概率统计习题
答案
八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案
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天津科技大学概率论与数理统计检测题4答案
,2一( 1(0.6, 0.1, 0.9; 2(; 3(1; 1,5e,0.3233
X012 4(. p0.8330.1520.015
解答如下:
1(; P(X,2),p(0),p(1),p(2),0.1,0.2,03,0.6
;( P(X,3),p(4),0.1P(X,4),1,P(X,4),1,0.1,0.9
2( P(X,2),1,P(X,2),1,p(0),p(1),p(2)
012222,2,2,2,2,1,e,e,e,1,5e,0.3233 ( 0!1!2!
,,,11/2kk,c,13(由,得( 1()2(),pk,c,c,c,,c,,,21(1/2),111kkk,,,
X4(取值为0、1、2,且
1055p(0),P(X,0),,,0.833 (第一次取到合格品), 1266
21010p(1),P(X,1),,,,0.152 (第一次取到废品,第二次取到合格品), 121166
21101p(2),P(X,2),,,,,0.15 (前两次都取废品)( 12111066
二(1(?; 2(?; 3(?.
解答如下:
1((本题属于第六节内容)
F(2),P(X,2),p(0),p(1),p(2),0.1,0.3,04,0.8(
A,{(1,3)(2,2)(3,1)} 2(A,{两次的点数和为4},则,则
3 ( P(X,4),P(A),36
446 3((二项分布),则( X~B(10,p)P(X,4),Cp(1,p)10
三(
232xx3,x1. ,则, X~B(3,)p(x),P(X,x),C()()(x,0,1,2,3)3555
3273 , p(0),(),,0.2165125
235412 p(1),3,()(),,0.432, 55125
233621p(2),3,()(),,0.288 , 55125
283p(3),(),,0.064 , 5125
X0123
X的概率函数为. 所以,2754368p125125125125
X2. 的所有可能取值为3,4,5(
11X,3:取出的3个球,号码分别只能为1,2,3,所以,,; PX,3,,,0.1310C5
2X,4:取出的3个球中,,只球号码是4,另外两个号码在1,2,3中任取2只,共有C3
2C33,,种,所以PX,4,,,0.3; 310C5
2X,5:取出3只球中,,只球的号码是5,另外两个号码在1,2,3,4中任取2只,有C4
2C64,,PX,5,,,0.6,,,,,,种,所以.(或PX,5,1,PX,4,PX,3,0.6) 310C5
X345X从而的概率函数为. p0.10.30.6
天津科技大学概率论与数理统计检测题5答案
1,,11,,,x,fx(),一(1(1, 0; 2(2, 0.3, ; 2,
,0, 其他,
,1,1,23(, ( 1,e,0.6321e,e,0.2325
解答如下:
,,,,,x,x,,a,11(由,得; 1,f(x)dx,aedx,,ae,a0,,0,,
由于连续型随机变量在一点取值的概率为零,所以( P(X,0),0
x,12,FFxx,1,,, 2(,,由于在点连续,所以(1)lim()limF(1),1F(x),,x,1x,1AA
2,,1A,2,即,得; F(1),F(1)A
0.8,10.2,1P(0.2,X,0.8),F(0.8),F(0.2),,,0.3 ; 22
1/2,,1,x,1,,,,1,x,1,1,x,1f(x),F(x), (注:也可写作)( ,0,其它.,
xx,,4414,1443(P(X,4),f(x)dx,edx,,e,1,e,0.6321; 0,,,,04
xx,,8818,1,244P(4,X,8),f(x)dx,ex,,e,e,e,0.2325 ( 4,,444
二(1(?; 2(?.
