NGUYN VN THN 9/2011 BI TP XC SUT V THNG K TON MclcI BITP 41 Tphp-Giitchthp 11.1 Tphp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Giitchthp. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22 Bincvxcsut 52.1 Binc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52.2 Xcsutcin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62.3 Xcsuthnhhc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72.4 Cccngthctnhxcsutcbn . . . . . . . . . . . . . . . . . . . . . . . . . 72.5 Cngthcxcsuty,cngthcBayes . . . . . . . . . . . . . . . . . . . . 113 Binngunhinvhmphnphi 144 Mtsphnphixcsutthngdng 234.1 PhnphiBernoulli,nhthc . . . . . . . . . . . . . . . . . . . . . . . . . . . . 234.2 PhnphiPoisson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 264.3 Phnphichun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 285 Lthuytmu 316 clngthamsthngk 346.1 clngtrungbnhtngth. . . . . . . . . . . . . . . . . . . . . . . . . . . . 346.2 clngtltngth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36MCLC 36.3 Tnghp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 377 Kimnhgithuytthngk 397.1 Sosnhkvngvimtschotrc . . . . . . . . . . . . . . . . . . . . . . . . 397.2 Sosnhhaikvng. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 427.3 Sosnhtlvimtschotrc . . . . . . . . . . . . . . . . . . . . . . . . . . 447.4 Sosnhhaitl. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45II BIGII 46PhnIBITPChng1Tphp-Giitchthp1.1 TphpBitp1.1. Cho dy tp hp A1, A2, . . . , An, . . ..Chng minh rng lun lun tn tidy tphpB1, B2, . . . , Bn, . . .,saocho:(a) CcBitngimtrinhau;(b)

i=1Ai=

k=1Bk.Bi tp1.2. ChngminhrngcchthcsauytngngnuAvBltphpconca:A B= , A B, B A.Bitp1.3. KhngnhchorngnuA, B, CltphpconcatphpsaochoA B CvB A C, thB= ,cngkhng?Bitp1.4. ChngminhrngnuA, B, Clcctphpconcatphp,saochoA B CvA C B, thA C= Bitp1.5. Tmbiuthcngincaccbiuthcsau:(a) (A B)(A C)(b) (A B)(A B);(c) (A B)(A B)(A B)(d) (A B)(A B)(A B)1.2Giitchthp 2(e) (A B)(B C)Bitp1.6. Hthcnotrongcchthcsauyng(a) A B C= A (B \ AB) (C \ AC)(b) A B= (A \ AB) B(c) (A B) \ A = B(d) (A B) \ C= A (B \ C)(e) ABC= AB(C B)(f) AB BC CA ABC(g) (AB BC CA) (A B C)(h) ABC A B(i) A BC= AC BC(j) A BC= C \ (C(A B))Bitp1.7. Chngminhrng:(a) A B A B= A(b) (A B)AB= AB BABitp1.8. Chngminh(a) NuA B= ABthA = B(b) A BC (A B)C(c) NuA1 A, B1 BvA B= thA1 B1= 1.2 GiitchthpBitp1.9. Mtlhngc50snphm.(a) Cbaonhiucchchnngunhincnglc5snphmkimtra?(b) Cbaonhiucchchnngunhinlnlt5snphm?Bitp1.10. Trongmththnginthoinib3s1.2Giitchthp 3(a) cbaonhiumycccchskhcnhau?(b) Cbaonhiumycs9cuicnccchscnliukhcnhau?Bi tp1.11. Mtlphcc40hcsinhgm20namv20n.Cbaonhiucchchiatrongminalpc10namsinhv10nsinh?Bi tp1.12. Numtngi c6i vkhcnhauv4i giykhcnhau. Cbaonhiucchkthpgiavvgiy?Bi tp1.13. NmngiA,B,C,D,Esphtbiutrongmthingh.Cbaonhiucchspxp:(a) NgiBphtbiusauA.(b) NgiAphtbiuxongthnltB.Bi tp1.14. C6hcsinhcspxpngivo6chghisthttrnmtbndi.Tmscchxp(a) 6hcsinhvobn.(b) 6hcsinhnyvobnsaocho2hcsinhA,Bngicnhnhau.(c) 6hcsinhnyngivobnsaocho2hcsinhA,Bkhngngicnhnhau.Bitp1.15. Mt lp c 40 hc sinh. Gio vin ch nhim mun chn ra mt ban cn s lp:1lptrng,1lpph,1thqu.Higiovinchnhimcbaonhiucchchnbancnslp?Bitp1.16. Mthpc8bi,6bitrng,4bivng.Ngitachnra6bithp.Hicbaonhiucchchnnu:(a) Khngyucugthm.(b) Phic2bi,2bitrng,2bivng.(c) Cng2bivng.Bitp1.17. Mt n cnh st khu vc c 9 ngi. Trong ngy cn c 3 ngi lm nhim v a im A, 2 ngi a im B cn 4 ngi trc ti n. Hi c bao nhiu cch phn cng?Bi tp1.18. Mttsnxutc12ngi, trongc4n, cnchiathnh4nhmunhau.Hytmscchphnchiasaochominhmc1n?Bitp1.19. Xp12hnhkhchln4toatu.Tmscchspxp:(a) Mitoac3hnhkhch.1.2Giitchthp 4(b) Mt toa c 6 hnh khch, mt toa c 4 hnh khch, 2 toa cn li mi toa c 1 hnh khch.Bitp1.20. Gism, n, rlccsnguyndng.ChngminhrngC0mCrnm + C1mCr1nm + + CrmC0nm= CrnBitp1.21. Chngminhrng(a) C1n + 2C2n + + nCnn= n2n1(b) 2.1.C2n + 3.2.C3n + + n(n 1)Cnn= n(n 1)2n2Bitp1.22. Chom, n, rlccsnguyndng.Chngminhrng(a)m

k=0Crnk= Cr+1n+1Cr+1nm(b)m

k=0(1)kCkn= (1)mCmn1Bitp1.23. Chngminhrng_C0n_2+_C1n_2+ + (Cnn)2= Cn2nBitp1.24. Chngminhrngn

k=02n!(k!)2[(n k)!]2= (Cn2n)2Chng2Bincvxcsut2.1 BincBitp2.1. Khinothcccngthcsau:(a) A + B= A(b) AB= A(c) A + B= ABHaiskinAvA + Bcxungkhckhng?Bi tp2.2. Mt chictuthygmmt bnhli, 4ni hi, 2tucbin. Gi A, Bi(i =1, . . . , 4), Cj(j= 1, 2) ln lt l cc s kin bnh li hot ng tt, ni hi th i hot ng tt,tucbinthjhotngtt.Bitrngtuhotngttkhivchkhibnhli,tnht1nihi vtnhtmttucbinuhotngtt. Gi Dlskintuhotngtt. HybiudinDvDquaA, Bi, Cj.Bitp2.3. C 4 sinh vin lm bi thi. K hiu Bi(i = 1, . . . , 4) l bin c sinh vin th i lmbithityucu.Hybiudinccbincsauy:(a) Cngmtsinhvintyucu.(b) Cngbasinhvintyucu.(c) Ctnhtmtsinhvintyucu.(d) Khngcsinhvinnotyucu.Bitp2.4. Xtphpth:Gieomtxcxc2ln.Mtkhnggianbincscpngviphpthtrn?2.2Xcsutcin 6Gi A: Tngsntchiahtcho3, B: Tr tuyti cahiusntlschn. BiudinA, B?Bitp2.5. ChoA, Blhaibincngunhinbit.TmbincXththc:X + A + X + A = BBi tp2.6. Xtphpth:Bnkhnghnchvo1biachonkhitrngbialnutinthdng.Biudinkhnggianbincscpcabinctrn.Chramthyccbinc.Bi tp2.7. Gieohaiconxcxccnivngcht.GiAilbincxyrakhisntmttrnconxcxcthnhtli(i = 1, . . . , 6),Bkbincxyrakhisntmttrnconxcxcthhailk(k = 1, . . . , 6).(a) HymtccbincA6B6, A3B5(b) Vitbngkhiuccbinc:A: hiugiasntmttrnconxcxcthnhtvthhai ctr stuytibngba.B:sntmttrnhaiconxcxcbngnhau.(c) Hychramtnhmyccbinc.2.2 XcsutcinBitp2.8. Mtnhmnngixpngunhinthnhmthngdi.(a) Tmxcsut2nginhtrcngcnhnhau.(b) Tmxcsut2ngingcchnhau2ngi.(c) Tmxcsut2ngingcchnhaurngi(0 < r < n 2).(d) Xttrnghpkhihxpthnhmtvngtrn.Bi tp2.9. Thangmycamttanh7tng,xutphtttngmtvi3ngikhch.Tnhxcsut:(a) Ttccngratngbn.(b) Ttccngramttng.(c) Mingiramttngkhcnhau.2.3Xcsuthnhhc 7Bi tp2.10. Cnqucucphnngunhinlnltvonhp,mihpcthchanhiuqucu.Khiphnbithpvcu,tmxcsutmihpchamtqucu.Bitp2.11. Cho mt l hng gm n sn phm trong c m sn phm xu. Ly ngu nhintlhngksnphm.Tmxcsutsaochotrongssnphmlyracngssnphmxu(s < k).Bi tp2.12. Tagieolintip4lnmtngtincningcht.Tmxcsutcaccbinc:(a) A:Chaimtsp.(b) B:Cbamtnga.(c) C:Ctnhtmtmtsp.Bitp2.13. Mihaisnphmcspngunhinvobahp.Tmxcsuthpthnhtcchabasnphm.Bitp2.14. Gieongthihaiconxcxcngchtcninlnlintip.Tmxcsutxuthintnhtmtlnhaimttrncngc6nt.2.3 XcsuthnhhcBi tp2.15. Mtthanhstthngcbthnhbakhcmtcchngunhin. Tmxcsutbakhctocthnhmttamgic.Bitrngthanhstdil(nvdi.)Bi tp2.16. (Bi tonButffon)Trnmtphngcccngthngsongsongcchunhau2a, gieongunhinmtcykimcdi 2l (l P(A), thP(B|A) > P(B)Bitp2.35. GisP(AB) = 1/4,P(A|B) = 1/8vP(B) = 1/2.TnhP(A).Bitp2.36. Bitrngtanhnctnhtmtmtngatrong3lntungngxuclp.Hixcsuttcc3mtngalbaonhiu?Bi tp2.37. Tungmtconxcschailnclpnhau.Bitrnglntungthnhtcsntchn.Tnhxcsuttngsnthailntungbng4.Bitp2.38. GisP(A) = P(B) = 1/4vP(A|B) = P(B).TnhP(AB).Bi tp2.39. Bnlintipvomtmctiuchonkhi cmtvinnutinri vomc tiu th ngng bn. Tm xc sut sao cho phi bn n vin th 6, bit rng xc sut trngchcamivinnl0.2vcclnbnlclp.Bi tp2.40. GisccbincA1, . . . , AnclpcxcsuttngngP(Ak)=pk(k=1, . . . , n).Tmxcsutsaocho:(a) khngmtbincnotrongccbincxuthin.(b) ctnhtmtbinctrongccbincxuthin.Tsuyracngthckhaitrintchn

