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1、数理统计课后习题答案 基本概念1解：设 为总体的样本 1）2）3）所以4）2.解：由题意得：i 0 1 2 3 4 个数 6 7 3 2 2 f xi0.3 0.35 0.15 0.1 0.1 因为 ，所以1 2 3 4 5, , , , X X X X X511 51 (1, ) ( , , ) (1 )i ix xiX B p f x x p p-= -55 5(1 )11(1 ) ,5x xiip p x x-= - =l ll ll5515 515 1!) , , () ( -=-= = exexx x f P Xiixi ix iL51 5511 1 ( , ) ( , , ) ,
2、, 1,.,5( )iX U a b f x x a xi b ib a b a= = =- -51 51, , 1,.,5( ) ( , , )0,a xi b ib a f x x =- = 其他( ) -= = =-=-5122 / 55125 121exp 221) , , () 1 , ( 2iiixx e x x f N Xippm L0110,( ) ,1,n k kkx xkF x x x xnx x+ 40, 00.3,0 10.65,1 2( )0.8,2 30.9,3 41, 4xxxF xxxx 3解：它近似听从均值为 172，方差为 5.64 的正态分布，即4解：因
3、k 较大(172,5.64) N( )55-5510 / 2- - - = = kXk P kXP k X Pm mm( ) ( ) ( ) ( ) ( ) ( )( )- 5 5 5 (1 5 ) 2 5 1 0.95 0.95P X k k k k k kkm F -F - =F - -F = F - =F =0 1 0.9 0.8 0.70.6 0.50.4 0.30.2 0.1 1 234 x y5解： 6解：7.解： 查卡方分位数表 c/4=18.31,c=73.248解：由已知条件得：由 相互独立，知 也相互独立，所以9.解: 1)2)3)4),5 1.65, 0.33 k k =
4、 = 查表( )-5250.8 53.8 1.1429 1.7143 (1.7143) ( 1.14296.3/6XP X P = - = - + - = + = - = = 则(1, ), 1 ( )i XY B p p F m = -iXiY1( , ), 1 ( ).ni XiY B n p p F m= -) 1 ( ,) 1 (,2p Np DX ESnp NpnDXX D Np EX X E - = =-= = = =lll = = = = = = DX ESn nDXX D EX X E2, ,( ) ( )12,12,2222a bDX ESna bnDXX Db aEX X
5、E-= =-= =+= =1 ,1,2= = = = = = DX ESn nDXX D EX X E m 10.解：1)2)11.解： 12.解：1)2) ( )2 2 212) 1 ( ) 1 ( ) 1 ( ) 1 ( s - = - = - = - = -=n DX n ES n S n E X X Enii( )2 222 4 22 21( 1) ( 1)( 1) , ( 1)niin S n SD X X D n S D n s cs s= - - = - = - ( )2412( 1)niiD X X n s= - = -pp p p p pnX Edt e dy e y dy
6、e y X nE Y Enn DY X E EY N X n Y n N Xty y2) (,2) 1 (222222| |21) () , 11, 0 ( ), 1 , 0 ( ), / 1 , 0 ( ) 102022 2= G = = = = = = = = =- +- +- + - Q 令pp p p p p2 1 1,2) 1 (222222| |21), 1 , 0 ( ) 21 102022 2= = G = = = = = =- +- +- + -niiniitx xX EnXnEdt e dx e x dx e x X EN X( )222 42/XE X E X En nm
7、m m - = - = ( )24 41 0 0.12/ 2/X XD En n n nm m - -= + = + 40 n 2 22 201 1, 22/ 2/ 2 2u uX Xu E u e du u e dun nm mp p+ +- - -= = = 3) 13.解： 14解：1） 且 与 相互独立2）2 222 20 0220 02 22 22 2 22 2 2 22 2 2 2(1) , 0.1,2/2 2 80010, 254.6, 255u uutue du ue duue d e dtXE X En n nn n np pp p pmmp pp p+ +- -+ +-=
8、= = =-G = - = = = ( ) ( )1 1 12 2 2/n X nP X P X Pnmm m - = - - = - 0.9752 1 0.95,2 2 21.96, 15.36, 162n n nnu n n =F -F - = F - = ( ) ( )( )1 12 221 1 1 1 11 1,n ni ii iY XY X a X na X an b b n bEY EX a S Sb b= = = - = - = - = - = 1 2 3 4 5 (0,2), (0,3) X X N X X X N + + +3 4 5 1 2 (0,1), (0,1)2 3X
