)2?a2DX
2(2.32)
L (C)=E(X-C)2=E
16.①连续型。 普照物
p50 -Th2.3证明过程
2令h(x)??x?Ex?
则
Eh(x)??h(x)?f(x)dx??????h(x)??2h(x)f(x)dx????22h(x)??2h(x)f(x)dx??h(x)??2h(x)f(x)dx
??2?h(x)??2f(x)dx??.p(h(x)??)22于是有ph(x)?????Eh(x) (*)
?2将h(X)=(X-EX)代入(*)得px?Ex??2
???E(x?EX)2?216
?Dx?2(证毕).
②离散型。
Eh(x)????2i????h(x)?p(x)??h(x)p(x)??h(x)?p(xi)??h(x)?p(x)iiiiiiih(xi)??2h(xi)??2h(xi)??2??
h(xi)??2?p(x)??i2p(h(xi)??)DX2于是ph(xi)??2???Eh(xi)?2同理将h (x)=(x-EX)代入得p(x?Ex??)?2
?2
17.解:设P表示能出厂。P=0.7+0.3×0.8=0.94
q表示不能出厂。Q=0.3×0.2=0.06
(1)X~b(n,0.94) X:能出厂数
KP(X=K)=Cn(0.94)K(0.06)n?K
n(2) P(X=n)=Cn(0.94)n(0.06)n?n=(0.94)
n
(3)Y~b(n,0.06) Y:不能出厂数。
11-P(Y=0)-P(Y=1)=1-Cn(0.06)1(0.94)n?1?Cn(0.06)0(0.94)n
0(4)EY=n×0.06,DY=n×0.06×0.94 18.解
X 0 1 2 ? n? pPg0 pq1 pq2 ?pqn? ?qqqnEX??n?pq?p?nq?p??.(其中注意应用:nq?) ?22p(1?q)(1?q)n?0n?0n?0nn??19.解:已知X~P(?) P(X?1)??1e??1!?2p(X?2)?2??2e??2!????2???1
EX=DX=?=1
EX2=(EX)2+DX=?2+?
P(x?3)??3e3!??e?11??.(其中??1) 66e2220.解:P:等车时间不超过2min的概率,X:等车时间
?2)? P?P(X q?1?p????f(x)dx??012dx? 553 52再会Y:等车时间不超过2分钟的人数
P(Y?2)?P(Y?3)?C3()()?C3()?21.解:设Y:利润
X:理赔保单如:X~b(8000,0.01)
17
25235132534444??0.35 2125125 Y=500×8000-40000X 由EX=np=8000×0.01=80
EY=4000000-40000×80=800000 22.解
??e??x,x?0(1)X~f(x)??
0,x?0? F(x)??x??f(x)dx???e??xdx???e??xd(??x)??e??x00xxx0??(e??x?1)?1?e??x
?1?e??x,x?0所以:F(x)??
x?0?0,
EX,DX推导见原习题解。 23.证明
?1?e??x,x?0 X~e (?)?F(x)??
x?0?0,P(X?r?sX?s)?p(X?r?s)1?p(X?r?s) ?p(X?s)1?p(X?s)1?(1?e??(r?s))e??r??s??r ???e??s??s1?(1?e)e=1-(1-e??r)?1?p(X?r)?p(X?r)
24.解:设X:表示元件寿命,X~e(1) 1000 Y:1000h不损坏的个数,当Y为2以上时系统寿命超过1000h,Y~b(3,p) P:1000h不损坏的概率。
P?P(X?1000)?1?p(X?1000)?eq?1?p?1?e?1,
多元件独立工作
22133p(Y?2)?P(Y?3)?C3Pq?C3p
1??10001000?e?1
?12?13?13 ?C23(e)(1?e)?C3(e)
?3e?2(1?e?1)?e?3 ?3e25.解:X~N(?,?2) EX?