k=1(1 pk)Bi tp2.41. CbatiuchphbinchovicchnmuamtchicxehiminolA:hp s t ng, B: ng c V6, v C: iu ha nhit . Da trn d liu bn hng trc y, tacthgisrngP(A) = 0.7,P(B) = 0.75,P(C) = 0.80,P(A+B) = 0.80,P(A+C) = 0.85,P(B +C) = 0.90vP(A+B +C) = 0.95,viP(A)l xcsut ngimua btkchn tiuchA,v.v. . . .Tnhxcsutcaccbincsau:(a) ngimuachntnhtmttrong3tiuch.(b) ngimuakhngchntiuchnotrong3tiuchtrn.(c) ngimuachchntiuchiuhanhit.(d) ngimuachnchnhxcmttrong3tiuch.2.5Cngthcxcsuty,cngthcBayes 112.5 Cngthcxcsuty,cngthcBayesBi tp2.42. GisP(B|A1)=1/2, P(B|A2)=1/4vi A1vA2lhai bincngkhnngvtothnhmthyccbinc.TnhP(A1|B).Bitp2.43. Mt hp ng 10 phiu trong c 2 phiu trng thng. C 10 ngi ln ltrtthm.Tnhxcsutnhncphnthngcamingi.Bitp2.44. Chaihpngbi.Hp1ng20bitrongc5biv15bitrng.Hp2ng15bitrongc6biv9bitrng.Lymtbihp1bvohp2,trnurilyramtbi.Tnhxcsutnhncbi?bitrng?Bi tp2.45. Trongmtvngdnc,c100ngithc30ngihtthucl.Bittlngi b vimhngtrongsngi htthucll60%, trongsngi khnghtthucll30%.Khmngunhinmtngivthyngibvimhng.(a) Tmxcsutngihtthucl.(b) Nungikhngbvimhngthxcsutngihtthucllbaonhiu.Bitp2.46. MttrungtmchnonbnhdngmtphpkimnhT.Xcsutmtngi ntrungtmmcbnhl0.8. Xcsutngi khmcbnhkhi phpkimnhdngtnhl0.9vxcsutngikhmkhngcbnhkhiphpkimnhmtnhl0.5.Tnhccxcsut(a) phpkimnhldngtnh.(b) phpkimnhchoktqung.Bitp2.47. Mt cp tr sinh i c th do cng mt trng (sinh i tht) hay do hai trngkhc nhau sinh ra (sinh i gi). Cc cp sinh i tht lun lun c cng gii tnh. Cc cp sinhi gi th gii tnh ca mi a c lp vi nhau v c xc sut l 0.5. Thng k cho thy 34%cpsinhiltrai;30%cpsinhilgiv36%cpsinhicgiitnhkhcnhau.(a) Tnhtlcpsinhitht.(b) Tmtlcpsinhithttrongscccpsinhiccnggiitnh.Bitp2.48. C 10 hp bi, trong c 4 hp loi I, 3 hp loi II, cn li l hp loi III. HploiIc3bitrngv5,hploiIIc4bitrngv6bi,hploiIIIc2bitrngv2bi.(a) Chnngunhinmthpvtlyhha1bi.Tmxcsutcbi.(b) Chnngunhinmthpvtly1bi th cbi trng. Tmxcsutbi lyrathucloiII.2.5Cngthcxcsuty,cngthcBayes 12Bi tp2.49. Chailsnphm,lthnhtc10snphmloiIv2snphmloiII.Lthhai c16snphmloi Iv4snphmloi II. Tmi ltalyngunhinmtsnphm. Sau , t 2 sn phm thu c ly h ha ra mt sn phm. Tm xc sut sn phmlyrasaucnglsnphmloiI.Bi tp2.50. C2lg. Lthnhtgm15con, trongc3congtrng. Lthhaigm20con,trongc4gtrng.Mtcontlthhainhysanglthnht.Sautlthnhttabtngunhinramtcon.Tmxcsutcongbtralgtrng.Bitp2.51. Bamytngsnxutcngmtloichitit,trongmyIsnxut25%,my II sn xut 30% v my III sn xut 45% tng sn lng. T l ph phm ca cc my lnltl0.1%; 0.2%; 0.4%.Tmxcsutkhichnngunhinra1snphmtkhoth(a) cchititphphm.(b) chititphphmdomyIIsnxut.Bitp2.52. Gi s 3 my M1, M2, M3sn xut ln lt 500, 1000, 1500 linh kin mi ngyvi t l ph phm tng ng l 5%, 6% v 7%. Vo cui ngy lm vic no , ngi ta ly mtlinhkincsnxutbi mttrong3mytrnmtcchngunhin, ktqulcmtphphm.TmxcsutlinhkinnycsnxutbimyM3.Bitp2.53. Ba khu pho cng bn vo mt mc tiu vi xc sut trng ch ca mi khul0.4; 0.7; 0.8.Bitrngxcsutmctiubtiuditkhitrngmtphtnl30%,khitrng2phtnl70%,cntrng3phtnthchcchnmctiubtiudit.Gismikhuphobn1pht.(a) Tnhxcsutmctiubtiudit.(b) Bitrngmctiub tiudit. Tnhxcsutkhuth3cnggpvothnhcng.Bitp2.54. HpIc10linhkintrongc3bhng.HpIIc15linhkintrongc4bhng.Lyngunhintmihpramtlinhkin.(a) Tnhxcsutc2linhkinlyrauhng.(b) Slinhkincnlitrong2hpembvohpIII.ThpIIIlyngunhinra1linhkin.TnhxcsutlinhkinlyrathpIIIbhng.(c) BitlinhkinlyrathpIIIlhng.Tnhxcsut2linhkinlyrathpIvIIlcbanulhng.Bi tp2.55. C3cahngI, II, IIIcngkinhdoanhsnphmY , trongth phncacahngI, IIInhnhauvgpi th phncacahngII. T lsnphmloi Atrong3cahnglnltl70%,75%v50%.Mtkhchhngchnngunhin1cahngvtmuamtsnphm.2.5Cngthcxcsuty,cngthcBayes 13(a) TnhxcsutkhchhngmuacsnphmloiA.(b) GiskhchhngmuacsnphmloiA,hikhnngngiymuaccahngnolnhiunht.Bitp2.56. Cho l mt php th ngu nhin vi 3 bin c s cp c th xy ra l A, BvC.Gistatinhnhvhnlnvclpnhau.TnhtheoP(A), P(B)xcsutbincAxuthintrcB.Chng3BinngunhinvhmphnphiBitp3.1. ChobinngunhinrircXcbngphnphixcsutchobibngsau:X 2 1 0 1 2P 1/8 2/8 2/8 2/8 1/8(a) TmhmphnphixcsutF(x).(b) Tnh P(1 X 1)v P_X 1hocX= 2_.(c) LpbngphnphixcsutcabinngunhinY= X2.Bitp3.2. BinngunhinrircXchmxcsutchobif(x) =2x + 125, x = 0, 1, 2, 3, 4(a) LpbngphnphixcsutcaX.(b) Tnh P(2 X< 4)v P(X> 10).Bitp3.3. GiXlbinngunhinrirccbngphnphixcsutsauX 1 0 3P 0.5 0.2 0.3(a) TnhlchchuncaX.(b) TnhkvngcaX3.15(c) TmhmphnphicaX.(d) TanhnghaY= X2+ X + 1.LpbngphnphixcsutcaY .Bitp3.4. BinngunhinXchmmtf(x)nhsauf(x) =___kx(2 x) khi1 < x < 20 nikhc(a) Xcnhgitrcakf(x)lhmmtcabinngunhinX.VikvatmctnhkvngvphngsaicabinngunhinX.(b) TmhmphnphiF(x)cabinngunhinX.(c) TmhmphnphiG(y)cabinngunhinY= X3.Bitp3.5. BinngunhinlintcXchmmtf(x) =___exkhix > 00 khix 0(a) Tnh P(3 X).(b) Tmgitrcaasaocho P(X a) = 0, 1.(c) XcnhhmphnphivmtxcsutcabinngunhinY=X.Bitp3.6. TnhP(X 8)nufX(x) =___196x3ex/2nux 00 nukhcBitp3.7. ChofX(x) =_2 x2vi _2 x _2TnhP(X< 0).Bitp3.8. BinngunhinXchmmtf(x) =___a exp_x2_khix 00 nikhcXcnh:16(a) Hngsa.(b) HmphnphixcsutF(x)(c) KvngvphngsaicabinngunhinX.(d) KvngvphngsaicabinngunhinY= (X/2) 1.Bitp3.9. ChoXlbinngunhinchmmtsaufX(x) =___c(1 x2) nu1 x 10 nu |x| > 1viclmthngsdng.Tm(a) hngsc(b) trungbnhcaX(c) phngsaicaX(d) hmphnphiFX(x).Bitp3.10. BinngunhinlintcXchmmtf(x) =___12x khi0 < x < 20 nikhcTmhmphnphivhmmtxcsutcaccbinngunhinsau:(a) Y= X(2 X).(b) Z= 4 X3.(c) T= 3X + 2.Bitp3.11. TnhphngsaicaXnupX(x) =___1/4 nux = 01/2 nux = 11/4 nux = 417Bi tp3.12. Tnhphnv mc25%(tclgitr x0.25saochoP(X 1(i) TmFY (0).(ii) TnhphngsaicaY .Bitp3.14. BinngunhinlintcXchmmtxcsutf(x) =___34x(2 x) khi0 x 20 nikhc(a) XcnhhmphnphixcsutF(x)cabinngunhinX.(b) Tnh E(X), Var (X)vtrungvcabinngunhinX.(c) tY=X,xcnhhmphnphivhmmtxcsutcabinngunhinY .18Bitp3.15. Tui th ca mt loi cn trng no l mt bin ngu nhin lin tc X(nvthng)chmmtf(x) =___kx2(4 x) khi0 x 40 nikhc(a) Tmhngsk.(b) TmF(x).(c) Tm E(X), Var (X)vMod(X).(d) Tnhxcsutcntrngchttrcmtthngtui.Bitp3.16. BinngunhinlintcXchmmtf(x) =___kx2e2xkhix 00 nikhc(a) Tmhngsk.(b) TmhmphnphixcsutF(x).(c) Tm E(X), Var (X)vMod(X).Bitp3.17. ChaithngthucAvB,trong:- thngAc20lgm2lhngv18ltt- thngBc20lgm3lhngv17ltt.(a) Lymithng1l.GiXlslhngtronghaillyra.TmhmmtcaX.(b) LythngBra3l.GiY lslhngtrong3llyra.TmhmmtcaY .Bitp3.18. Mtthngng10lthuctrongc1lhng.Takimtratngl(khnghonli)chotikhiphthinclhngthdng.GiXlslnkimtra.TmhmmtcaX.Tnhkvngvphngsai.Bitp3.19. Mtbinngunhinlintcchmmtxcsutsau:fX(x) =___cxex/2nux 00 nux < 019(a) Tmhngsc.