习题一、基本概念1．解：设12345,,,,XXXXX为总体的样本1）51151~(1, ) (,,)(1)iixxiXBpf xxppL555(1)11(1),5xxiippxx2）55155151!!),,()(~exexxxfPXiixiixi3）5155111~( ,) (,,),,1,...,5()iXUa bfxxaxib ibabaL所以5151,,1,...,5()(,,)0,axib ibaf xxL其他4）5122/55125121exp221),,()1,(~2iiixxexxfNXi2.解：由题意得：i 0 1 2 3 4 个数6 7 3 2 2 fxi0.3 0.35 0.15 0.1 0.1 因为0110,( ),1,nkkkxxkFxxxxnxx，所以40,00.3,010.65,12( )0.8,230.9,341,4xxxFxxxx3．解：它近似服从均值为172，方差为 5.64 的正态分布，即(172,5.64)N4．解：55-5510/2--kXkPkXPkXP因 k 较大-555(15)2510.950.95PXkkkkkkk,51.65,0.33kk查表5．解：-5250.853.81.14291.7143(1.7143)( 1.14296.3/ 6XPXP）0 1 0.9 0.8 0.7 0.6 0.5 0.4 0.30.2 0.1 1 2 3 4 x y 0.9564(10.8729)0.8293 6．解：~(20,0.3),~(20,0.2),~(0,0.5),0.30.30.3YNZNYZYZNP YZP YZP YZ设与 相互独立，0.30.42430.42430.50.50.51(0.4243)(1(0.4243))22(0.4243)YZYZPP22 0.66280.67447.解：101010222111~(0,4),~(0,1),211110.05,0.95444444iiiiiiiiXXNNcccPXPXPX则查卡方分位数表c/4=18.31,c=73.24 8．解：由已知条件得：(1,),1()iXYBppF:由iX 互相独立，知iY 也互相独立，所以1( ,),1().niXiYB n ppF:9.解: 1) )1(,)1(,2pNpDXESnpNpnDXXDNpEXXE2) DXESnnDXXDEXXE2,,3) 12,12,2222abDXESnabnDXXDbaEXXE4) 1,1,2DXESnnDXXDEXXE10.解：1) 22212)1()1()1()1(nDXnESnSnEXXEnii2)222242221(1)(1)(1),~(1)niinSnSDXXD nSDn2412(1)niiDXXn11.解：nXEdtedyeydyeyXnEYEnnDYXEEYNXnYnNXtyy2)(,2)1(222222||21)(),11,0(),1,0(~),/1,0(~)1020222令211,2)1(222222||21),1,0(~)211020222niiniitxxXEnXnEdtedxexdxexXENX12.解：1) 22242/XE XE XEnn244 100.12 /2 /XXDEnnnn40n2) 2222011,22 /2 /22uuXXu Eueduuedunn22222002200222222222222222(1),0.1,2/2 280010,254.6,255uuutueduueduuede dtXE XEnnnnnn3) 111222/nXnPXPXPn0.975210.95,2221.96,15.36,162nnnnunn13.解：11222111 1111,nniiiiYXYXaXnaXanbb nbEYEXaSSbb14．解：1）12345~(0, 2),~(0,3)XXNXXXNQ34512 ~(0,1),~(0,1)23XXXXXNN且122XX与3453XXX相互独立1111,,2.23cdn2）2345222212 ~(2),~(1)3XXXXX22122234523~(2,1),,2,123XXFcmnXXX15.解：设1(1, )pFn，即()1...
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