?2?2e?3
?????(x??)2x??e?dx,(令y?) 2?2?2??X18
????12???(???y)e1?y22dy
? ?2?26.解
?1???y22??edy?u
p?p(x?19.6)?p(x??19.6)?p(x?19.6)
?2?1??0(??19.6?0?)?0.05 10?? n=100
Y:误差绝对值大于19.6的次数 Y~b(100,0.05)
a=P(Y≥3)=1-P(Y=0)-P(Y=1)-P(Y=2)
用泊松分布近似计算:?=np?100?0.05?5 a=1-P(Y=0)-P(Y=1)-P(Y=2)
50e?551e?552e?5?1-???1?18.5e?5 0!1!2!27.解:
设C:损坏,
则由题意:p(cX?200)?0.1 p(c200?X?240)?0.01
p(cX?240)?0.2
p(X?200)??0(200?220)??0(?0.8)?0.2119 25p(200?X?240)?2?0(0.8)?1?0.5762
p(X?240)?p(X?200)?0.2119
所以:P(C)=0.2119×0.1+0.5762×0.01+0.2119×0.2=0.06931 而由贝叶斯定理有:
p(200?X?240c)? ?p(c,200?x?240)
p(c)0.5762?0.01?0.083
0.0693128.解:设数学成绩为:X,X~N(70,100),由题意:
P(X?a)?5%
即1?p(X?a)?5%?p(X?a)?0.95
19
a?70)?0.95 10a?70 =1.645
10 ?0( a=70+10×1.645=86.45分 29.
X20? 22? 24? 26? p0.1 0.4 0.3 0.2
Y100? 121? 144? 169? p0.1 0.4 0.3 0.2
30.解: 令Y=?X+β
FY(Y)?P(Y?y)?p(?x???y)?p(X?'fY(y)?FY'(y)?FX(y???)?Fx(y??) xy???)?fx(y???)?1?
1y???1??,a??b ??b?a? ??0,其他?1??,ax???y?bx??即 =?b??a?
?0,其它?也即Y在[a?+β,b?+β]上服从均匀分布。
?1?,?1?x?131.解: 令Y=X2, X~fx(x)??2
?其他?0,FY(y)?p(Y?y)?p(X2?y)?p(?y?X?y)?Fx(y)?Fx(?y)
‘ fY(y)?FY(y)?Fx(y)?Fx(?y)?fx(y)’‘12y?fx(?y)??12y
?1?1,〈0y?1,0?y?1?? 即:fY(y)??4y ??4y?0,?0,其他其他??FY(y)??y?yfX(x)dx??y0y11 dx?x0y?222 20
???即: FY(y)?????0,y?0y,0?y?1 21,y?1?e?x,x?032.解: x~e(1)~fX(x)??
?0,x?0 Y=ax+β
FY(y)?p(Y?y)?p(?X???y)?p(X?'fY(y)?Fy(y)?FX('y???
)?Fx(y???)
y???)?fX(y???)?1?
??1???y???e,=???0,?(y??)y???0??1e?,=????其他0,?1y?? 其他FY(y)?Fx(y???)?1?ey???y???
1?(y??)???,??1?e?0,?1?(y??)??0=?1?e?,???0,其他?y??
其他
?1?33.解: 令X:直径 f(x)~?b?a??0 Y:体积 Y?a?x?b其他
4x?()3 32baEY??g(x)f(x)dx??????ba4x31411?()?dx????x432b?a3b?a32?11??(b4?a4)24(b?a)
?1(b?a)(b2?a2) 2434.解:
FY(y)?p(Y?y)?p(?y?x2?X?y)?p(X??2?y)??20y1??e?dx
?x ???20e?d(?)??ex??x?y?20
yy???2?2???e?1??1?e . (y?0)
??所以:
21
??y F?2Y(y)??1?e,y?0
??0,y?0所以: dFyY(y)1dy?fy)?2e?Y(2 (y?0)
所以:
?yg(y)??1??fY(y)?22e,y?0 所以:Y~?2(2)
??0,y?0
.35.解 X~e(2)
所以: f?2e?2x,x?0?1?e?2x,x?0X(x)?? F?0,x?0X(x)??