解答如下:
,,,,/22A,11,f(x)dx,Asinxdx,,Acosx,A 1(由,得. 0,,,,0
2(根据连续型随机变量分布函数的性质,P(a,X,b),F(b),F(a). 三(
021111,,xx01. (1)由,得; ,,,,,,,fxdxaedxdxaeaa,1(),,,,,,,0,,2422
x(2). F(x),f(t)dt,,,
xx11txx,0 当时,, Fxftdtedte,,,()(),,,,,,22
xx0111xt02,,x当时,, ,,,,,Fxftdtedtdt()(),,,,,,,02424x0211tx,2当时,. Fxftdtedtdt,,,,()()1,,,,,,,024
1,xex, 0,,2,1x,Fxx(), 02,,,,所以,随机变量X的分布函数为 . ,24,
1, 2x,,
,,
1). (3PXF(1)1(1)10.8161,,,,,,,,2e
2. (1)由随机变量分布函数的性质,有
,,lim()0FxAB,,,,x,,,1111,2,,,,AB, 解得,于是Fxx()arctan. ,,,22,,lim()1FxAB,,,x,,,,,2
,X(2)由于在的可导点处,得随机变量的概率密度为 F(x)F(x),f(x)
111, f(x),(,arctanx),x,(,,,,,). 2,,2(1,x)1111111,,1,,,,,,,PXfxdxdx(1)()arctan()x(3). ,12,,,,11,,1x442,,
,,,,,,110002,,,,,PXfxdxdx(1500)()10003. (1),,,. 2,,,,15001500xx3,,1500
2YB~(4), (2) 各元件工作相互独立,寿命大于1500小时的元件数, 3
2804(1)1(0)1(0)1(1)所求概率为. p,PY,,,PY,,,p,,,,4381
或者: p,P(Y,1),p(1),p(2),p(3),p(4)4444
2121212180132231404()()6()()4()()()() . ,,,,,,,,3333333381
天津科技大学概率论与数理统计检测题6答案
21一( 1(; ,,,,,99
1/4,0,x,2,0,y,2,,32( ; f(x,y),,40,其它.,
,4,3 3(. (1,e),e,0.0489
,y,e,0,x,1,y,0,4( f(x,y),,0,其它.,
解答如下:
1121111(2),,,,1,,(1),,,,pp,,1(,, XX69183333
11111(1),,,pp(2),,,p(3),,, ,,, YYY632189
212,,p(2,2),p(2)p(2),(,,),, 由,得; XY939
211,,p(2,3),p(2)p(3),(,,),, 由,得( XY9318
112112,,,,,,,,,,,, (或由,则)( 333399X~U(0,2)、Y~U(0,2)2(由,则
1/2,0,x,2,1/2,0,y,2,,,f(x),f(y), ,,XY0,其它;0,其它;,,
YX 因为与相互独立,所以
1/4,0,x,2,0,y,2,,, f(x,y),f(x)f(y) ,XY0,其它.,
113(1)()dddd(41) ( PX,Y,,fx,yxy,xy,,,,,,,444x,y,1x,y,1
YX3(由于与相互独立,所以
2,, P(X,2,Y,1),P(X,2)P(Y,1),f(x)dx,f(y)dyXY,,1,,
2,,,2x,3y,2x2,3y,,,4,3( ,2edx,3edy,[,e],[,e],(1,e),e,0.048901,,01
4( 因为与相互独立,所以 XY
,y,e,0,x,1,y,0, f(x,y),f(x)f(y),,XY0,其它.,
二(
y1,,,21,y,0,02,,,x,,,e2()fx,(),fy1. 由 ; 及 2,,XYy,0.其他.,0,,,0,,
YX随机变量与相互独立,得
y, ,02,0,,,xy12,e, , 4fxyfxfy(,)()(),,,XY其他,0,,
所以
yy,,22,,11,,22PXYfxydxdydxedyedx()(,),,,,,x,,,,,00x42xy, xx,2, 121,22 )[]10.6321.,,,,,,,edxee0,02
,,,,11kk,41(,),,,dxfxydykdxxydy2. (1)由,得( ,,,,,,,,004
,,,,(2)f(x),f(x,y)dy; f(y),f(x,y)dx( XY,,,,,,
,,,,x,0x,1当f(x),f(x,y)dy,0dy,0或时,, X,,,,,,
