北航概率统计、概统随A2008-2009(1)考试卷A卷及答案A和B卷
北京航空航天大学
BEIHANG UNIVERSITY
2008,2009 学年 第一学期期末
考试统一用答
册
考试课程 概率统计 ,09J70040,
概率统计与随机过程A,09J70050,
概率论与数理统计
班 级_____________ 学 号 _____________
姓 名______________ 成 绩 _________
考场教室_________ 任课教师_________
题号 一 二 三 四 五 六 七[七] 八[八] 总分
分数
阅卷人 校对人
A
2009年1月 16 日10,30—12,30
一、单项选择题(每小题3分,满分18分)
412X,X1、设是来自正态总体的样本,设, XXXX,,,N(0,),,i12344i,1
2,,当 ( )时, 概率最大。 P{1,X,2}
6363(A) , (B) , (C) , (D) 。 ln22ln2ln22ln2
,,1,(1)01xxx,,,,,(1-),2、 设总体的密度函数为,其中,X,,0fx(;),,,0,其它,
是来自总体X的样本,则参数的矩估计量为( )。 ,XXX,,,12n
X2X2XX(A) , ( B) , (C) , ( D) 。 1,X2,X2,X1,X
2222ˆˆ3、设是来自正态总体的样本,当( )时,是 N(,),,,,,,XX,,Xcc,1n
nn11222ˆ,,,XXX,X()的无偏估计,其中, 。 ,,,iinn,i,11i
1111(A) , (B) , ( C) , ( D ) 。 ,,nnn,1n,1
2E|X,,|,4、设随机变量,则[ ]. X~N(,,,)
2,(A) 0, (B) , (C) , (D) . ,,
2,
5、两人约定在某地相会,假定每人到达的时间是相互独立的,且到达时间在中午12时
到下午1时之间服从均匀分布,则先到者等待10分钟以上的概率为[ ].
25254711(A) , (B) , (C), (D). 36725236
22,6、设是总体的样本,,已知,下列几个作为的估计量中, X,X,,,,,XN(,,,)12n
较优的是[ ].
nn112222ˆˆ,,(X,X),,(X,X)(A) , (B) , ,,1i2inn,1,,i1i1
n,n1112222ˆˆ,,(X,,),,(X,,)(C) , (D) . ,,34iin,1ni,1i,1
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二、填空题(每小题3分,满分18分)
1、有个白球与个黑球任意地放入两个袋中,每袋装个球.现从两袋中各取一 nnn
,则所取两球颜色相同的概率为 。 球
2、在无线电通讯中,由于随机干扰,当发出信号为“0”时,收到信号为“0”、不清
和1的概率分别为0.7,0.2,0.1; 当发出信号为 1时,收到信号为1、不清和0
的概率分别为0.9,0.1和0.如果在发报过程中发出0和1的概率分别是0.4和
0.6。当收到信号为不清时,原发信号是1的概率为 。 3、三门火炮同时炮击一敌舰(每炮发射一弹).设击中敌舰一、二、三发炮弹的概率 分别为0.6、0.5、0.3,而敌舰中弹一、二、三发时被击沉的概率分别为0.3、 0.6、0.9.则敌舰被击沉的概率为 。
F(x)Y4、 设随机变量X的分布函数为,随机变量服从两点分布:
P{Y,a},p,P{Y,b},1,p,(0,p,1)Y,并且X与相互独立,
Z,X,Y则随机变量的分布函数 。 F(z),Z
225、设是来自正态总体的样本,未知,已知。 N(,),,,,XXX,,,12n00
X,,对给定置信水平1,,(01,,,),满足 , Pab{}1,,,,,2/n,0
,,00()ab,PXbXa{}1,,,,,,ab, 即满足的实数有无穷多组, ,,
nn
当 ,b, 时,就可使得的置信水平为1,,的置信区间,a,
,,,,00a,b 的长度最短。(用
正态分布的分位点
示出所求的即XbXa,,,,,nn,,
可。)
2x,x,?,x6、设总体,为的样本. ,,X~N,,,X12n
nnn1n112222x,x 记 ,这里规定 ,. s,(x,x)s,(x,x),i,i,i22nn,,,1,ii1,1i
2,在未知方差, 检验假设:时, H,,,00
选取检验用的统计量是 。
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三、(满分12分)对同一目标进行三次独立射击,第一、二、三次射击的命中概率分别为
0.4,0.5,0.7,试求:(1)在这三次射击中,恰好有一次击中目标的概率;
(2)至少有一次击中目标的概率.