?0,x?0FXY(y)?p(Y?y)?p(1?e?2?y)?p(e?2X?1?y)?p(?2X?ln(1?y)) ?p??X??1?1?2ln(1?y)???FX(?2ln(1?y))
??2(?1?2ln(1?y)??1?e,0?y?1??1,y?1 ??1?(1?y),0?y?1?y,0?y?11,y?1 ???1,y?1 ???0,y?0?0,y?0?0,y?0????y,0?y?1F(y)???1,y?1 f?1,0?y?1Y?Y(y)???0,y?0?0,其他
?1?12?(lnx??)36.解:由已知参考(p)知f????(lnx),x?0?2?266~67X(x)e,?x ???0,x?0?2??x?0,当前价格X0?10元。 (lnx?u)2 EX????2?20x?fX(x)dx????102??e?dx
依据 p67-eg2.31 EX?e???22?15?e2???2?152
??e?2?1?4?e?2229225?225
22
x?0x?0
DX?e2???(e??1)?4
229?2??ln??225??225???ln?229?2(lnx?u)?1??ln229?ln225?2?2?,x?0 其中 fX(x)??2??xe1??ln225?ln229?0,x?02?222
连续复合年收益率 r=lnx-lnx0=lnx-ln10
FR(r)?P(R?y)?p(lnX?ln10?y)?p(lnX?y?ln10)?p(X?10?ey)
??10?ey12??x0e?(lnx??)22?2dx lnx??)2?lnx??2?????所以:
10?ey10?2?e?(lnx??)22?2d(2) 令?v
ln10?y????22?e?vdv?2??0(ln10?y??2?)
FR(y)?2??0(y?ln10??2?)
fr(y)?Fr'(y)????
注:对数正态分布与对数正态分布的矩,包括中心矩,原点矩等,如EX,DX均不作要求属于超纲内容,Black-Scholes
期权定价公式一般是作为研究经济现象工具也属于超纲内容,因而本题不作要求。 37.证明:
FY(y)?P(Y?y)?p(X?y)?p(X?y2) 显然当y≤0时,P(X?y)?0,所以 fY(y)?0 y>0时,FY(y)?p(X?y)?y22?y20f(x)dx
dFY(y)d?0f(x)dxfY(y)???2y?f(y2)(复合函数求导方法)
dydy所以:
?2y?f(y2), fY(y)??0,?y?0 y?038.解:X密度函数fX(x), Y=ax+b
23
y?b FY(y)?p(Y?y)?p(aX?b?y)?p(X?)??afX(x)dx , a?0
??adFY(y)d?fY(y)??dy固当a>0时, fY(y)?y?ba??y?bfX(x)dxdy?1y?b?fX() aa
1y?b?fX() aa1y?b) 当a<0时,fY(y)???fX(aa
习题三答案
1. 证明:F?x2,y2??F?x1,y2??F?x2,y1??F?x1,y1? ?P?x1<x?x2,y?y2??P?x1<x?x2,y?y1?
?P?x1<x?x2,y1<y?y2??0由概率的非负性,知上式大于等于零,故得正. 2. 解:①, x2 x1 0 1 2 0 1/56 10/56 10/56 1 5/56 20/56 10/56 PX2?x2j 3/28 15/28 5/14 ??P?X1?x1i? 3/8 5/8 1 ②. P?X1?0,X2?0???1?P{x2?0|x1?0}?P?x1?0?
?1??1??C27??3203205????821?8?56?14 ?P?X1?X2??P?x1?0,x2?0??P?x1?1,x2?1??21 56P?x1x2?0??1?P?x1x2?0??1?P?x1?1,x2?1??P?x1?1,x2?2?
?1?5102613??? 145656283. 解:①由概率的性质
??又
??0??0?x0?????3x?4yf?x,y?dxdy??0kedxdy?1?k1?12 01????x?3x?4yg?x,y?dxdy??0?0k2edxdy?1?k2?21
????0② fX?x?? fY?y???????3x?4y01kedy?k1e?3x????3x?4y01kedx?k1e?4y1?3e?3x x>0 41??4e?4y y>0 324
当y>0时 gX?x??当x>0时 gY?y????x0k2e?3x?4ydx?21?3xe?e?7x x>0 4????yk2e?3x?4ydx?7e?7y y>0
4. 解:①
?1?,0?x?2,0?y?1f?x,y???2
?0,其他?