,,10,x,1当时, fxfxydyxydyx()(,)42,,,X,,,,0
,,,,f(y),f(x,y)dx,0dx,0yy,,01或当时,, Y,,,,,,
,,1当01,,y时,. fyfxydxxydx()(,)42y,,,Y,,,,0
2,x2,y01,,,x01,,,y,,所以, ; fx(),fy(),,,XY其他.0,0,其他.,,
(3)由于,所以随机变量与相互独立( YXfxfyfxy()()(,),XY
0.5120.5(4) (0.5)2d; PY,,yy,y,0,04
10.5111(0.5)d4d2d (或); PY,,xxyy,xx,,,,00044
10.2312120.2 ( P(X,0.5,Y,0.2), dx4xydy,x,y,,,0.030.50,,0.50425
(或 P(X,0.5,Y,0.2),P(X,0.5),P(Y,0.2)
10.2312120.2, 2xdx,2ydy,x,y,,,0.03 )( 0.50,,0.50425
天津科技大学概率论与数理统计检测题7答案
Y01237Z,40123一(1(; ; P0.30.20.10.20.2P0.20.20.10.20.3
W01425( P0.10.40.30.2
22X,1,3,1135X,1125102(; ( P0.20.10.10.30.3P0.10.30.30.3解答如下:
Y,X,21(取值为,且 0、1、2、3、7
,, P(Y,0),P(X,,2),0.3P(Y,1),P(X,,1),0.2
,, P(Y,2),P(X,0),0.1P(Y,3),P(X,1),0.2
; P(Y,7),P(X,5),0.2
Z,,X,1取值为,4、0、1、2、3,且
,, P(Z,,4),P(X,5),0.2P(Z,0),P(X,1),0.2
,, P(Z,1),P(X,0),0.1P(Z,2),P(X,,1),0.2
; P(Z,3),P(X,,2),0.3
2W,X0、1、4、25取值为,且
, P(W,0),P(X,0),0.1
P(W,1),P(X,,1),P(X,1),0.2,0.2,04,
P(W,4),P(X,,2),0.3P(W,25),P(X,5),0.2,(
2X,1,3、,1、1、3、52.取值为,且
,, P(2X,1,,3),P(X,,1),0.2P(2X,1,,1),P(X,0),0.1
,, P(2X,1,1),P(X,1),0.1P(2X,1,3),P(X,2),0.3
; P(2X,1,5),P(X,3),0.3
2取值为,且 X,11、2、5、10
2 , P(X,1,1),P(X,0),0.1
2 , P(X,1,2),P(X,,1),P(X,1),0.2,0.1,0.3
22 ,( P(X,1,5),P(X,2),0.3P(X,1,10),P(X,3),0.3
二(1. ? 2. ?(
解答如下:
y,1y,1F(y),P(Y,y),P(3X,1,y),P(X,),F()(( 1Y33
1/6,0,x,6,,f(x),2(由于X~U(0,6),则 于是 ,X0,其它.,
y,3 , F(y),P(Y,y),P(X,3,y),P(X),y,3),f(x)dxYX,,,
1/6,0,y,3,6,,,f(y),[F(y)],f(y,3), 所以, ,YYyX0,其它.,
1/6,,3,y,3,,, ,0,其它.,
333三(1.解:由 F(y),P(Y,y),P(1,X,y),P(X,1,y),P(X,(1,y))Y
33 ,1,P(X,(1,y)),1,F((1,y)),y,(,,,,,,)X
3223,f(y),[F(y)],,f((1,y)),3(1,y),(,1),3(1,y)f((1,y))得 YYyXX
23(1,y),,y,(,,,,,) ( 6,[1,(1,y)]
6y, 2.解: F(y)P(Yy)P(64Xy)P(X),,,,,,,Y4
6y6y,,, 1P(X)1F(),,,,,X44
1,x,,0,,,由于 fx(),,,X
,,其他0,
16,y,6,16,yy,,0,,,,,,(),[()],[1,()],(),故 fyFyFf,44YYyXyX,444,0,其它,
1,,,6,4,,y,6,, ,4,
,0,其它.,
Y3.解 设的分布函数为,则 Fy()Y
y,8y,8PX(),F() =PYy(),=PXy(28),,==, Fy()XY22
Y 于是的概率密度函数为
y,8y,81y,8,,f(y),[F(y)],f(),(),f() YYyXyX22221y,8y,8,,,,0,,4,, ,2162
,0,其它,
y,8,,,8,y,16,, ,32
,0,其它.,