2x,,0,x,,,2四、(满分12分) 已知随机变量的概率密度为 , X(),fx,,
,0,其它,
试求的分布函数和概率密度. Y,sinXF(y)f(y)YY
五、(满分8分)设随机变量X的二阶矩存在,
2证明:当时,的值最小,最小值为DX. k,EXE(X,k)
2六、(满分12分)设总体,从此总体中取一容量为4的样本,设X~N(0,3)X,X,X,X1234
22, Y,a(X,2X),b(3X,4X)1234
(1)求服从的分布;(2)求服从的分布; 3X,4XX,2X3412
2a,bY(3)试决定常数,使随机变量服从分布. ,
七、(满分8分)(此题学过1-9章和11-13章的学生做,仅学过1至9章的学生不做)
XYZ(t),Xsin,t,Ycos,t设 ,其中是常数, 与是相互独立的随机变量, ,
X~N(0,1)且, , 试求: Y~U[,3,3]
222EYEXE[Z(t)]E[Z(t)Z(t,,)](1),; (2),,; E[Z(t)]
Z(t)(3)问是否为广义平稳过程,
[七]、(满分8分)(此题仅学过1至9章的学生做;学过1至9章和11-13章的学生不做)
有甲、乙两炮向同一目标轮流射击,直至有一炮击中目标为止.甲、乙两炮击中的概率分别
YX为0.3和0.7,规定甲炮先射.以和分别表示甲、乙两炮所用炮弹数.
EX,EYYX(1)试写出的分布律,求的分布律; (2)求。
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八、(满分12分)(此题学过1至9章和11-13章的学生作,仅学过1至9章的学生不做)
四个位置:1,2,3,4在圆周上逆时针排列.粒子在这四个位置上随机游动.粒子从任何一个
12X(n),j位置,以概率逆时针游动到相邻位置; 以概率顺时针游动到相邻位置;以表示时33
j(j,1,2,3,4)刻粒子处在位置, n
{X(n),n,1,2,?}试作:(1)写出齐次马尔可夫链的状态空间;
{X(n),n,1,2,?}(2)求齐次马尔可夫链的一步转移概率矩阵;
(2)P(3) 求两步转移概率矩阵; (4)求该齐次马尔可夫链的平稳分布.
[八]、(满分12分)(此题仅学过1至9章学生做,学过1-9章和11-13章学生不做)
X,X,,,,,X,,,,设是相互独立的随机变量序列,且其分布律为 12n
121(n,1,2,,,,); P{X,n},,P{X,,n},,P{X,0},1,,nnnn,1n,1n,1333
n12(n,1,2,,,,)EX记Y,X,。 (1)求,,; (2)求, ; DXEYEXDY,nninnnnn,1i
limP{|Y|,,},1(3)证明: 对任给,,0,成立。 nn,,
概率统计 ,09J70040,、 概率统计与随机过程A,09J70050,
概率论与数理统计 考试卷
A、B卷答案及评分细则 (2009-1-16 )
A卷 :一、单项选择题(每小题3分,满分18分)
1、B;2、D;3、A; 4、C;5、A;6、C 。
二、填空题(每小题3分,满分18分)
12CCn,1n,1n2P(A),1、, 或 . P(A),,22n,12n,1Cn2
30.752、 ;3、; ,0.428577
,pF(z,a),(1,p)F(z,b)4、 ; F(z),P{X,Y,z}Z
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,,,1,15、,,z,z,,z; a,,,,(1)b,,,(1),,,,1,122222
x,,x,,00,,6、T,,n,1~tn,1 。 ss
n,1
B卷 :一、单项选择题(每小题3分,满分18分)
1、C;2、A;3、C ; 4、B;5、D;6、A;
二、填空题(每小题3分,满分18分)
,pF(z,a),(1,p)F(z,b) 1、 ; F(z),P{X,Y,z}Z
,,,1,1,,z,z,z2、,; a,,,,(1)b,,,(1),,,1,,122222
x,,x,,00,,T,,n,1~tn,13、; ss
n,1
12CCn,1n,1n24、P(A),,或 . P(A),,22n,12n,1Cn2
30.755、; 6、; ,0.428577
三、(满分12分)
i,1,2,3解 设第次射击时击中目标,,相互独立, A,A,A,Aii123
; P(A),0.4,P(A),0.5,P(A),0.7123
B,AAA,AAA,AAA(1) 设恰好有一次击中目标,则, B,11231231231