?xy?,0?x?2,0?y?1xyF?x,y???01/2dxdy??2 ?0?0,其他?
②
fX?x???101121dy? 0?x?2;fY?y???0d?1,0?y?1
222x0?x?2?1?y?xyFX?x???0dx??2 ;FY?y???01dx??
2其他0??0? ③
0?y?1
其它
Py<x?2???12111?y<x22dxdy?1??0?x22dxdy?3
5.
?2f?x,y???
?00?x?y?1
其它0?x?1其它其它
fX?x???x1?2??1?x?2dy??
?0?2y2dx??
?0
fY?y???y0?y?1
00,其它??x?y?1?,0?x?2,1?y?2?0?2其它?1???fx,y?x? 6.①F?x,y??? ,0?x?2,y>2??2?20?x?2,1?y?2?y?1,x>2,1?y?2?1,x?2,y>2?②
G?x,y??PX2?x,Y2?y?P?x<X?x,?y<Y?y?(xy?x)/2,0?x?4,1?y?4?0,x?0,或y?1??x G(x,y)??,0?x?4,y?42?y?1,x>4,1?y?4??1,x>4,y?4?????
讨论如下:
25
③
13??21P?x<1,y>?=??dxdy??10?3/2dxdy?1/4
22?D2?7. 证明:P?x1<X?x2?P?y1<Y?y2???Fx?x2??Fx?x1??Fy?y2??Fy?y1? ?Fx,y?x2,y2??Fx,y?x2,y1??Fx,y?x1,y2??Fy?x1,y1? ?P?x1<X?x2,y1<Y?y2? 故独立得证. 8. ??x1|x2?1 0 1 31 2 3P x9. 解:① P ②Py?yj?ij?PiPji???i?1pipji ③Pij?xpipjix?ipipjix
111P?P??0 P 1322124421??p?y??1?p?x?0? 故不独立. ② P11410. 解:① P11?11. x y y1 y2 y3 pix x1 1/24 1/8 1/8 3/8 1/2 1/12 1/4 1/3 p1,p2,?,pn1/4 3/4 1 x2 pjy1/6 p11,p12,?,p1np21,p22,?,p2n独立p1,p2,?,pn?pp,?p12m12. 证明:必要性: 由:
??????pm1,pm2,?,pmn?秩为1
p1,p2,?,pnp11,p12,?,p1np11,p12,?,p1np21,p22,?,p2n秩为1p21,p22,?,p2n?充分性,若
??????pm1,pm2,?,pmnpm1,pm2,?,pnm 26
?p?x?x2y?y1??p?x?x2y?y2??p?x?x2y?y3???p?x?x2y?yn??p?x?xy?y??p?x?xy?y??p?x?xy?y???p?x?xy?y??1112131n??
.............??p?x?xmy?y1??p?x?xmy?y2???p?x?xmy?yn?? 从上式可得x与y独立. 13. 解:① X2?Bx?c?0,
有实根的概率 B2?4c?0,PB2?4C?0
???P?2,1??P?3,1??P?3,2??P?4,4??P?4,2??P?4,1??P?4,3?
?P?5,1??P?5,2??P?5,3??P?5,4??P?5,5??P?5,6??P?6,1??P?6,2??P?6,3??P?6,4??P?6,5???P?6,6?
=
19 36② PB?4C?0?P?2,1??P?4,4??2??1 1814. 解:fx1y1?x,y??f?x,y? fy1?y?x>0 其它k1e?3x?4y?3e?3x?? 当 y?0时, fx1y1?x,y??k1?4y?0e3 当 y<0时, fx1y1?x|y??0x????,???
?4e?4y,y?0 当 x?0时, fy1|x1(y|x)=f1(x,y)/fX1(x)??
?0,y?0 当 x<0时 fy1x1?xy??0??<y<??
f2?x,y??3e?3x?3y??② 当 y?0时, fx2y2?x,y??fy2?y??0Y<0时,fx2x?y x?yy2(x,y)?0 x?(??,??)
