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1、注意:这是第一稿(存在一些错误)第四章概率论习题_偶数.doc2 方案一:平均年薪为 3 万方案二:记年薪为 X,则(1.2)0.2p X=,(4.2)0.8p X=1.2 0.24.2 0.83.63EX=+=故应采用方案二4()1228p X=,()1314p X=,()3428p X=,()157p X=,()5628p X=,()3714p X=,()124p X=,113153123456786281428728144EX=+=。6 不会8()()22002,2xxxxXfxf x y dyedyex=,0 x=880818tteETt edtedt=+=。14(1)1a=时,()2
2、142150nCpC=,()111142151nC CpC=1111421524153nC CEC=2a时,()21522150aCpC=,()111522151aaC CpC=,()222152aCpC=11215222151521215aaaC CCaECC=+=由243E=得,10a=。(2)()4510599154C CpC=,()5410599155C CpC=,()6310599156C CpC=,()7210599157C CpC=,()8110599158C CpC=,()9010599154C CpC=4554637281901051051051051051059999999
3、1515151515154567896C CC CC CC CC CC CECCCCCC=+=。16 记Y为进入购物中心的人数,X为购买冷饮的人数,则()()()1m kkkXYmm kpXkpYm C pp=()1!mm kkkmm keC ppm=()()1!mm kkm keppmkk=()01!kkmmmeppkm=()()()101!kpmpmpepekm=()!kppek=故购买冷饮的顾客人数服从参数为p的泊松分布,易知期望为p。18()2222112151522221515154442601233363aaaaCC CCDCCC=+=.20()0EXxf x dx=,22122x
4、DXEXx edx=112xE Xx edx=()2221112xD XE XE Xx edx=22()()()()101012p Xp Xp Yp Y=(1)()()()()10,11,11,0p XYp XYp XYp XY+=+=+=()()()()()()011110p Xp Yp Xp Yp Xp Y=+=+=34=(2)()()()()()()()()01101000101YE Xp Xp Yp Xp Y=+=+()()()()()()01111011110p Xp Yp Xp Y=+=()()()()()()22111YYYD XE XE X=()()21YE X=()()()(
5、)()()012201 0001 01p Xp Yp Xp Y=+=+()()()()()()012211 1011 11p Xp Yp Xp Y=+=12=24()22,0,0,0.xXexfxx=()44,0,0,0.yYeyfyy=()21,0,0,0.xXexFxx=()41,0,0,0.yYeyFyy=(1)min,ZX Y=()()()()()()61,0,1110,0.zZXYezFzp ZzFzFzz=故Z服从参数为6=的指数分布,故16EZ=,136DZ=。故()1DZCv ZEZ=。(2)max,ZX Y=()()()()()()2411,0,0,0.zzZXYeezFzp
6、 ZzFz Fzz=()0712ZEZzdFz=,()()22204933144144ZDZEZEZz dFz=,故()337DZCv ZEZ=。(3)ZXY=+,34EZEXEY=+=,516DZDXDY=+=,()53DZCv ZEZ=26(1)()()1,2Xfxf x y dy=,1x()0E X=,()122211123DXEXEXx dx=,同理()()1,2Yfyf x y dx=,1y()0E Y=,13DY=()()111111cov,()()()149X YE XYE X E Yxyxy dxdy=+=(),cov,13X YX YDXDY=,故X和Y正相关。又()()(,
7、)XYf x yfx fy,故X和Y不独立。(2)()()1122222222221111cov,()()()()()()1049XYE X YE XE YE X YD X D Yx yxy dxdy=+=故0=,即X和Y不相关。又()()2222,XYFx yp Xx Yy=(),pxXxyYy=(),yxyxf t v dtdvxy=所以()22,111,()()44XYfx ym xn yxyxy=,故2X和2Y相互独立。28(1)不会写(2)()()()111111cov,cov,cov,nnijijiijjX XxXXXD Xnnnn=(3)()0011cov,cov,nkkkkij
8、ij nS TXX+=+=()0011cov,nkkijij nXX+=+=()()()0000000111111cov,cov,cov,nnknkkkkijijijij ni nj ni nj kXXXXXX+=+=+=+=+=+=+001kii nDXkn=+=,1kkiiDSDXk=,001nkkjj nDTDXk+=+=,()0cov,kkkSknkDSDT=。30(1)()()()20,00005p XYp YXp X=,()()()10,11005p XYp YXp X=,()()()11,00115p XYp YXp X=,()()()11,11115p XYp YXp X=,(
9、)305p X=,()215p X=,()305p Y=,()215p Y=,()()()0,000p XYp Xp Y=,故X和Y不独立。(2)()cov,()()()X YE XYE X E Y=()()()()00,00,11,0p XYp XYp XY=+=+=()()()111,11125p XYp Xp Y+=故X和Y正相关。32(1)由0=知,X和Y不相关,等价于X和Y相互独立。()0,1XN,()1,4YNEaEXbEYb=,EaEYbEXa=,22224Da DXb DYab=+=+,22224Da DYb DXab=+=+,224bab+=+和224aab=+分别为和的标准
10、化变量。()()cov,cov,aXbY aYbX=()()()()()22cov,cov,cov,abX YabX XY Y=+()5ab DXDYab=+=()2222cov,544abDDabab=+(2)12=时,()cov,1X YDXDY=,EaEXbEYb=,()22222cov,42Da DXb DYabX Yabab=+=+则()2242DababCvEb+=(3)因EaEYbEXa=,()22222cov,42Da DYb DXabX Yabab=+=+()22,42N aabab+故定义知的中位数为a,众数为a。(4)()()()()()22cov(,)cov,22abX Yab DXDYabab=+=+故2ba=或2ab=时,和不相关。又正态分布的独立性与相关性相同,故2ba=或2ab=时,和独立且不相关,否则不独立且相关。