(b) TmhmphnphixcsutFX(x).(c) TmtrungbnhcaX(d) TmlchchuncaX.(e) TmMed(X).Bitp3.20. GiXltuithcaconngi.MtcngtrnhnghincuchobithmmtcaXlf(x) =___cx2(100 x)2khi0 x 1000 khix < 0hayx > 100(a) Xcnhhngsc.(b) TnhkvngvphngsaicaX.(c) Tnhxcsutcamtngictuith 60(d) Tnhxcsutcamtngictuith 60,bitrngngihinnay50tui.Bitp3.21. Mt thit b gm 3 b phn hot ng c lp vi nhau, xc sut trong khongthigiantccbphnhngtngngbng0.2;0.3;0.25.GiXlsbphnbhngtrongkhongthigiant.(a) LpbngphnphixcsutcaX.(b) VitbiuthchmphnphicaX.(c) TnhP(0 < X 4)theohaicch.Bi tp3.22. Mtmu4snphmcrtrakhnghonli t10snphm. Bitrngtrong10snphmnyc1thphm.Tnhxcsutthphmctrongmu.Bitp3.23. Mtcihpcha100transistorloiAv50transistorloiB.(a) Cctransistorcrtralnlt, ngunhinvchonli, chonkhi lyctransistorloiButin.Tnhxcsut9hoc10transistorcrtra.(b) Slngcctransistortnhtphirtra,ngunhinvchonli,lbaonhiunutamunxcsutlycchloiAnhhn1/3?Bitp3.24. GiXlslnmtnhtxuthinsaubalntungmtconxcxc.(a) LpbngphnphixcsutcaX.20(b) Tnhxcsutctnhtmtlncmtnht.(c) Tnhxcsutctiahailnmtnht.(d) TnhEX, V ar(X)Bitp3.25. Xttrchi,tungmtconxcxcbaln:nucbalnc6ntthlnh6ngn,nuhailn6ntthlnh4ngn,mtln6ntthlnh2ngn,vnukhngc6ntthkhnglnhght.MilnchiphingAngn.Hi:(a) Albaonhiuthngichivluvdihuvn(giltrchicngbng).(b) Albaonhiuthtrungbnhmilnngichimt1ngn.Bi tp3.26. Mththnganninhgmc10thnhphnhotngclplnnhau.Hthnghotngnutnht5thnhphnhotng.kimtrahthngchotnghaykhng, ngi ta kim tra nh k 4 thnh phn c chn ngu nhin (khng hon li). H thngc bo co l hot ng nu t nht 3 trong 4 thnh phn c kim tra hot ng. Nu thtschc4trong10thnhphnhotng,thxcxuththngcbocolhotnglbaonhiu?Bi tp3.27. Trongmttrchi nmphi tiu, ngi chi hngvmttmbialncvmtvngtrncbnknh25cm.GiXlkhongcch(theocm)giauphitiucmvobiavtmvngtrn.GisrngP(X x) =___cx2nu0 x < 251 nux 25viclmthngsno.(a) Tnh(i) hngsc(ii) hmmt,fX(x),caX(iii) trungbnhcaX(iv) xcsutP(X 10|X 5).(b) Ngichismt1(nv:ngnng)chomilnphngvthng___10 nuX r1 nur < X 2r0 nu2r < X< 25Vigitrnocarthstintrungbnhngichitcbng0.25?21Bitp3.28. ChoXlmtilngngunhincphnphixcsutnhsauX 0 1 2 3 4 5 6 7P 0 a 2a 2a 3a a22a27a2+ a(a) Xcnha(b) TnhP(X 5),P(X< 3).(c) TnhknhnhtsaochoP(X k) 12Bitp3.29. ChohmmtcabinngunhinXcdng(a)f(x) =___Ax khix [0, 1]0 khix/ [0, 1](b)f(x) =___Asin x khix [0, ]0 khix/ [0, ](c)f(x) =___Acos x khix [0,12]0 khix/ [0,12](d)f(x) =___Ax4khix 10 khix < 1HyxcnhA.TmhmphnphixcsutcaX.TnhEX, V ar(X)nuc.Bitp3.30. ChobinngunhinlintcXchmphnphiF(x) =___0 khix < 2a + b sin x khi 2 x 21 khix >2via, blhngs.22(a) Tmavb.(b) Viavbtmccua),tnhhmmtf(x)caXvMod(X), Med(X), P(X>4)Bitp3.31. Cho Xv Yl hai bin ngu nhin c lp v c phn phi xc sut tng nglX 1 0 1 2P 0.2 0.3 0.3 0.2Y 1 0 1P 0.3 0.4 0.3TmphnphixcsutcaX2,X + Y .Tnhkvng,phngsaicaX,X + Y .Bitp3.32. Mt mu gm 4 bin ngu nhin X1, X2, X3, X4c lp vi nhau tng i mt.MibinngunhinXi,i = 1, . . . , 4chmmtnhsau:f(x) =___2x khi0 < x < 10 nikhctY= max{X1, X2, X3, X4}vZ= min{X1, X2, X3, X4}.TmhmmtcaY vZ.Bitp3.33. Cho FXl hm phn phi xc sut ca bin ngu nhin X. Tm hm phn phixcsutcabinngunhinY=___X|X|nuX = 01 nuX= 0Bitp3.34. Tmhmphnphica12(X +|X|)nuhmphnphicaXlFX.Bi tp3.35. GisXchmphnphi lintc F(x). Xc nhhmphnphi caY= F(X).Bi tp3.36. GisF(x)lhmphnphicabinngunhindnglintcX,ctnhchtP(X< t + x|X> t) = P(X< x) vix, t > 0ChngminhrngF(x) = 1 exvix > 0.Chng4Mtsphnphixcsutthngdng4.1 PhnphiBernoulli,nhthcBi tp4.1. C8000snphmtrongc2000snphmkhngttiuchunkthut.Lyngunhin(khnghonli)10snphm.Tnhxcsuttrong10snphmlyrac2snphmkhngttiuchun.Bi tp4.2. Khitimtruynmtloihuytthanh,trungbnhcmttrnghpphnngtrn1000trnghp.Dngloihuytthanhnytimcho2000ngi.Tnhxcsut(a) c3trnghpphnng,(b) cnhiunht3trnghpphnng,(c) cnhiuhn3trnghpphnng.Bitp4.3. Gistlsinhcontraivcongilbngnhauvbng12.Mtgianhc4ngicon.Tnhxcsut4acongm2traiv2gi.1traiv3gi.4trai.Bitp4.4. Mtnhmysnxutvitlphphml7%.(a) Quanstngunhin10snphm.Tnhxcsuti) cngmtphphm.ii) ctnhtmtphphm.4.1PhnphiBernoulli,nhthc 24iii) cnhiunhtmtphphm.(b) Hi phi quansttnhtbaonhiusnphmxcsutnhnctnhtmtphphm 0.9Bi tp4.5. Tlmtloibnhbmsinhtrongdnslp=0.01.Bnhnycnschmsc c bit lc mi sinh. Mt nh bo sinh thng c 20 ca sinh trong mt tun. Tnh xc sut(a) khngctrnghpnocnchmsccbit,(b) cngmttrnghpcnchmsccbit,(c) cnhiuhnmttrnghpcnchmsccbit.TnhbngquylutnhthcridngquylutPoissonsosnhktqukhitaxpxphnphinhthcB(n; p)bngphnphiPoissonP(np).Bitp4.6. TlctringhngcvinAtrongmtcucbucl60%.Ngitahikin 20 c tri c chn mt cch ngu nhin. Gi Xl s ngi b phiu cho A trong 20 ngi.(a) Tmgitrtrungbnh,lchchunvModcaX.(b) TmP(X 10)(c) TmP(X> 12)(d) TmP(X= 11)Bi tp4.7. GistldncmcbnhAtrongvngl10%. Chnngunhin1nhm400ngi.(a) Vitcngthctnhxcsuttrongnhmcnhiunht50ngimcbnhA.(b) Tnhxpxxcsutbngphnphichun.Bi tp4.8. MtmysnxutrasnphmloiAvixcsut0.485.Tnhxcsutsaoctrong200snphmdomysnxutractnht95snphmloiA.Bi tp4.9. Davosliutrongqukh, taclngrng85%ccsnphmcamtmysnxutnolthphm.Numynysnxut20snphmmigi,thxcsut8hoc9thphmcsnxuttrongmikhongthigian30phtlbaonhiu?Bitp4.10. Xc sut trng s l 1%. Mi tun mua mt v s. Hi phi mua v s lin tiptrongtithiubaonhiutunckhngthn95%hyvngtrngstnht1ln.4.1PhnphiBernoulli,nhthc 25Bi tp4.11. Trongtrchi"bucua cbaconxcsc,miconcsumthnhl:bu,cua, hu, nai, tmvg. Gischai ngi, mtngi chi vmtngi lmci. Numivn ngichi cht mt (mt trong cchnh:bu,cua, hu,nai,tm v g) sau khi chinhiu vn th ngi no s thng trong tr chi ny. Gi s thm mi vn ngi chi t 1000 nu thng s c 5000 , nu thua s mt 1000 . Hi trung bnh mi vn ngi thng s thngbaonhiu?Bitp4.12. Cbalgingnhau:hailloiI,milc3bitrngv7bien;mtlloiIIc4bitrngv6bien.Mttrchictranhsau:Mivn,ngichichnngunhinmtlvlyrahai bi tl. Nulycnghai bi trngth ngi chi thng,ngclingichithua.(a) NgiAchitrchiny,tnhxcsutngiAthngmivn.(b) GisngiAchi10vn,tnhsvntrungbnhngichithngcvsvnngiAthngtinchcnht.(c) Ngi Aphi chi tnhtbaonhiuvnxcsutthngtnhtmtvnkhngdi0,99.Bitp4.13. ChoXvY lhaiilngngunhinclp.(a) GisX B(1,15), YB(2,15). Lpbngphnphi xcsutcaX+ Y vkimtrarngX + Y B(3,15)(b) GisX B(1,12), YB(2,15). TmphnbxcsutcaX+ Y . ChngminhrngX + Y khngcphnbnhthc.Bitp4.14. Hai cu th nm bng vo r. Cu th th nht nm hai ln vi xc sut trngr ca mi ln l 0.6. Cu th th hai nm mt ln vi xc sut trng r l 0.7. Gi Xl s lntrngrcachaicuth.LpbngphnphixcsutcaX,bitrngktqucacclnnmrlclpvinhau.Bitp4.15. Buindngmtmytngcachtrnbthphnloitngkhuvc gi i, my c kh nng c c 5000 b th trong 1 pht. Kh nng c sai 1 a ch trnbthl0,04%(xemnhvicc5000bthnyl5000phpthclp).