?4e?4y0?y?x?f(x,y)?4x??当X≥0时,fy2x2(yx)?2 (1?e)fx2z(x)?0其它?当X<0时,
fy2x2(yx)=0,y?(??,??)
215.解 1.由S(D)=4?1/2?7
27
得X与Y的联合密度函数为f(x,y)???27(x,y)?D
其它?0?2/70?x?y?12. 由于0≤y≤1时,f(x,y)??
0其它?从而 0≤y≤1时,fy(y)??????f(x,y)dx??y?102dx?2(y?1)
77
??27又 1≤y≤3时,f(x,y)????0从而 1≤y≤3时,fy(y)?y?1?x?2其它2y?1?????f(x,y)dx??2/7dx?2(3?y); 7又当Y<0时,Y>3时,f(x,y)=0,从而fY(y)=?????f(x,y)dx?0
2y?1)/70?y?1?(?综上得: fy(y)??(23?y)/71?y?3
?0其它???270?y?x?1此外,0≤X≤1时,f(x,y)??
?其它?0从而 0≤X≤1时,fx(x)??????f(x,y)dx??x?102dy?2(x?1)
77
当 1≤X≤2时,f(x,y)??7从而 1≤X≤2时,fx(x)?当X<0或X>2时,f(x,y)=0 从而fx(x)???2??0x?1?y?x?1其它x?1?????f(x,y)dy??2dy?4
7x?17?????f(x,y)dy?0
2x?1)/70?x?1?(?41?x?2 综上得:fx(x)??7?0其它?16. 证明:
??1Da?x?b又?(x)?y??(x) (1)f(x,y)??
?0其它?fY(y)??ba?b?a1dx???DD??0a?x?b,?(x)?y??(x)其它
1f(x,y)D当a?x?b,?(x)?y??(x)时,fxy(x|y)??b?afY(y)
28
?D1 b?a1f(x,y)D?fxy(x|y)??b?afY(y)故fxy(x|y)为均匀分布
?1??(x)?y??(x),a?x?b ??b?a其它D??0又 fX(x)????(x)(x)??(x)??(x)1dy???DD?0?a?x?b其它
fyx11?f(x,y),a?x?b,?(x)?y??(x)?D ?????(x)??(x)fX(x)?(x)??(x)?0?D故 fyx(yx)为均匀分布.
17. 解: f?x,y??fy|xyxfx?x? fy?y?????????f?x,y?dx
故fxy?x,y??fyx?yx?f?x??????f?x,y?dx
18. f?x,y??fX?x??fY?y? 故独立. g?x,y??gX?x?gY?y? 不独立. 19. f?x,y??fX?x?fY?y? 不独立. 20. f?x????10?x?1
0其它?2?1?y?e2 f?y???2??0????y???,0?x?1其它
21. 若?x,y?服从二元正态分布,则
f?x,y??12????21??2?x?u1?22?12?e21??2?1???x?u?2(x?u1)(y?u2)?y?u2?2??2????22???2??12??1??1f?x??e2??1
?1f?y??e2??229
?y?u2?22?22
因为 f?x?,f?y?均无参数ρ,故可见,f?x,y?不能由fX?x?,fY?y?决定.
?fX?x??????22. 解:
1e2??x22x2?y2?2?1?sinxsiny?dy
1?1?sinxsiny?dy?e2?y2?2
1?e2??????e?y22?x2?????e?y22dy1?e2?23. x2?21,fY?y??e2? ???y???
fX?x?,fY?y?均服从正态分布,但是f?x,y?不服从正态分布.
? 0 1 2 3 4 P 1/9 4/9 4/9 0 0 ? 0 1 2 -1 -2 P 3/9 2/9 1/9 2/9 1/9 ? ? -2 -1 0 1 2 0 1 2 3 4 0 0 1/9 0 0 0 1/9 0 1/9 0 1/9 0 1/9 0 1/9 0 1/9 0 1/9 0 0 0 1/9 0 0 1/9 2/9 3/9 2/9 1/9 p????i? 1/9 2/9 3/9 2/9 1/9 1 p????j? 24. ? ? 25.证明:
2 3 4 5 6 7 8 9 10 11 12 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 当n=2时,P?X1+X2=k???P?Xi?0kk1?i,X2?k?i? ?i,??p?X2?k?i?