P(B),P(AAA),P(AAA),P(AAA)于是 1123123123
,P(A)P(A)P(A),P(A)P(A)P(A),P(A)P(A)P(A) 123123123,0.4,0.5,0.3,0.6,0.5,0.3,0.6,0.5,0.7,0.36;„„„„„„.6分
A,(2) 设至少有一次击中目标,则, A,A,A,A123
P(A),P(A,A,A),1,P(A)P(A)P(A)故 123123
,1,0.6,0.5,0.3,0.91.„„„„„„„„„„„„„„„„„„„„..6分
四、(满分12分)
A6-5
解 , F(y),P{Y,y},P{sinX,y}Y
y,1,P{S},1(1)当时, ; F(y),P{Y,y},P{sinX,y}Y
y,,1,P(,),0(2) 当时, ; F(y),P{Y,y},P{sinX,y}Y
,1,y,0(3) 当时, ; ,f(x)dx,0F(y),P{Y,y},P{sinX,y}Y,sin,y
0,y,1(4)当 时, F(y),P{Y,y},P{sinX,y}Y
arcsiny,xx222 , ,arcsiny,f(x)dx,dx,dx,,,22,sinx,y0arcsiny,,,,
0,y,0,
,2;„„„„„„„„„„„„„„„„„„„„„8分 F(y),arcsiny,0,y,1,Y,,1,y,1,
21,,0,,1y,2,(),(2)所以 fy . „„„„„„„„„„„„„„„„„„4分 1,y,Y
,0,其它,
五、(满分8分)
22证明 由 E(X,k),E[(X,EX),(EX,k)]
22 ,E[(X,EX),2(X,EX)(EX,k),(EX,k)]
22 , ,E(X,EX),(EX,k)
2,„„„„„„„„„„„„„„„„„„6分 ,E(X,EX),DX
2DX知当k,EX时,的值最小,最小值为.„„„„„„„„„„„„„„2分 E(X,k)
222222(用, f(k),E(X,k),EX,2EX,k,k,EX,(EX),(k,EX)
或利用导数求极值也可.)
六、(满分12分)
2i,1,2,3,4解 因为 X~N(0,3),,且相互独立, X,X,X,Xi1234
2,DX,3; EX,0ii
2(1)因为,, D(X,2X),5,3E(X,2X),01212
2所以,„„„„„„„„„„„„„„„„„„„„„„„4分 X,2X~N(0,5,3)12
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2(2)由,, D(3X,4X),25,3E(3X,4X),03434
2得;„„„„„„„„„„„„„„„„„„„„„„„4分 3X,4X~N(0,15)34
11(X,2X)~N(0,1)(3), (3X,4X)~N(0,1)12341535
112222[(X,2X)]~(1),故, [(3X,4X)]~,(1)12341535
112222 由分布的可加性,得, (X,2X),(3X,4X)~,(2),123445225
11Ya,,b,由所给的表达式,即知 ;„„„„„„„„„„„„„„4分 45225
七、(满分8分)(此题学过1-9章和11-13章的学生做,仅学过1至9章的学生不做)
2解.(1)由题设条件得 ; EX,0,DX,1,EX,1
2; EY,0,EY,1,DY,1
E[XY],EX,EY,0 ; „„„„„„„„„„„„„„„„„„„„„„„„2分
E[Z(t)],sin,t,EX,cos,t,EY,0(2),„„„„„„„„„„„„„„„„1分 E[Z(t)Z(t,,)]
2,[sin,t,cos,(t,,),cos,t,sin,(t,,)]E[XY] ,sin,t,sin,(t,,),EX
2 ,cos,t,cos,(t,,),EY
,cos,t,cos,(t,,),sin,t,sin,(t,,) , „„„„„„„„„„„„2分 ,cos,,
2 , „„„„„„„„„„„„„„„„„„„„„„„„„„„„„1分 E[Z(t)],1
2E[Z(t)],0E[Z(t)Z(t,,)](3)因为,,,(不依赖于), E[Z(t)],1t,cos,,
Z(t)所以 是广义平稳过程. „„„„„„„„„„„„„„„„„„„„„„„„2分 八、(满分12分)(此题学过1至9章和11-13章的学生作,仅学过1至9章的学生不做)
S,{1,2,3,4}解.(1)依题意 ,状态空间 , „„„„„„„„„„„„„„„3分
A6-7