(a) Tnhsbthtrungbnhmiphtmycsai.(b) Tnhsbthtinchcnhttrongmiphtmycsai.(c) Tnhxcsuttrongmtphtmycsaitnht3bth.Bi tp4.16. Mtbi thi trcnghimgmc10cuhi, mi cuc4phngntrli,trong ch c mt phng n ng. Gi s mi cu tr li ng c 4 im v cu tr li saibtr2im.Mtsinhvinkmlmbibngcchchnngunhinmtphngnchomicuhi.4.2PhnphiPoisson 26(a) Tnhxcsuthcsinhnyc4im.(b) Tnhxcsuthcsinhnybimm.(c) GiXlscutrling,tnhE(X)vV ar(X).(d) Tnhscusinhvinnyckhnngtrlinglnnht.Bitp4.17. Cc sn phm c sn xut trong mt dy chuyn. thc hin kim tra chtlng, mi gi ngi ta rt ngu nhin khng hon li 10 sn phm t mt hp c 25 sn phm.Qutrnhsnxutcbocoltyucunuckhngqumtsnphmlthphm.(a) Nu tt c cc hp c kim tra u cha chnh xc hai th phm, th xc sut qu trnhsnxutcbocotyucutnht7lntrongmtngylmvic8gilbaonhiu?(b) SdngphnphiPoissonxpxxcsutctnhtrongcu(a).(c) Bitrnglnkimtrachtlngcui cngtrongcu(a), qutrnhsnxutcbocotyucu.Hixcsutmu10snphmtngngkhngchathphmlbaonhiu?4.2 PhnphiPoissonBi tp4.18. Mttrungtmbuinnhnctrungbnh3cucinthoi trongmipht.Tnhxcsuttrungtmnynhnc1cuc,2cuc,3cucgitrong1pht,bitrngscucgitrongmtphtcphnphiPoisson.Bitp4.19. TnhP(X 1|X 1)nuX P(5)Bitp4.20. ChoX,Y lccbinngunhinclp,X P(1),Y P(2)(a) TnhxcsutP(X + Y= n)(b) TnhxcsutP(X= k|X + Y= n)Bi tp4.21. Mtcahngchothuxetnhnthyrngsngi nthuxetvongy th by cui tun l mt i lng ngu nhin Xc phn phi Poisson vi tham s = 2.Giscahngc4chict.(a) Tmxcsutkhngphittc4chictucthu.(b) Tmxcsutttc4chictucthu.(c) Tmxcsutcahngkhngpngcyucu.4.2PhnphiPoisson 27(d) Trungbnhcbaonhiutcthu.(e) Cahngcnctnhtbaonhiutxcsutkhngpngcnhucuthubhn2%Bitp4.22. Mt tng i bu in c cc cuc in thoi gi n xut hin ngu nhin, clpvinhauvctctrungbnh2cucgitrong1pht.Tmxcsut(a) cng5cucinthoitrong2pht,(b) khngccucinthoinotrongkhongthigian30giy,(c) ctnht1cucinthoitrongkhongthigian10giy.Bitp4.23. Cc cuc gi in n tng i tun theo phn phi Poisson vi mc trn mipht.Tkinhnghimcctrongqukh,tabitrngxcsutnhncchnhxcmtcucgi trongmtphtbngbalnxcsutkhngnhnccucgi notrongcngthigian.(a) GiXlscucginhnctrongmipht.TnhxcsutP(2 X 4).(b) Taxt100khongthigianmtphtlintipvgiUlskhongthigianmtphtkhngnhnccucgiinno.TnhP(U 1).Bi tp4.24. Ti mtimbnvmybay, trungbnhtrong10phtc4ngi nmuav.Tnhxcsut:(a) Trong10phtc7nginmuav.(b) Trong10phtckhngqu3nginmuav.Bitp4.25. Cc khch hng n quy thu ngn, theo phn phi Poisson, vi s lng trungbnh5ngimipht.Tnhxcsutxuthintnht10khchhngtrongkhongthigian3pht.Bi tp4.26. Skhchhngnquythungntuntheophnphi Poissonvi thams = 1trongmikhong2pht.Tnhxcsutthigianinkhikhchhngtiptheoxuthin(tkhchhngtrc)nhhn10pht.Bitp4.27. S lng nho kh trong mt ci bnh quy bt k c phn phi Poisson vi thams. Hi gitr lbaonhiunutamunxcsutcnhiunhthai bnhquy, trongmthpc20bnh,khngchanhokhl0.925?Bitp4.28. Mt trm cho thu xe Taxi c 3 chic xe. Hng ngy trm phi np thu 8 USDcho1chicxe(btkxeccthuhaykhng).Michiccchothuvigi20USD.Gi s s xe c yu cu cho thu ca trm trong 1 ngy l i lng ngu nhin c phn phiPoissonvi = 2.8.4.3Phnphichun 28(a) Tnhstintrungbnhtrmthuctrongmtngy.(b) Giibitontrntrongtrnghptrmc4chicxe.(c) Theobn,trmnnc3hay4chicxe?Bi tp4.29. Tac10mysnxut(clpnhau), mi mysnxutra2%thphm(khngtchun).(a) Trungbnhcbaonhiusnphmcsnxutbi myutintrckhi ntorathphmutin?(b) Talyngunhinmtsnphmtmi mysnxut. Hi xcsutnhiunhthai thphmtrong10snphmnylbaonhiu?(c) Lmlicu(b)bngcchsdngxpxPoisson.(d) Phi lyratnhtbaonhiusnphmcsnxutbimyutinxcsuttctnhtmtthphmkhngnhhn1/2(gisrngccsnphmlclpvinhau)?4.3 PhnphichunBitp4.30. Ccktqucabikimtrachsthngminh(IQ)chocchcsinhcamttrng tiu hc cho thy im IQ ca cc hc sinh ny tun theo phn phi chun vi cc thams l = 100 v 2= 225. T l hc sinh c im IQ nh hn 91 hoc ln hn 130 l bao nhiu?Bi tp4.31. Gischiudi X(nv tnhm)camtni xebtk tuntheophnphichunN(, 0.012).(a) Mtngi nngshumtchicxehi caocpcchiudi lnhn15%chiuditrungbnhcamtchuxe.Hitlchuxecthsdnglbaonhiu?(b) Gisrng=4. Hi chiudi caxelbaonhiunutamunchcancthsdng90%chuxe?Bi tp4.32. ngknhcamtchi titmydomtmytintngsnxutcphnphichunvitrungbnh = 50mmvlchchun= 0.05mm.Chititmycxemltyucunungknhkhngsaiqu0.1mm.(a) Tnhtlsnphmtyucu.(b) Lyngunhin3snphm.Tnhxcsutctnhtmtsnphmtyucu.4.3Phnphichun 29Bitp4.33. Trng lng X(tnh bng gam) mt loi tri cy c phn phi chun N(, 2),vi = 500(gam)v2= 16(gam2).Tricythuhochcphnloitheotrnglngnhsau:(a) loi1:trn505gam,(b) loi2:t495n505gam,(c) loi3:di495gam.Tnhtlmiloi.Bi tp4.34. MtcngtykinhdoanhmthngAdnhspdngmttrong2phngn kinh doanh. K hiu X1l li nhun thu c khi p dng phng n th 1, X2l li nhunthu c khi p dng phng n th 2. X1, X2u c tnh theo n v triu ng/ thng) vX1 N(140, 2500),X2 N(200, 3600).Nubitrng,cngtytntivphttrinthlinhunthuctmthngkinhdoanhAphittnht80triung/thng.HychobitcngtynnpdngphngnnokinhdoanhmthngA?Vsao?Bi tp4.35. Nghincuchiucaocanhngngitrngthnh,ngitanhnthyrngchiucaotuntheoquylutphnbchunvitrungbnhl175cmvlchtiuchun4cm.Hyxcnh:(a) tlngitrngthnhctmvctrn180cm.(b) tlngitrngthnhcchiucaot166cmn177cm.(c) tmh0,nubitrng33%ngitrngthnhctmvcdimch0.(d) gii hnbinngchiucaoca90%ngi trngthnhxungquanhgitr trungbnhcan.Bi tp 4.36. Ta quan tm n tui th X(theo nm) ca mt thit b. T kinh nghim trongqukh,taclngxcsutthitbloinycnhotngttsau9nml0.1.(a) TaaramhnhsauchohmmtcaXfX(x) =a(x + 1)bvi x 0tronga > 0vb > 1.Tmhaihngsa,b.(b) Nutaaramtphnphichunvitrungbnh = 7choX,thgitrthamslbaonhiu?(c) Taxt10thitb loi nymtcchclp. Tnhxcsut8hoc9thitb loi nyctuiihotngthn9nm.4.3Phnphichun 30Bi tp4.37. EntropyHcamt binngunhinlintc Xc nhnghalH=E[ln fX(X)]vifXlhmmtxcsutcabinngunhinXvlnllogarittnhin.TnhentropycabinngunhinGaussvitrungbnh0vphngsai2= 2.Chng5LthuytmuBi tp5.1. Sliuvchiucaocaccsinhvinn(nv:inch)trongmtlphcnhsau:62 64 66 67 65 68 61 65 67 65 64 63 6768 64 66 68 69 65 67 62 66 68 67 66 6569 65 70 65 67 68 65 63 64 67 67(a) Tnhchiucaotrungbnhvlchtiuchun.(b) Trungvcachiucaosinhvinlpnylbaonhiu?Bitp5.2. Chobdliusau:4.2 4.7 4.7 5.0 3.8 3.6 3.0 5.1 3.1 3.84.8 4.0 5.2 4.3 2.8 2.0 2.8 3.3 4.8 5.0Tnhtrungbnhmu,phngsaimuvlchtiuchun.Bitp5.3. Chobdliusau:43 47 51 48 52 50 46 4945 52 46 51 44 49 46 5149 45 44 50 48 50 49 50Tnhtrungbnhmu,phngsaimuvlchtiuchun.Bitp5.4. Xtbiuthcy=