9、 X X X XN N+ + +1 22X X +3 4 53X X X + +1 11 1, , 2.2 3c d n = = =( )23 4 52 2 2 21 2 (2), (1)3X X XX X c c+ +15.解：设 ，即 16解：17.证明：1）( )( )2 21 22 23 4 523 (2,1), , 2, 123X XF c m nX X X+= = =+ +1(1, )pF n a-= ( ) 1 ( ) 1 P F p P F p a a a = - - = -( ) ( ) 12 ( ) 2( ) 12P T P T pP T ppP Ta aaa - - =
10、- = - = -122112( )( ) (1, )pp pt nt n F naa- = = =( )( ) ( )( )( ) ( )( )( ) ( )1 2 1 21 2 1 22 2 2 2 21 2 1 2 1 22 21 2 1 22 2 2 21 2 1 2 1 2 1 221 22 21 2 1 2 (0,2), (0,2), (0,1), (0,1)2 2( ) (1),( ) ( ) (2)2 2 2221 0.1,2X X X XX X N X X N N NX X X X X XX X X X tP t PX X X X X X X XX X tPX X X Xcc
11、 c+ -+ -+ + -+ + + = + + - + + - + = - = + + - =0.9 (1,2)8.532tF = =2 21 1 12 22121221 1( ) 0, ( ) , (0, )1( 1)1( 1)1 11n n nnnn nE X X D X X X X Nn nnS nX XnX X nnt nn S nSns scsss+ + + +- = - = -+- = -+ -又 2）3）18. 解：19.解20.解：21. 解：1)因为 ，从而2 21 1 11 1( ) 0, ( ) , (0, )n n nn nE X X D X X X X Nn ns
12、s+ + + +- = - = -2 21 1 11 1( ) 0, ( ) , (0, )n nE X X D X X X X Nn ns s- - = - = -( ) ( )( ) 62 , 47 . 61 , 96 . 1 25 . 0 , 975 . 0 25 . 0, 95 . 0 1 25 . 0 2 25 . 0/25 . 0 25 . 0975 . 0 = F - F =- - = -n n u n nn nnXn P X Psms m , 0,1, , ( ) , ( ) ,0, , 1,X U a bx ax a b x af x F x a x b b ab ax a
13、bx b - = = 1(1) ( )(1 ( ) ( )nf x n F x f x- = -111( )1( ) , , 0, , 1( ) , , ( ) ( ( ) ( )0, , nnnnb an x a bb a b ax a bx an x a bf x n F x f x b a b ax a b- = - - = = - -( ) ( ) ( ) ( ) ( )( ) ( )5 5(1) (1)1 1515 5 55 55 5(5)1 110 1 10 1 10 1 1 10121 1 121 (1 ( 1) 1 (1 1 (1) 1 (1) 0.57851215 15 1.
14、5 (1.5) 0.9332 0.70772i ii iiiiii iP X P X P X P XXPXP X P X P= = = = - = - = - - - = - - - = - -F - = - - +F = -F =- = = SSPSSP FSS( )( ) 94 . 0 05 . 0 99 . 0 57 . 37 85 . 10) 20 ( ), 1 , 0 ( ), , 0 ( 22012 22220122= - = = -=-=cc csmsmsms mPXXNXN Xii iiii25. 解：1)2)26.解：1)2) 3) 27解：28.解：( )895 . 0 1
15、 . 0 995 . 0 58 . 381965 . 11 ), 19 ( 192222222012= - = =-=scs sSPSX Xii( ) 4532 . 0 7734 . 0 2 2 1 ) 75 . 0 ( 2 1431435 / 20803 80 = - = + F - = - =-= - U PXP X P( ) ( ) 05 . 0 1 975 . 0 2 1 064 . 2 1 064 . 25 / 2674 . 7803 80 = + - = - =-= - T PXP X P8413 . 0 120472 . 4472 . 4=-=-= + s ss a XPa XP
16、 a X P2 2 2 2 2 22 2 2 2 223 1 32 2 2 2 2 2 2SP S P S P S Ps s s s ss ss - = - - = = 22199.5 28.5 0.95 0.05 0.9SPs = -=-=-cc cT PcT Pc SXPc SXP cXSPm mm22cov( , )( , )( ) ( )1( ) ( )1cov( , ) ( )1( , )1i ji ji ji ji j i j i ji jX X X Xr X X X XD X X D X XnD X X D X XnX X X X E X X X X X X X Xnr X X
