第一章3.解：因为iixayc所以iixacy11niixxn1111niiniiacynnacyn1niicaynacy所以xacy成立因为2211nxiisxxn22122111niiiniiniiacyacyncycyncyyn又因为2211nyiisyyn所以222xyscs成立6.解：变换1027iiyx*ixiy-35-91234im234111liiiymyn13529312434101.52710yx=2211lyiiismyyn222212351.5391.54121.5341.510440.252214.4025100xyss7解：身高154:158158:162162:166166:170170:174174:178178:182组中值156160164168172176180学生数101426281282*11liiixmxn115610160141642617212168281768180210016622*11liiismxxn2222222110156166141601662616416628168166100121721668176166218016633.448解：将子样值重新排列（由小到大）-4，，，，，0，0，，，，，，172181203.2147.211.2ennenMXXRXXMXX9解：121211121211nnijijnxnxnnxnn112212nxnxnn12221121nniisxxnn1212221122111122121222222111222112212122222211221122112212121222211211122121nniinnijijxxnnxxnxnxnnnnnsxnsxnxnxnnnnnsnsnxnxnxnxnnnnnnnnnxnnsnsnn22212211222122222112212112212122121222212121122212122nnxnxnxnnnsnsnnxnnxnnxxnnnnnnxxnsnsnnnn12.解：ixP:iExiDx1,2,,in1122111111nniiiinniiiinEXExExnnnnDXDxDxnnnn13.解：,ixUab:2iabEx212ibaDx1,2,,in在此题中1,1ixU:0iEx13iDx1,2,,in112111101113nniiiinniiiiEXExExnnDXDxDxnnn14.解：因为2,iXN:0iXE1iXD所以0,1iXN:1,2,,in由2分布定义可知222111nniiiiXYX服从2分布所以2Yn:15.解：因为0,1iXN:1,2,,in1230,3XXXN:12303XXXE12313XXXD所以1230,13XXXN:2212313XXX:同理2245613XXX:由于2分布的可加性，故22212345612333XXXXXXY:可知13C16.解：（1）因为20,iXN:1,2,,in0,1iXN:所以22121niiXYn:11122YYyFyPYyP220yfxdx211'221YYyfyFyf因为2122202200nxnxexnfxx所以21122202200nynnYyeynfyy（2）因为20,iXN:1,2,,in0,1iXN:所以22221niiXnYn:22222220nyYnYnyFyPYyPfxdx222'22YYnynfyFyf故221222202200nnnynnYnyeynfyy（3）因为20,iXN:1,2,,in10,1niiXNn:所以22311niiXYnn:22333210ynYYFyPYyPyfxdxn233'2211YYyfyFyfnn22110200xexfxxx故23210200ynYeyfynyy（4）因为20,iXN:1,2,,in所以1224210,11niiniiXNnXYn::224224442210'2211yYYYyFyPYyPfxdxyfyFyf故24210200yYeyfyyy17.解：因为Xtn:存在相互独立的U，V0,1UN:2Vn:使UXVn221U:则221UXVn由定义可知21,Fn:18解：因为20,iXN:1,2,,in10,1niiXNn:221nmiinXm:所以1112211nniiiinmnmiiininXmXnYtmXnXm:（2）因为0,1iXN:1,2,,inm221221niinmiinXnXm::所以221122211,niniiinmnmiiininXmXnYFnmXnXm:19.解：用公式计算20.010.019090290U查表得0.012.33U代入上式计算可得20.01909031.26121.2620.解：因为2Xn:2En22Dn由2分布的性质3可知0,12XnNn:22XncnPXcPnn22212222limcnntnXncncnPedtnnnn故2cnPXcn第二章1.0000,0()0,0()()1()111xxxxxexfxxExfxxdxxedxxeedxex令从而有1x2.111121).()(1)(1)1111kkxxExkpppkpppp令1p＝X所以有1pX２）．其似然函数为1`11()(1)(1)nixiinXnniLPPppp1ln()ln()ln(1)niiLPnpXnp1ln1()01niidLnXndppp解之得11niinpXX３．解：因为总体Ｘ服从Ｕ（a，b）所以2122!2!!()1233niiabnEXrnrXXXXabSXSbXS222（a-b）（）D（X）=12令E（X）=D（X）=S，1S=na+b2（）a4.解：（1）设12,,nxxxL为样本观察值则似然函数为：111()(),01,1,2,,ln()lnlnlnln0nniiiniiiniiLxxinLnxdLnxdL（-1）解之得：11lnlnniiniinxnx（2）母体X的期望10()()1Exxfxdxxdx而样本均值为：11()1niiXxnExXXX令得5.。解：其似然函数为：1111111()2(2)1ln()ln(2)10niiixnxniniiniiLeeLnxx令得：（2）由于00011222111()()()xxxxnniiiixxEedxedxxeedxEExExnnnn以11niixn为的无偏估计量。6.解：其似然函数为：(1)(1)()()(1)!(1)!11kknnkxnxikiLxexeiikkiiln()ln(1)ln()11nnLnkkXXiiii1ln()0niidLnkdX解得1niinkkXX1(),0,fxx７．解：由题意知：均匀分布的母体平均数220，方差1212)0(222用极大似然估计法求得极大似然估计量似然函数：ninL11)(niiiixx1)(maxmin0选取使L达到最大取niix1max由以上结论当抽得容量为6的子样数值，，，，，，时2.2即,1.124033.0122.22.2...