21,,00,,33,,12,,00,,33,,()Pp(2)转移概率矩阵 ,„„„„„„„„3分 ,,ij4,41200,,33,,21,,00,,33,,
2121,,,,0000,,,,3333,,,,1212,,,,0000(2)2,,,,3333P,P (3) ,,,,,12120000,,,,3333,,,,2121,,,,0000,,,,3333,,,,
45,,00,,99,,45,,00,,99 ;„„„„„„„„„„„„„3分 ,,,5400,,99,,54,,00,,99,,
(,,,),(,,,)ppppppppP,12341234(4) , ,,,,,1pppp1234,
21,,00,,33,,12,,00,,33(p,p,p,p),(p,p,p,p), 12341234,,1200,,33,,21,,00,,33,,
12p,p,p, 12433
21p,p,p, 21333
21p,p,p, 32433
A6-8
12 , p,p,p41333
p,p,p,p,1。 1234
1解之得 . p,p,p,p,12344
1111(p,p,p,p)故平稳分布为. „„„„„„„„„„„„„3分 ,(,,,)12344444
[七]、(满分8分)(此题仅学过1至9章的学生做;学过1至9章和11-13章的学生不做)
X,Y解 (1) 以和分别表示甲乙在第轮射击中击中目标.显然是离散型随机变量. ABiii
根据题意知
,; {X,1},A,ABP{X,1},P(A),P(AB),P(A),P(A)P(B),0.79111111111
, {X,2},ABA,ABAB1121122
;„ P{X,2},[P(A),P(A)P(B)]P(A)P(B),0.79,0.2122211{X,k},AB?ABA,AB?ABAB, 11k,1k,1k11k,1k,1kk
k,1P{X,k},[P(A),P(A)P(B)]P(AB?AB),0.79,(0.21), kkk11k,1k,1
k,1k,1,2,?X于是,的分布律为, . „„„„„„„„„2分 P{X,k},0.79,(0.21)
0,1,2,?Y的可能取值为:,, ; {Y,0},AP{Y,0},P(A),0.311
,; {Y,1},AB,ABAP{Y,1},P(AB),P(ABA),0.5531111211112P{Y,2},P(ABAB,ABABA)112211223
,[P(AB),P(ABA)]P(AB),0.553,0.21, 2222312
„
k,1P{Y,k},[P(AB),P(ABA)]P(AB?AB),0.553,0.21, kkkkk,111k,1k,1于是,
Y的分布律为
k,1P{Y,0},0.3k,1,2,?,, . „„„„„„„„„2分 P{Y,k},0.553,(0.21)
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,,,,k1,k,1EX,k,0.79,(0.21),0.79k,(0.21)(2) ,,k1k,1,
11 ;„„„„„„„„„„„„„„„„„„„2分 ,0.79,,,1.265820.79(1,0.21)
,,,,k1,k,1EY,0,0.3,k,0.553,(0.21),0.553k,(0.21) ,,k1k,1,
10.5330.7,0.553,, 。„„„„„„„„„„„„„2分 ,,0.8860722(1,0.21)(0.79)0.79
[八]、(满分12分)(此题仅学过1至9章学生做,学过1-9章和11-13章学生不做)
证明 (1)由数学期望和方差的性质及条件,有
11, EX,,n,,n,,0,0nn,1n,133
112n222()()0, ,,,,,,,EXnnnn,1n,1n,1333
2n2,„„„„„„„„„„„„„„„„„„„„„„„„„3分 ,,DXEXnnn,13
nn11EY,E(X)2)(, ,EX,0,ni,inn,1ii,1
nnn1i121,„„„„„„„„„„„„„„„„4分 ,DX,DY,D(X),,i,nii22,1nn3nii,1,1,1i
nn1111(3); ,,,DXnDY,D(X),i,ni22nnnni,1,1i
对任意,由契比雪夫不等式,得 ,,0
,P{|Y,EY|,,} 1,P{|Y|,,}nnn
DY1n, ,,,,1122,n,于是成立 。„„„„„„„„„„„„„„„„„„„„„„„„„5分 limP{|Y|,,},1n,,n
A6-10