ni=1(xia)2.Vianothytgitrnhnht?32Bi tp5.5. Xtyi= a + bxi,i = 1, . . . , nva, blcchngskhc0.Hytmmilinhgiaxvy,sxvsy.Bi tp5.6. Gistacmucngmccgitrquantrcx1, x2, . . . , xnvtnhctrungbnhmuxnvphngsaimus2n.Quantrcthmgitrth(n + 1)lxn+1,gixn+1vs2n+1lnltltrungbnhmuvphngsaimungvimuc(n + 1)quantrc.(a) Tnhxn+1theoxnvxn+1.(b) Chngtrngns2n+1= (n 1)s2n +n(xn+1xn)2n + 1Bi tp5.7. Tbngccsngunhinngi talyra150s. Ccscphnthnh10khongnhsau:xi1 11 21 31 41 51 61 71 81 9110 20 30 40 50 60 70 80 90 100ni16 15 19 13 14 19 14 11 13 16Xcnhtrungbnhmuvphngsaimu.Bitp5.8. Khostthunhpcacngnhnmtcngty,chobibngsau(nvngnng).Thunhp [500, 600] [600, 700] [700, 800] [800, 900] [900, 1000] [1000, 1100][1100, 1200]Sngi 2 10 15 30 25 14 4Xcnhthunhptrungbnh,lchchun.Bitp5.9. olnghuyttngca8ngimnhkho,tac2, 863, 372, 752, 623, 503, 253, 123, 15Hyxcnhccctrngmu.Bi tp5.10. Quanstthigiancnthitsnxutmtchititmy,tathucsliuchobngsau:33Khongthigian(pht) Slnquanst20-25 225-30 1430-35 2635-40 3240-45 1445-50 850-55 4Tnhtrungbnhmux,phngsaimus2.Bitp5.11. odicamtloitrcxe,tacktquNhm 18.4-18.6 18.6-18.8 18.8-19 19-19.2 19.2-19.4 19.4-19.6 19.6-19.8ni1 4 20 41 19 8 4Hytnhditrungbnhvphngsaimu.Chng6clngthamsthngk6.1 clngtrungbnhtngthBi tp6.1. Trntpmugm100sliu,ngitatnhcx=0.1s=0.014.Xcnhkhongtincy95%chogitrtrungbnhtht.Bi tp6.2. Chnngunhin36cngnhncax nghipth thylngtrungbnhl380ngn/thng. Gislngcngnhntuntheophnphi chunvi =14ngnng. Vitincy95%,hyclngmclngtrungbnhcacngnhntrongtonxnghip.Bi tp6.3. oscbnchulccamtloi ngth nghim, ngi tathucbsliusau4500, 6500, 5200, 4800, 4900, 5125, 6200, 5375T kinh nghim ngh nghip, ngi ta cng bit rng sc bn c phn phi chun vi lchchun= 300.Hyxydngkhongtincy90%choscbntrungbnhcaloingtrn.Bi tp 6.4. Sn lng mi ngy ca mt phn xng l bin ngu nhin tun theo lut chun.Ktquthngkca9ngychota:27, 26, 21, 28, 25, 30, 26, 23, 26Hyxcnhcckhongtincy95%chosnlngtrungbnh.Bitp6.5. QuanstchiucaoX(cm)camtsngi,taghinhnx(cm) 140-145 145-150 150-155 155-160 160-165 165-170Sngi 1 3 7 9 5 2(a) Tnhxvs26.1clngtrungbnhtngth 35(b) clngtincy0.95Bitp6.6. imtrungbnhmntonca100thsinhdthivotrngAl5vilchchunl2.5.(a) clngimtrungbnhmntoncatonththsinhvitincyl95%.(b) Visaisclngimtrungbnhcua)l0.25im,hyxcnhtincy.Bitp6.7. Tuith ca mt loibng n cbit theo quy lut chunvi lch chun100gi.(a) Chn ngunhin 100 bng n thnghim,thy mibng tuith trung bnh l 1000gi.HyclngtuithtrungbnhcabngnxnghipAsnxutvitincyl95%.(b) Vidungsaicaclngtuithtrungbnhl15gi,hyxcnhtincy.(c) dungsaicaclngtuithtrungbnhkhngqu25givitincyl95%thcnphithnghimtnhtbaonhiubng.Bi tp6.8. Khi lngccbaobtm ti mtcahnglngthctuntheophnphichun.Kimtra20bao,thykhilngtrungbnhcamibaobtml48kg,vphngsaimus2= (0.5kg)2.(a) Vi tincy95%hyclngkhi lngtrungbnhcamtbaobtm thuccahng.(b) Vidungsaicaclngcua)l0.284kg,hyxcnhtincy.(c) dungsaicaclngcua)khngqu160gvitincyl95%,cnphikimtratnhtbaonhiubao?Bi tp6.9. ongknhcamtchititmydomtmytintngsnxut,taghinhncsliunhsau:x 12.00 12.05 12.10 12.15 12.20 12.25 12.30 12.35 12.40n 2 3 7 9 10 8 6 5 3vinchstrnghptnhtheotnggitrcaX(mm).(a) Tnhtrungbnhmuxvlchchunscamu.(b) clngngknhtrungbnhtincy0.95.6.2clngtltngth 36(c) Numunsaisclngkhngqu = 0.02mmtincy0.95thphiquansttnhtmytrnghp.Bitp6.10. NgitaoionNa+trnmtsngivghinhnlicktqunhsau129, 132, 140, 141, 138, 143, 133, 137, 140, 143, 138, 140(a) Tnhtrungbnhmuxvphngsaimus2.(b) clngtrungbnhcatngthtincy0.95.(c) Numunsaisclngtrungbnhkhngqu = 1vitincy0.95thphiquanstmugmtnhtmyngi?Bitp6.11. Quansttuithx(gi)camtsbngndoxnghipAsnxut,taghinhnx 1000 1100 1200 1300 1400 1500 1600 1700 1800n 10 14 16 17 18 16 16 12 9vinchstrnghptheotnggitrcax.(a) Tnhtrungbnhmuxvlchchunmus.(b) clngtuithtrungbnhcabngntincy0.95.(c) Nu mun sai s c lng khng qu = 30 gi vi tin cy 0.99 th phi quan st mugmtnhtmybngn?Bitp6.12. Chiudicamtloisnphmcxutkhuhnglotlbinngunhinphnphi chunvi =100mmv2=42mm2. Kimtrangunhin25snphm. Khnng chiu di trung bnh ca s sn phm kim tra nm trong khong t 98mm n 101mm lbaonhiu?6.2 clngtltngthBitp6.13. Trc bu c, ngi ta phng vn ngu nhin 2000 c tri th thy c 1380 nging h mt ng c vin K. Vi tin cy 95%, hi ng c vin thu c ti thiu bao nhiuphntrmphiubu?Bi tp6.14. Mtloibnhctltvongl0.01.Munchngtmtloithucchiunghim (ngha l h thp c t l t vong nh hn 0.005) tin cy 0.95 th phi th thuctrntnhtbaonhiungi?6.3Tnghp 37Bi tp6.15. clngxcsutmcbnhganvitincy90%vsaiskhngvtqu2%thcnphikhmtnhtbaonhiungi,bitrngtlmcbnhganthcnghimchobng0,9.Bi tp6.16. Gisquanst100ngi thyc20ngi b bnhstxuthuyt. Hyclngtlbnhstxuthuyttincy97%.Numunsaisclngkhngqu3%tincy95%thphiquansttnhtbaonhiungi?Bi tp6.17. Mtloithucmiemiutrcho50ngibbnhB,ktquc40ngikhibnh.(a) clngtlkhibnhpnudngthuciutrvitincy0.95v0.99.(b) Nu mun sai s c lng khng qu 0.02 tin cy 0.95 th phi quan st t nht mytrnghp?Bitp6.18. Tamunclngtlvinthucbscmptrongmtlthucln.(a) Numunsai sclngkhngqu0.01vi tincy0.95th phi quansttnhtmyvin?(b) Quanstngunhin200vin, thyc18vinb stm. Hyclngptincy0.95.(c) Khi,numunsaisclngkhngqu0.01vitincy0.95thphiquansttnhtmyvin?Bitp6.19. Munbittrongaocbaonhiuc,ngitabtln2000con,nhduxonglith xungh.Saumt thigian,ngita btln 500con v thy c20 conc cnh ducalnbttrc.Davoktquhyclngscctronghvitincy95%.Bi tp6.20. cthdoncslngchimthngnghtivnnhmnh,ngichbt89con, emeokhoenchochngri thi. Saumtthi gian, ngbtngunhinc120convthyc7conceokhoen.Hydonschimgipngchvntincy99%.6.3 TnghpBitp6.21. Cnth100qucam,tacbsliusau:Khilng(g) 32 33 34 35 36 37 38 39 40Squ 2 3 15 26 28 6 8 8 46.3Tnghp 38(a) Hyclngkhilngtrungbnhccqucamtincy95%.(b) Camckhilngdi34gccoilcamloi2.Tmkhongclngchotlloi2vitincy90%.Bitp6.22. emcnmtstricyvathuhoch,tacktqusau:X(gam) 200-210 210-220 220-230 230-240 240-250Stri 12 17 20 18 15(a) Tmkhongclngcatrnglngtrungbnhcatri cyvi tincy0.95v0.99.(b) Numunsai sclngkhngqu=2gamtincy99%th phi quansttnhtbaonhiutri?(c) TricyckhilngX 230gamcxpvoloiA.HytmkhongclngchotlpcatricyloiAtincy0.95v0.99.Numunsaisclngkhngqu0.04tincy0.99thphiquansttnhtmytrnghp?Chng7Kimnhgithuytthngk7.1 SosnhkvngvimtschotrcBi tp7.1. Gimcmtx nghipchobitlngtrungbnhca1cngnhnthucxnghipl380ngn/thng.Chnngunhin36cngnhnthylngtrungbnhl350ngn/thng, vi lch chun s = 40. Li bo co ca gim c c tin cy c khng, vi mc cnghal = 5%.Bitp7.2. Trongthpnin80,trnglngtrungbnhcathanhninl48kg.Nayxcnhlitrnglngy,ngitachnngunhin100thanhninotrnglngtrungbnhl50kgvphngsaimus2=(10kg)2.Thxemtrnglngthanhninhinnayphichngcthayi,vimccnghal1%?Bi tp7.3. Mtcahngthcphmnhnthythi gianvaquatrungbnhmtkhchhngmua25ngnngthcphmtrongngy.Naycahngchnngunhin15khchhngthytrungbnhmt khchhngmua24ngnngtrongngyvphngsai muls2=(2ngnng)2.Vimcnghal5%,kimnhxemcphiscmuacakhchhnghinnaythcsgimsthaykhng.Bitrngscmuacakhchhngcphnphichun.Bitp7.4. i vi ngi Vit Nam, lng huyt sc t trung bnh l 138.3 g/l. Khm cho 80cng nhn nh my c tip xc ho cht, thy huyt sc t trung bnh x = 120 g/l; s = 15 g/l.T kt qu trn, c th kt lun lng huyt sc t trung bnh ca cng nhn nh my ho chtnythphnmcchunghaykhng?Ktlunvi = 0.05.Bi tp7.5. Trongiukinchnnui bnhthng, lngsatrungbnhca1conbl14kg/ngy.Nghingiukinchnnuikmilmcholngsagimxung,ngitaiutrangunhin25convtnhclngsatrungbnhca1controng1ngyl12.5vlchchuns=2.5. Vi mcngha=0.05. hyktluniunghi ngni trn. Githitlngsabl1binngunhinchun.7.1Sosnhkvngvimtschotrc 40Bitp7.6. Tin lng trung bnh ca cng nhn trc y l 400 ngn /thng. xt xemtin lng hin nay so vi mc trc y th no, ngi ta iu tra 100 cng nhn v tnh cx = 404.8ngn/thngvs = 20ngn/thng.Vi = 1%(a) Nulpgithit2phavgithit1phathktqukimnhnhthno?