17、X Xnss- - - =- - = - =- - = - - - = - - - = -习题二、参数估计1 解：矩估计所以 ，2解：1） 无解，依定义：2）矩法：( )2 2 21212) 1 ( 2 ) 1 ( , ) 1 (, 21), 2 , 2 ( ss m- = - = - = - = = + =+n ES n ET S n Y Y TX YnY N X X YY Yniinii i n i i令( )1 3.40.1 0.2 0.9 0.8 0.7 0.76 6X = + + + + + =( ) ( )1 111 1ln ln( 1) lnn nni ii iniiL x xL
18、 n xa aa aa a= = = + = + = + + 121ln ln 01ˆ 1 0.2112lnniiniid nL xdnxaa aa= = + =+= - - =3077 . 012 1ˆ ,212) 1 ( ) 1 (1101021=-= =+=+ = + = +XXXxdx x EX aaaaa aaa1 211 2ˆ ˆ , 11lnniiX nXXa a= - = = - +- 1 2ˆ ˆ 0.3079, 0.2112 a a 3077 . 0 2ˆ,21= = = = X X EX qq11 1l
19、n 0nniLnLq qq= = - =21ˆmaxii nX q =21 1ˆ ˆ1.2, 0.4722 12EX DXq q= = = =极大似然估计：31）解：矩法估计：最大似然估计： 2)解：矩估计：最大似然估计： 3）解： 矩估计：联立方程： 极大似然估计：依照定义，4) 解： 矩估计：，不存在 2 2ˆ ˆ1.1, 0.18332 12EX DXq q= = = =11 1ˆ, EX XXll= = =11 1,ln lnnii in n xx nii iL e e L n L xlll l l=-= = = = - 21
20、11ˆln 0,ni niiid n nL xd Xxll l= - = = = ( ) X P lX X EX = = =1ˆ, l l1,ln lnix nxnniii iL e e L n nx xx xl ll ll l- -= = = - + - 2ˆln 0,d nxL n Xdll l= - + = =( )2,2 12b a a bEX DX- += =( )2*2*2*221ˆ 32ˆ3a X Mb X Ma bXb aM= -+ =-= + 1 1ˆˆ min , maxi ii n i na X b X
21、 = =00ln EX dx xxqq+= =，无解；故，依照定义，5)解： 矩法：即极大似然估计： 无解，依定义有：7)解：矩法： 2 21 11,ln ln 2 lnn nnii ii iL L n xx xqq q= = = = - ln 0nLa q= =(1)ˆX q =( )/0( ) (1) (2)x txEX e dx t e dta baa b a bb+ +- - -= = + = G + G X a b = + =2 2 2 20( ) (1) 2 (2) (3)tEX t e dt a b a ab b+-= + = G + G + G2 2 2 2 2212
22、 2 ( )iM Xna ab b a b b = + + = + + = =22 22 2 2 2 *2* *1 11ˆˆ ,iM X X X MnX M Mba b= - = - = - =2 2* 2 * 21 11 11 1ˆˆ ( ) , ( )n ni ii iX M X X X M X Xn na b= = - = - - = = - ( )( )/11 1 1exp ,ln lninx ninL e nx n L n nxa bab a bb b b b- - -= = = - - = - - + 2ln 0, ln ( ) 0n n n
23、L L x aa b b b b = = = - + - = a(1) (1)ˆˆ ,L LX X X X a b a = = - = -2 22 23 2 2230 0 04 2 2 2(2)x xtx x xEX e dx e d te dt Xq qq qq q p q p p p+ + +- -= = = = G = ˆ2MX pq = 极大似然估计：8)解：矩法： 极大似然估计： 4 解：记 则 ；5.解：222 2223 31 14 4iix nxn nii ii ixL e x eq qq p q p- -= = = = 222ln ln4 3 ln
24、 ln lniixL n n n x q pq= - - + -2233 2ˆln 2 0,3iL ixnL xnqq q q= - + = =2 22 2 2 22 22 2 02 22 2 22 2 30( 1) (1 ) (1 ) (1 )(1 ) (1 )1 2 21x x xx x xxxd dEX x xd dd dq Xdq dq qq q q q q qq qq q qq q -= = = - - = - = - -= = = = =- 2ˆMXq =2 2 2 21( 1) (1 ) (1 ) ( 1)ln 2 ln ( 2 )ln(1 ) ln( 1)i