??i?k2 ??P?Xi?0k1 ??i?0i?1i!??e??1(k?i)!?ik?i?1?2
?e?(?1??2)?i?ok
i!(k?i)!30
(?1??2)k?(?1??2) ? ,k=0,1,2,?? ek!由数学归纳法,设对n-1成立,即
?Xi?1n?1i服从参数为
??i?1n?1i的泊松分布,因为Xn服从参数为?n的泊松分布,故
?Xi?1n?1i?Xn服从参数为??i??n的泊松分布,即对n成立.
i?1n?126.
??e??xi,xi?0 i=1,2. ,Z=X1+ X2 fi(xi)??xi?0,?0Z>0时,
f(Z)??f1(z?t)f2(t)dt???e??(z?t)??e??tdt
00zz ??2e??z ??ze2?z0dt
??z
所以fZ(z)?fx?x(z)??120,z?0 2??z??ze,z?0?27.由题意知X1~N(4,3), X2~N(2,1).且两者相互独立.
X1?X2X-X2,Y=1,且X和Y服从正态分布 22EX1?EX24?2??3 故EX?221111DX?DX1?DX2??3??1?1
4444EX1?EX24?211EY???1 , DY?DX1?DX2?1
2244由X1=X+Y, X2=X-Y得X=故X~N(3,1),Y~N(1,1) 即fX(x)?1e2??(x?3)22 , fY(y)?1e2??(y?1)22
28. 用数学归纳法进行推广,与25题类似. Xi~N(?i,?i2),i=1,2??n.
?xi?1ni~N(??i,??i2)
i?1i?1nn1??,(x,y)?G29.f(x,y)??2
??0(x,y)?G当0<z?2时,
31
FZ(z)?P??X??z????f(x,y)dxdy ?Y?G0???1???zy1?? ????dx?dy????dx?dy
???0??2??zy2? ?(0??11zy01dx)dy?0 21zydy 02z ?
4??当z>2时,FZ(z)?1?1 z???0?0z?0???1?1故FZ(z)??z0?z?2 fZ(z)???4?4?1?1?1z?2???z?z230.
P?XY?0??0.3?0?0.3?0.2?0.8 P?XY??1??0.1 P?XY?1??0.1
z?00?z?2 z?2XY 0 P -1 1 0.8 0.1 0.1 X 0 P 1 ?0?0.8?(?1)?0.1?1?0.1?0 EXYE(X?Y)?EX?EY
=0×0.6+0.4×1+(-1) ×0.4+0×0.2+1×0.4
=0.4 31 由题意知 fX(x)??所以:
0.6 0.4 Y -1 P 0 1 0.4 0.2 0.4 ?11?x?2,其它?0 fY(y)???11?y?2,其它?0
?11?x?2,1?y?2 f(x,y)??0其它?E?max(X,Y)????????????max(x,y)?1dxdy
?? ?21?221max(x,y)dxdy
2??1211xdxdy??21?2121ydxdy
2 x>y y>x ?
?(?x1xdy)dx??(?ydy)dx
x32
??21(x2?x)dx??211(4?x2)dx 221x32x22x32 ?1?1?2x1?326
8137???2? 33265 ?
3 ?32.证明: P(X=ai,Y=
j)=Pij, i,j=1,2…….
P(X= ai)=Pi?
P(Y=bj)=P?j?? j=1,2……
E(X?Y)???(ai?bj)?Pij
i?1j?1 ???a?P???biiji?1j?1i?1j?1??ii?j????j?Pij
??ap??bi?1j?1??p?j
?EX?EY
EXY???aibjpij??aipj??bj?pj?EX?EY
i?1j?1i?1j?1???33.