(b) Gingcua,vix = 406ngn/thngvs = 20ngn/thng.Bi tp7.7. Mtmynggi ccsnphmckhi lng1kg. Nghi ngmyhotngkhng bnh thng, ngi ta chn ra mt mu ngu nhin gm 100 sn phm th thy nh sau:Khilng 0.95 0.97 0.99 1.01 1.03 1.05Sgi 9 31 40 15 3 2Vimcngha0.05,hyktlunvnghingtrn.Bitp7.8. Trng lng trung bnh khi xut chung mt tri chn nui trc l 3.3 kg/con.Nmnayngitasdngmtloithcnmi,cnth15conkhixutchungtacccsliunhsau:3.25, 2.50, 4.00, 3.75, 3.80, 3.90, 4.02, 3.60, 3.80, 3.20, 3.82, 3.40, 3.75, 4.00, 3.50Githittrnglngglilngngunhinphnphitheoquylutchun.(a) Vimcngha = 0.05.Hychoktlunvtcdngcaloithcnny?(b) Nutri chnnui bocotrnglngtrungbnhkhi xutchungl3.5kg/conth cchpnhnckhng?( = 0.05).Bitp7.9. ocholesterol(nvmg%)chomtnhmngi,taghinhnlicChol. 150160 160-170 170-180 180-190 190-200 200-210Sngi 3 9 11 3 2 1Chorngcholesteroltuntheophnphichun.(a) Tnhtrungbnhmuxvphngsaimus2.(b) Tmkhongclngchotrungbnhcholesteroltrongdnstincy0.95.(c) Ctiliuchobitlngcholesteroltrungbnhl0= 175mg%.Gitrnycphhpvimuquanstkhng?(ktlunvi = 0.05).7.1Sosnhkvngvimtschotrc 41Bi tp7.10. Quanstshoahngbnratrongmtngycamtcahngbnhoasaumtthigian,ngitaghicsliusau:Shoahng(o) 12 13 15 16 17 18 19Sngy 3 2 7 7 3 2 1Githitrngshoabnratrongngycphnphichun.(a) Tmtrungbnhmux,phngsaimus2.(b) Sau khi tnh ton, ng ch ca hng ni rng nu trung bnh mt ngy khng bn c 15o hoa th chng th ng ca cn hn. Da vo s liu trn, anh (ch) hy kt lun gipngchcahngxemcnntiptcbnhaykhngmcngha = 0.05.(c) Gi s nhng ngy bn c t 13 n 17 o hng l nhng ngy bnh thng. Hy clngtlcanhngngybnhthngcacahngtincy90%.Bi tp7.11. Mtx nghipcmtsrtlnccsnphmbngthpvi skhuyttttrungbnhmisnphml3.Ngitacitincchsnxutvkimtra36snphm.Ktqunhsau:Skhuyttttrnsnphm 0 1 2 3 4 5 6Ssnphmtngng 7 4 5 7 6 6 1Gisskhuytttcaccsnphmcphnphichun.(a) Hyclngskhuyttttrungbnhmi snphmsaukhi ci tin, vi tincy90%.(b) Hychoktlunvhiuqucaviccitinsnxutmcngha0.05.Bi tp7.12. nhgitcdngcamtchnbi dngmduhiuquanstlshngcu.Ngitamshngcuca20ngitrcvsaukhinbidng:xi32 40 38 42 41 35 36 47 50 30yi40 45 42 50 52 43 48 45 55 34xi38 45 43 36 50 38 42 41 45 44yi32 54 58 30 60 35 50 48 40 50Vimcngha = 0.05,cthktlungvtcdngcachnbidngny?7.2Sosnhhaikvng 42Bi tp7.13. Gistamunxcnhxemhiuqucachnkingivivicgimtrnglngnhthno.20ngiqubothchinchnking.Trnglngcatngngitrckhinking(Xkg)vsaukhinking(Ykg)cchonhsau:X 80 78 85 70 90 78 92 88 75 75Y 75 77 80 70 84 74 85 82 80 65X 63 72 89 76 77 71 83 78 82 90Y 62 71 83 72 82 71 79 76 83 81Kimtraxemchnkingctcdnglmthayitrnglnghaykhng( = 0.05).7.2 SosnhhaikvngBi tp7.14. Mtnhphttrinsnphmquantmnvicgimthigiankhcasn.V vyhai cngthcsncemthnghim. Cngthc1lcngthccccthnhphnchunvcngthc2cthmmtthnhphnlmkhmi cchorngslmgimthigiankhcasn.Tccthnghimngitathyrng1= 2= 8pht.10vtcsnvicngthc1 v10vtkhccsnvicngthc2.Thigiankhtrung bnhcatngmulx1= 121phtvx2= 112pht.Nhphttrinsnphmcthrtraktlungvnhhngcathnhphnlmkhmi?Vimcngha5%.Bitp7.15. Tc chy ca hai loi cht n lng c dng lm nhin liu trong tu v trcnghincu.Ngitabitrnglchchuncatcchycahailoinhinliubngnhauvbng3cm/s. Hai mungunhinkchthcn1=20vn2=20cthnghim;trungbnhmutcchylx1=18cm/svx2=24cm/s.Vimcngha=0.05hykimnhgithuythailoichtnlngnyccngtctchy.Bi tp7.16. Theodigicphiuca2cngtyAvBtrongvng31ngyngitatnhcccgitrsaux sCngtyA 37.58 1.50CngtyB 38.24 2.20GithitrnggicphiucahaicngtyAvBlhaibinngunhinphnphitheoquylutchun.Hychobitnghakvngcaccbinngunhinnitrn?HychobitcskhcbitthcsvgicphiutrungbnhcahaicngtyAvBkhng?Vimcngha = 5%7.2Sosnhhaikvng 43Bitp7.17. Hm lng ng trong mu ca cng nhn sau 5 gi lm vic vi my siu caotnochaithiimtrcvsau5gilmvic.Tacktqusau:Trc: n1= 50 x = 60mg% sx= 7Sau: n2= 40 y= 52mg% sy= 9.2Vi mcngha=0.05, cthkhngnhhmlngngtrongmusau5gilmvicgimihaykhng?Bi tp7.18. Trngcngmtginglatrnhaitharungnhnhauvbnhailoiphnkhcnhau.nngythuhochtacktqunhsau:Thathnhtlymu1000bnglathyshttrungbnhcamibnglx = 70htvsx= 10.Tha th hai ly mu 500 bng thy s ht trung bnh mi bng l y= 72 ht v sy= 20.HiskhcnhaugiaXvY lngunhinhaybncht,vi = 0.05?Bi tp7.19. sosnhtrnglngtrungbnhcatrssinhthnhth vnngthn,ngitathcntrnglngca10000chuvthucktqusauy:Vng Schuccn Trnglngtrungbnh lchchunmuNngthn 8000 3.0kg 0.3kgThnhth 2000 3.2kg 0.2kgVimc ngha = 0.05cthcoitrng lngtrungbnh catrssinhthnh thcaohnnngthnhaykhng?(Githittrnglngtrssinhlbinngunhinchun).Bi tp7.20. sosnhnnglchctonvvtlcahcsinh, ngi takimtrangunhin8embnghaibitonvvtl.Ktquchobibngdiy(Xlimton,Y liml):X 15 20 16 22 24 18 20 14Y 15 22 14 25 19 20 24 16Gi s Xv Yu c phn phi chun. Hy so snh im trung bnh gia Xv Y , mc ngha5%.Bitp7.21. Haimycsdngrtncvoccbnh.Ngitalymungunhin10bnhdomythnhtv10bnhdomythhaithcktqusau:7.3Sosnhtlvimtschotrc 44My1 16.03 16.01 16.04 15.96 16.05 15.98 16.05 16.02 16.02 15.99My2 16.02 16.03 15.97 16.04 15.96 16.02 16.01 16.01 15.99 16.00Vimcngha = 0.05cthnirnghaimyrtncvobnhnhnhaukhng?Bitp7.22. nghincunhhngcamtloithuc,ngitacho10bnhnhnungthuc. Ln khc h cng cho bnh nhn ung thuc nhng l thuc gi. Kt qu th nghim thucnhsau:Bnhnhn 1 2 3 4 5 6 7 8 9 10Sgingcthuc 6.1 7.0 8.2 7.6 6.5 8.4 6.9 6.7 7.4 5.8Sgingvithucgi 5.2 7.9 3.9 4.7 5.3 5.4 4.2 6.1 3.8 6.3Gi s s gi ng ca bnh nhn tun theo phn phi chun. Vi mc ngha 5%, hy kt lunvnhhngcaloithuctrn.Bitp7.23. Quan st sc nng ca b trai (X) v b gi (Y) lc s sinh (n v gam), ta cktquTrnglng 3000-3200 3200-3400 3400-3600 3600-3800 3800-4000Sbtrai 1 3 8 10 3Sbgi 2 10 10 5 1(a) Tnhx,y,s2x,s2y.(b) SosnhcckvngX,Y(ktlunvi = 5%).(c) Nhphai muli. Tnhtrungbnhvlchchuncamunhp. Dngmunhpclngscnngtrungbnhcatrssinhtincy95%.7.3 SosnhtlvimtschotrcBi tp7.24. Trongmtvngdncc18btraiv28bgimcbnhB.Hirngtlnhimbnhcabtraivbgicnhnhaukhng?(ktlunvi = 0.05vgisrngslngbtraivbgitrongvngtngngnhau,vrtnhiu).Bitp7.25. Mt my sn xut t ng vi t l chnh phm l 98%. Sau mt thi gian hotng, ngi tanghi ngtltrnb gim. Kimtrangunhin500snphmthyc28phphm,vi = 0.05hykimtraxemchtlnglmviccamyccncnhtrchaykhng?7.4Sosnhhaitl 45Bitp7.26. o huyt sc t cho 50 cng nhn nng trng thy c 60% mc di 110 g/l.S liu chung ca khu vc ny l 30% mc di 110 g/l. Vi mc ngha = 0.05, c th ktluncngnhnnngtrngctlhuytsctdi110g/lcaohnmcchunghaykhng?Bitp7.27. TheomtnguntinthtlhdnthchxemdncatrnTivil80%.Thmd 36 h dn thy c 25 h thch xem dn ca. Vi mc c ngha l 5%. Kim nh xem nguntinnycngtincykhng?Bi tp7.28. Mtmysnsuttng,lcutlsnphmloiAl20%.Saukhipdngmtphngphpcitinsnxutmi,ngitaly40mu,mimugm10snphmkimtra.Ktqukimtrachobngsau:SsnphmloiAtrongmu 1 2 3 4 5 6 7 8 9 10Smu 2 0 4 6 8 10 4 5 1 0Vimcngha5%.Hychoktlunvphngphpsnsutny.Bitp7.29. T l ph phm ca mt nh my trc y l 5%. Nm nay nh my p dngmtbinphpkthutmi.nghincutcdngcabinphpkthutmi,ngitalymtmugm800snphmkimtravthyc24phphm.(a) Vi = 0.01.Hychoktlunvbinphpkthutminy?(b) Nunhmybocotlphphmsaukhipdngbinphpkthutmil2%thcchpnhnckhng?( = 0.01).7.4 SosnhhaitlBitp7.30. Trong90ngidngDDTngabnhngoidathc10nginhimbnh;trong100ngikhngdngDDTthc26ngimcbnh.HirngDDTctcdngngabnhngoidakhng?(ktlunvi = 0.05)Bi tp7.31. Ngitaiutra250ngixAthyc140nviutra160ngixBthyc80n.Hysosnhtlnhaixvimcngha5%.Bi tp7.32. pdnghai phngphpgieoht. TheophngphpAgieo180htth c150 ht ny mm; theo phng php Bgieo 256 ht th thy c 160 ht ny mm. Hy so snhhiuqucahaiphngphpvimcngha = 5%.Bitp7.33. Theoditrnglngcamtstrssinhtimtsnhhsinhthnhphvnngthn, ngi tathyrngtrongs150trssinhthnhphc100chunnghn3000gam,vtrong200trssinhnngthnc98chunnghn3000gam.Tktquhysosnhtltrssinhctrnglngtrn3000gamthnhphvnngthnvimcngha5%.PhnIIBIGIITphp-GiitchthpGiibi1.1. TalpdyB1, B2, . . . , B2nhsau:B1= A1, B2= A2\ A1, . . . , Bn= An\n1_k=1Ak(a) TachngminhBi Bj= (i = j),gisi < j.Gisa Bi= Ai\