25、nx n nx ni iiiL x xL n nx n xq q q qq q- -= - - = - -= + - - + - 2 2 2ˆln 0,1Ln nx nLXqq q q -= - = = -1 111 211 2( , , ) (1 ) (1 )ln ( , , ) ln (1 )ln(1 )n ni ii i i iy y ny ynninL p y y y p p p pL p y y y ny p n y p= =-= = - = -= + - -1 2( , , ) 0(1 )ny p dL p y y y ndp p p-= =-ˆ pY =0 0
26、1, ; 0,i i i iy x a y x a = = (1, )iY B p1,ln lninx n nxiL e e L n nxl ll l l l- -= = = - 6 解：因为其寿命听从正态分布，所以极大似然估计为：依据样本数据得到：。由此看到，这个星期生产的灯泡能运用 1300 小时的概率为 0 7.解：由 3.2)知所以平均每升氺中大肠杆菌个数为 1 时，出现上述状况的概率最大。8 1）解： 2）解：， 9 解：由极大似然估计原理得到 10 解：应当满意：711 1 20000ˆln 0, , 201000 1000i iid nL nx X xvd Xll l=
27、 - = = = = =1ˆ0.05Xl = =2 211ˆ ˆ , ( )niix xnm s m= = -2ˆ ˆ 997.1, 17235.811 m s = =( ) . 1 50 / 1 * 4 2 * 3 10 * 2 20 * 1 17 * 0ˆ= + + + + = = XLl 0.950.95, 0.95ˆ( ) 0.95,x Ap x A pAA U sm ms smms - - = =-F = = +0.95ˆˆ , x A U X m = = +2 2ˆ ˆ ,
28、 x S m s = =122( 1)( 1)xun Snamss-0.95ˆ( ) 0.95,AA u S XSmf-= = +4 41 1 11 ;3 6 1049762 0; 3 0pp pqq q= = = = = =当 时，当 时， 当 时，ˆ1 q =ˆq 结果取决于样本观测值11.解：无偏， 方差最小 所以：12、1）解：2） 13解：( ) ( ) ( ) ( )1 2 1 20,11 1ˆ, ; max , ; max ;0 , ;1 ,n nn n i ii iL x x x L x x x f x f xq qq q= = = =
29、0,10.511max 1,2nniixq = = ( )1 2,nx x x2 2 2 2 21 11 1 1 1 5ˆ ˆ ( ) ( ) , ( ) ( )6 3 36 9 18E D a a a a a a a s s s s s = + + + = = + + + =2ˆ ( 2 3 4 )/5 2 E a a a a a a a = + + + = 2 23 3 11 4 1ˆ ˆ ˆ ( ) ,4 16 4E D D a a a a a a a s s a = + + + = = = 1 2ˆ ˆ ,
30、 a a3ˆ a2 3 1ˆ ˆ ˆ D D D a a a ( )212 2 2 2 2 2 21 111ˆ 1 2( 1)2 2( 1)( ) 2( 1) ni i i iiE Eknx x x x n nk ks s m m s s-+ +=- + = - + - - = =2( 1) k n = -1, 21 10,nkk k ii i ixn ny x x x Nn n ns= - - = - = - 令222( 1)2( 1)122( 1)xnnix nE y e dxn nnk nss spp sp-+ - = = =- = -
31、14 证明： 151）解： 是的 无偏估计 2）解： 可以看出 最小。16解：22 2 2 2 2 2 2 2 2( ) E X cS EX cES DX E X c cnss m s m - = - = + - = + - =1cn =( )( )1, 1,1 1( )( )n nk kk k i k k ii i i ix yn nx x y y x yn n n n= = - - - = - - 21, 1, 1, 1,2 2 2 2( 1) ( 1)( 1)n n n nk i k i k kk k i k k i k k i k k ii in y x n x y x ynx yn
32、n n n= = = = - -= - - + ( )( )2 22 22( 1) ( 1)2( 1) ( 1)( 2) ( 1) ( 1)i in nE x x y y Exy ExEyn nn Exy n n ExEy n nExy ExEyn n n- - - - = - + - - - -+ = -1 ( 1) ( 1)ˆ( ) cov( , )1n nEZ n Exy ExEy Exy ExEy X Y Zn n n- -= - = - = =-( )22 2 2 2 21 11 1 1( ) ( ( ) ( )1 1 1n ni i ii iES E X X E X nX