Y 0 1 2
………….. 19 19C1C204020C6020 20C0C204020C60
1182020C19C1C19 4012C20C20C20C401920CP40201840C20CEX?0?1??C220?????19?C20?20?4020C40202020 ………….. C60CCC 206060602020C60C60C60 =8
34.因为可以看成是9重见利试验,EX=np=9?1?()?
935.参考课本P84的证明过程. 36.
X2 ??825??0 1 2 pi1 XX1 0 1 15521 5628285655535 5614285633
pX2j 31510 282828Cov(x1, x2)=E X1 X2-Ex1 ×Ex2
153535-? 28562815 = -
224 =
?12e?3x?4y37.f(x,y)???0?21e?3x?4yg(x,y)???0所以:
x?0,y?0
其它x?y?0
其它?3e?3x fX1(x)???0?4e?4y fY1(y)???0x?0 x?0y?0 y?0
?21?3x?(e?e?7x)gX2(x)??4?0?y?0 其它x?0其它
?7e?7y gY2(y)???01因为f(x,y)?fX1?fY(y),所以X,Y独立. 故Cov(x1,y1)?EX1Y1?EX1Y1?0
Cov(x2,y2)?EX2Y2?EX2?EY2
? ???0??x??x0xy?21e?3x?4ydydx????021?3x1(e?e?7x)dx? 471?3x?4yxy?21edydx?1? ?0?07811?? ? 4974938. X1?X?Y,DX1?DX?DY?2cov(X,Y) 所以 cov(X1,Y2)?DX1?DX?DY3?1?11??
222因为 X1,Y1 独立。
所以 Cov(X1,Y1)=0
(也可以计算:Cov(X1,Y1)= Cov(X+Y,X-Y) = Cov(X,X-Y)+Cov(Y,X-Y)
= Cov(X,X)-Cov(X,Y)+Cov(y,x)-Cov(Y,y)
34
=1?39.
11+?1=0) 22?1? f(x,y)?????0 fX(x)?x2?y2?1其它
?1?x21?1?x2?dy?21?x2? fY(y)??1?y21?1?y2?dx?21?y2?
所以: f(x,y)?fx(x)fy1y 所以:X,Y不独立。 ?x,y?40.
cov(x,y)EXY?EX?EY=……… ?DXDYDXDY?XY?cov(X,Y)EXY?EX?EY??DXDYDXDY44225 ?66112?2257541.
①?rA,rB?cov(rA,rB)61??
DrADrB16?92 ②rP?xrA?(1?x)?rB
Drp?x2DA?(1?x)2DB?2x(1?x)DA ?16x?9(1?x)?2x(1?x)?4?3? ?13x?6x?9
222DB??AB
1 23232 ?13(x?)?13?2?9
1313当x?39108?时,Drp最小为9?
131313min(DrA,DrB)?min(16,9)?9
当13x?6x?0时,即0?x?42.解
①设投资组合的收益率为rp,则 rp?xrA?(1?x)rB
26时,Drp?min(DrA,DrB) 13Dp?x2DA?(1?x)2DB?2x(1?x)DADB??AB
35
所以: Dp?x2DA?2x(1?x)DA22DB??AB?(1?x)2DB??AB?(1?x)2DB?(1?x)2DB?AB
2 ?(xDA?(1?x)DB?AB)2?(1?x)2DB(1??AB)
当x=1 时,Dp?DA?0
22x?1时,?AB?1故?AB?1,即1??2?0,由D?0得D?(1?x)D(1??BPBAB)?0 AB所以,对任意x,有DP?0,所以,任意组合P都有风险。 ② 若
?AB=1,当?AB=1时
DB
Dp?x2DA?(1?x)2DB?2x(1?x)DA ?(xDA?(1?x)DB)2 设投资组合中数为X,则 xDA?(1?x)DB?0
即x?DBDB?DA,1?X??DADB?DA
此时Dp=0
当?AB??1时,DP?x2DA?(1?x)2DB?2x(1?x)DADB ?xDA?(1?x)DB选投资组合中权数x,使得
xDA?(1?x)DB?0 x???