i1k=1Ak,tcla A1, a/ Ak(k = 1, . . . , i 1),vvya

j1k=1Ak.Suyraa/ Bj.VyBi Bj= (i = j)(b)_i=1Ai=_k=1BkGisa i=1Ai, tc ltnti ch sj nosaochoa Aj. Nua Bjtha

j=1Bj.Nua

j1i=1 Ai,gii1lchsnhnhtsaochoa Ai1.Khi,a Bi1,tcla

j=1Bj.Vy_i=1Ai _k=1Bk.Ngcli,gisa

j=1Bj,suyratntijsaochoa Bj,tcla Aj, a/

j1k=1Ak.Do,a

i=1Aj.Vy_k=1Bk _i=1AiGiibi1.3. Khng phi khi no cng ng. V d, xt A, B, Cl cc tp con khc rng cavrinhautngimt,khiA B CvB A CnhngB = Giibi1.5.(a) (A B)(A C) = A BC(b) (A B)(A B) = A48(c) (A B)(A B)(A B) = AB(d) (A B)(A B)(A B) = (e) (A B)(B C) = AB AC B BC= B AC B(A C) = B ACGiibi1.7.(a) A B A B= AB AB= A(B B) = A(b) (A B)AB= (A B)(A B) = AB BAGiibi1.9.(a) C550= 2118760(b) A550= 254251200Gii bi 1.11. utintachn10namsinhtrong20namsinh,thcC1020cch.Sau,chn 10 n sinh trong 20 n sinh th c C1020cch. Theo quy tc nhn, s cch phn chia thayuculC1020C1020Giibi1.13.(a) Cmi honv 5ngi nyslmtcchspxpthtphtbiuAtrcBhocBtrcA.MscchxpAtrcBbngviscchxpBtrcAvchcnichAvBtrong1honv5ngi.Do,scchxpngiBphtbiusauAl5!2= 60(b) TaxemABlmtnhmvtatinhnhhonv bnphntsau: AB,C,D,E. Nhvy,scchxpngiAphtbiuxongthnltngiBl4! = 24Giibi1.15. utintachnlptrngvc40cchchn.Tiptheotachnlpphvc39cchchn.Cuicngtachnthquthc38cchchn.Do,scchchnbancnslpl40.39.38 = 59280cch.Giibi1.17. Scchc3ngilmnhimvaimAlC39.Scchc2ngiaimBlC26Scchc4ngilinlC44Theoquytcnhn,scchphncnglC39.C26.C44= 1260Giibi1.19.49(a) utintaxp3trong12hnhkhchlntoath1, th cC312cch. Sau, taxp3trong9hnhkhchcnlilntoath2thcC39cch.Tiptheo,taxp3trong6hnhkhch cn li ln toa th 3 th c C36cch. Cui cng, ta xp 3 trong 3 hnh khch cn liln toa th 4 th c C33cch. Theo quy tc nhn, s cch xp s l C312.C39.C36.C33= 369600cch.(b) utintaxp6hnhkhchvotoath1vcC612cch.Sau,taxp4hnhkhchvotoath2vcC46cch.Tiptheo,taxp1hnhkhchlntoath3vcC12cch.Cui cng, taxp1hnhkhchlntoath4vcC11cch. Tuynhintacthxem4toatul4nhmvtacthhonv4nhmny.Scchhonvl4!.Do,scchxpthayucul4!.C612.C46.C12.C11= 665280cch.Giibi1.21.(a) Tac,(1 + x)n=n