33、 EX nEXn n n= = - = - = - - - 22 2 2 21 ( ) 1n nn nss m m s = + - + = - 2S 2s( )22 2 4 2 2 2 421 2 22( 1), ,1 1nD S n DS E S DSn ns s ss- = - = - = = - - ( ) ( ) ( )( ) ( ) ( )22 2 2 2 2 2 2 41 1 1222 2 2 2 2 2 2 42 2 22 1( )2( )1nE S D S E SnE S D S E Sns s s ss s s s- = - + - =- = - + - =+( )22 22
34、E S s -比较有效 17.解： 18.解： 11 1(1)0 01 110 044 3 (1 ) (1 )34 4(1 ) (1 )3 1n nn nx x nE X n dx t tdtnt tdt t tdtnqqq qq qq- -= - = - = - - - = = + ( )11( ) ( )0 04 4 4 4 4( )3 3 3 3 3 1n nn nx x n nE X EX n dx t tdtnqq qqq q-= = = = =+ (1) ( )3,4 4nEX EXqq = =2122 2 2 2 2 2(1)0 01 1 1 13 1 3 (1 ) 3 3 5
35、2 10x xEX dx t t dtqq q qq q = - = - = + - = 2122 2 4 2 2( )0 01 33 3 35 5nx xEX dx t dtqq q qq q = = = = 2 22 2 2(1) (1) (1) (1)34 16 16( ) 16( )10 16 5D X DX EX E Xq qq = = - = - =22 2 2 2( ) ( ) ( ) ( ) (1)4 16 16 16 3 9 3( ) ( ) 43 9 9 9 5 16 15 5n n n nD X DX EX E X D Xqq q = = - = - = =( )43nX
36、21 22, 2 D D q q s = =1 1 2 2 1 2 1 2 1 2 2 1( ( ) 1 1 E c c c c c c c c c c q q m m m m + = + = + = + = = - ）， ，( ) ( )22 2 2 2 2 21 1 2 2 1 2 1 1( 2 2 2 1 D c c c c c c q q s s s + = + = + - ）( )22 21 1 1 11 2 12 1 3 2 1 min2 1 212*3 3 3c c c cc c c+ - = - + -= - = = - = ，11! !ixnn nxii iL e ex xl
37、 lll- -= =ln ln ln !iL n nx x l l =- + - 是有效估计量，19.解： 留意：T 是有效估计量，201）解：2）T 是有效估计量，( ) ln , ( ) ,d nx n nL n x c EX EXdl l ll l l l= - + = - = = =T X =1( )DTc nll= =1 11L ,ln ln ( 1) lni inn n niix x L n x q q q q- -= = = + - 1 1 1ln ln ln , lni i id nL x n x T xd n n q q q = + = - - - = - 1 1 11 1
38、100 0 01ln ln ln lniE X x x dx xdx x x x dxq q q qqq- -= = = - = - 1ETq=1011lnln 0xxx x xx xq qq qqq- - -= = = - =-221( ) 1( )gDTc n nqqq q-= = =-11(1 ) (1 ) ,ln ln ( )ln(1 )inx n nx niL p p p p L n p nx n p- -= - = - = + - -1ln ,1 1d n nx n nL x T Xdp p p p p -= - = - - = - - 1 11 1 1(1 p) (1 p)n n
39、k k kk k kdEX EX kp p k p qdq- -= = = = - = - = 201 11nkkd d pp q pdq dq p p p= = = =-( )1nc pp= -是相合估计量。21.解：T 是有效估计量22.1）解： 2）所以是有效估计量 3）2 22( ) ( ) 1 ( ) 1( ) ,(1 p ( )0,(nc p g p g pI p DTn p c npDX qDXn npqq -= = = =-= = ）T X =1ln ln,1 1innx nxiLq qq qq q= = = - - ( ) ( ) ( ) ln ln ln ln 1 ln L
40、 n nx q q q = - - +( )ln 1ln ,ln 1 1 lnd n n nxL xd nq q q qq q q q q q q - += - + = - - - - ( )( )ln 1,1 lnT X gq q qqq q- += =-( )( )11 120 00ln ln ln ln 1 ,1 1 1 ln 1 lnlnx xx xx xEX EX dx x dxq q q q q q q qq qq q q q q qq - += = = = - = - - - - ( ) ( )111 12 2ln ln (n )ln(1 )xi xinxi n xiiLL xi xiq qq qq q- -= = - = - = + - - n1ln1 (1 ) n1ˆTxi xid nL xi