2DBDA?DB , 1?x?DADA?DB,此时DP?0
③不卖座即0<x<1,能在0<x<1上得到比证券A和B的风险都小的投资组合,意味着Dp的最小值在0<x<1达到。 由
?DP?2?DA?2(1?x)DB?(2?4x)DADB??AB?0 ?xDB?DADB?ABDA?DB?2DADB?AB
所以 x?所以 DA?DB?2DADB?AB?DA?DB?2DADB?2DADB(1??AB)
?(DA?DB)2?2DADB(1??AB)?(DA?DB)2?0 故为使0<x<1则
36
?DB?DADB?AB?0? ?
??DB?DADB?AB?DA?DB?2DADB?AB解得:
?AB?DBDA且?AB?DADBDADBDBDA
如果DA?DB,则上述等价于?AB?
如果DB?DA,则上述等价于?AB? 综上:
当?AB?min(DA,DB)max(DA,DB)时,可在不卖座的情况下获得比DA和DB都小的风险投资组合。
43. ①Er=0.1×(-3%)+1%×0.105+2%×0.175+3%×0.26+4%×0.125+5%×0.13+6%×0.065+7%×0.04=2.755% ②P(r=-3%/rf=1.5%)=
P(r??3%,rf?1.5%)p(rf?1.5%)?0.025?0.05 0.50.050.1?0.1 P(r=2%/rf=1.5%)=?0.2 0.50.50.150.075?0.3 P(r=4%/rf=1.5%)=?0.15 P(r=3%/rf=1.5%)=0.50.50.050.025?0.1 P(r=6%/rf=1.5%)=?0.05 P(r=5%/rf=1.5%)=0.50.50.025?0.05 P(r=7%/rf=1.5%)=
0.5 P(r=1%/rf=1.5%)=
所以:E(r/ rf=1.5%)=0.05×(-3%)+0.1×1%
+0.2×2%+0.3×3%+0.15×4%+0.1×5%+0.05×6%+0.05×7%=3% 44. fYX(yx)? EYX?x)?f(x,y)8xy2y ,0<x≤y≤1 ??fX(x)4x(1?x2)1?x2???????yfYX(yx)dy??y?x12ydy 1?x212311?x?x22??yx?? (0<x<1) 1?x231?x3?22x2??所以: E?YX?x)???331?x?0?46.看成伯努利试验,X~b(120,
0P(A)?1-[C120(0?x?1
其它1019120)()20201)→X~P(6)泊松分布or X~ N(6,5.7),A=“X≥10” 20119119119129?C120()1()19?C120()2()18????C120()9()11] =采用泊松分
202020202020布或正态分布近似计算
=0.0465(二项分布计算结果)
P(A)=0.022529+0.011264+0.005199+0.002228+0.000891+0.000334+0.000118+0.000039
+0.000012+0.000004+0.000001=0.042619---泊松分布
37
P(A)=1-F(10)=1-Ф0[(10-6)/σ] σ2=5.7
=1-Ф0[1.675415827737….]=1-0.95352 or 0.95244=0.04648 or 0.04756 显然,本题正态分布比泊松分布更准确。
47.X=开动生产的机床数 X~b(200,0.7) 所以X~N(140,42) 设以95%以上概率保证正常生产机器为K台,则 P(X≤K)≥0.95??(K)??0(K?140)?0.95 42所以 (K?140)?1.65?K?150.7 所以K=151台 42所以 各电能≥K×15=2265个电能单位,以95%保证机器都正常 48.Xi~U??0.5,0.5?,EXi?0,DXi?1 X1, X2相互独立, 12X??xi~N(0,300?i?13001)?X~N(0,25),X的均值为0,所以密度函数关于原点对称 12(1)P(?xi?1300i?15)?p(x?15)?2(?(??)??(15))?2?2?(15)
=2(1??0(3))?0.0027 (2)X=X1+?+XN~N(0,
n) 12 P(x?10)?p(?10?x?10)?2?(10)?1?2?(10)?1?0.9 n12 ?0(1012所以 n=440.
n)?0.95?1012n?1.65?n?440.77
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