k=0Cknxk(7.1)Lyohmcpmtca(7.1),tacn(1 + x)n1=n

k=0kCknxk1(7.2)Thayx = 1vobiuthc(7.2)tacpcm.(b) Lyohmcp2ca(7.1),tacn(n 1)(1 + x)n2=n

k=0k(k 1)Cknxk2(7.3)Thayx = 1vobiuthc(7.3)tacpcm.Giibi1.23. p dng bi (1.20) bng cch thay n bng 2n, rbng n v m bng n. Ta c,C0nCnn+ C1nCn1n+ + CinCnin+ + CnnC0n= Cn2nDoCin= Cnini = 0, . . . , nnntacpcm.BincvxcsutGiibi2.1.(a) Tac,A A + B= AsuyraA = vB= .Thlitathyng.VyA = , B= (b) Tac,A AB= AsuyraA = vB= .Thlitathyng.VyA = , B= (c) Tac, A A + B=AB B A + B=AB A, tclA B A. Do, A=B.Thlithyng.VyA = BTac,A.A + B= A(A B) = (AA)B= .VyA, A + Bxungkhc.Giibi2.3.(a)Gi A:"Cngmtsinhvintyucu"Tac,A = B1B2B3B4 + B1B2B3B4 + B1B2B3B4 + B1B2B3B4(b)Gi B:"Cngbasinhvintyucu"Tac,B= B1B2B3B4 + B1B2B3B4 + B1B2B3B4 + B1B2B3B4(c)Gi C:"Ctnhtmtsinhvintyucu"Tac,C= B1 + B2 + B3 + B4(d)Gi D:"Khngcsinhvinnotyucu"Tac,D = B1B2B3B451Giibi2.5. Tac:X + A + A + A = BX.A + XA = BX(A + A) = BX = BX = BGiibi2.7.(a) MtccbincA6B6, A3B5A6B6:sntmttrnchaiconxcxcul6A3B5:s ntmt trnconxcxcthnhtl 3v trnconxcxcthhail5.(b) VitbngkhiuccbincA, B.A = {A1B4, A2B5, A3B6, A4B1, A5B2, A6B3}B= {A1B1, A2B2, A3B3, A4B4, A5B5, A6B6}(c) Mtnhmyccbincl{A, A}Giibi2.9.Gi A:TtccngratngbnB:TtccngramttngC:Mingiramttngkhcnhau(a) Xcsutttccngratngbn, P(A) =163.(b) Xcsutttccngramttng, P(B) =663.(c) Mingiramttngkhcnhau, P(C) =6 5 463.52Giibi2.11.Gi A:TrongssnphmlyracngssnphmxuMicchlyngunhinksnphmtlhngnsnphmlmtthpchpkcanphnt.Do, || = CknCCsmcchlyrassnphmxutmsnphmxutronglhng.CCksnmcchlyrak ssnphmtttn msnphmtttronglhng.Theoquytcnhn, |A| = CsmCksnm.Do,P(A) = |A|||=CsmCksnmCknGiibi2.13.Gi A:Hpthnhtccha3snphmMicchxpngunhinmtsnphmvomttrong3hplmtcchchnngunhinmttrong3hp.Do,scchxp12snphmngunhinvo3hplschnhhplpchp12ca3phnt,tcl || = A123= 312.Scchxp3snphmchohpthnhtlC312.Scchxp9snphmchohpthhaivbalschnhhplpchp9ca2phnt,tclA92= 29.T,theoquytcnhn, |A| = 29C312Do,P(A) = |A|||=29C312312= 0.212Giibi2.15.Gi A:Baonstbratothnhmttamgic.Gix, y, l (x + y)ldi3khccbngunhin.Khi, = {(x, y)|x > 0, y> 0, x + y< l}tothnhtamgicthx, yphitha:___x + y > l (x + y)x + l (x + y) > yy + l (x + y) > x___x + y >l2y 0Giibi3.7. TathyrngfX(x)lmthmixngquatrcOx.DoP(X< 0) = 1/2.Giibi3.9.(a) TatmctiukinfX(x) 0vimix Rv_fX(x)dx = 1.Tac_fX(x)dx =_11c(1 x2)dx =4c3Suyrac =34.Tathyrngvigitrc =34thfX(x) 0vimix R.Vyc =34(b) TacEX=_xfX(x)dx =_1134x(1 x2)dx = 068(c) TacEX2=_x2fX(x)dx =_1134x2(1 x2)dx =15Do,V ar(X) = EX2(EX)2=15(d) TacFX(x) =_xfX(t)dt.Vix < 1thFX(x) = 0.Vi 1 x 1thFX(x) =_x134(1 t2)dt = 14x3+34x +12Vix > 1thFX(x) = 1.VyFX(x) =___0 nux < 114x3+34x +12nu1 x 11 nux > 1Giibi3.11. TacpX(x) =___1/4 nux = 01/2 nux = 11/4 nux = 2DoE_X_=3

i=1xipX(xi)= 0.14+ 1.12+ 2.14= 1E(X) =3

i=1xipX(xi)= 0.14+ 1.12+ 4.14=3269Do,V ar(X) = EX _E_X__2=32 12=12Giibi3.13.(a) Vix < 0hocx > 4thfX(x) = F

X(x) = 0.Vi0 < x < 1thfX(x) = F

X(x) = 1/2.Vi1 < x < 4thfX(x) = F

X(x) = 1/6.Vix = 0taclimxx+FX(x) FX(0)x= limx0+x/2x= 1/2limxxFX(x) FX(0)x= limx0+0 0x= 0Do,FXkhngcohmtix = 0.Vix = 1taclimxx+FX(1 + x) FX(1)x= limx0+1+x6+13 12x= 1/6limxxFX(1 + x) FX(1)x= limx01+x212x= 1/2Do,FXkhngcohmtix = 1.Vix = 4taclimxx+FX(4 + x) FX(4)x= limx0+1 1x= 0limxxFX(4 + x) FX(4)x= limx04+x6+13 1x= 1/6Do,FXkhngcohmtix = 4.VyfX(x) =___1/2 nu0 < x < 11/6 nu1 < x < 40 nux < 0hocx > 4(b) Do0.5 < 0.75 < 1nn1 < x0.75< 4.T,tacphngtrnhFX(x0.75) =x0.756+13= 0.75Giiratacx0.75= 2.5.70(c) TacEX =_xfX(x)dx=_1012xdx +_4116xdx= 1.5(d) TacE(1/X) =_1xfX(x)dx=_1012xxdx +_4116xxdx= (e) Tac(i) FY (0) = P(Y< 0) = P(X 1) = F(1) = 0.5.(ii) TatnhEY = P(X 1) + P(X> 1)= F(1) + 1 F(1)= 1 2F(1)= 0E(Y2) = P(X 1) + P(X> 1)= 1DoV ar(Y ) = E(Y2) (EY )2= 1 02= 1Giibi3.15.(a) Tatmktiukinf(x) 0vimix Rv_f(x)dx = 1.Tac,_f(x)dx = k_40x2(4 x)dx = k643Suyrak =364.Tathyrngvik =364thf(x) 0vimix R.Vyk =364.71(b) F(x) =_xf(y)dy.Nux < 0thF(x) = 0Nu0 x 4thF(x) =364_x0y2(4 y)dy=116x33256x4Nux > 4thF(x) = 1VyF(x) =___0 x < 0116x33256x40 x 41 x > 4(c) Tac,EX =_xf(x)dx=_40364x3(4 x)dx= 2.4EX2=_x2f(x)dx=_40364x4(4 x)dx= 6.4V ar(X) = EX2(EX)2= 0.64Xtg(x) =364x2(4 x), 0 x 4.Tac,g

(x) =364x(8 3x)Tathyg(x)idutdngsangmkhiquagitrx =83.Dog(x)tccitigitrny.VyMod(X) =83(d)Gi A:Cntrngchttrcmtthngtui.TacA = {X< 1}vP(A) = P(X< 1) = F(1) =116 3256= 0.050872Giibi3.17.(a)Gi A:NhnclhngtthngAB:NhnclhngtthngBX:SlhngtronghaillyraTacXlyccgitr0,1,2vA, Blccbincclp.TacP(X= 0) = P(A B) = P(A)P(B) =18201720= 0.765P(X= 1) = P(AB + AB) = P(A)P(B) + P(A)P(B) =2201720+1820320= 0.22P(X= 2) = P(AB) = P(A)P(B) =220320= 0.015T,tacbngphnphixcsutX 0 1 2P 0.765 0.22 0.015vhmmtcaXf(x) =___0.765 khix = 00.22 khix = 10.015 khix = 20 khix = 0, 1, 2(b)Gi Y :slhngtrong3llyratthngBTac,P(Y= k) =Ck3C3k17C320, k = 0, 1, 2, 3vtanhncbngphnphixcsutY 0 1 2 3P 0.596 0.358 0.045 0.00173vhmmtcaYf(y) =___0.596 khiy= 00.358 khiy= 10.045 khiy= 20.001 khiy= 30 khiy = 0, 1, 2, 3Giibi3.19.(a) TiukinfX(x) 0vimix Rv_fX(x)dx = 1,tatnhcc 0v_fX(x)dx =_0cxex/2dx = 4cSuyrac = 1/4(b) Theonhngha,FX(x) =_fX(x)dx.Nux < 0thFX(x) = 0.Nux 0thFX(x) =_014tet/2dt = 1 12(x + 2)ex/2.VyFX(x) =___0 nux < 01 12(x + 2)ex/2nux 0(c) TacEX=_xfX(x)dx =_014x2ex/2dx = 4.(d) TacEX2=_x2fX(x)dx=_014x3ex/2dx= 24V ar(X) = EX2(EX)2= 8(X) =_V ar(X)= 2274(e) DoXlbinngunhinlintcnnviMed(X) = mtaphicP(X m) = 1/2,tclF(m) = 1 12(m + 2)em/2=12.Ttatmcm = 1.5361.Giibi3.21.(a) Tac,X 0 1 2 3P 0.42 0.425 0.14 0.015(b) Tac,F(x) =___0 khix 00.42 khi0 < x 10.845 khi1 < x 20.985 khi2 < x 31 khix > 3(c) Tacthtnhbngmttronghaicchsau:P(0 < X 4) = P(X= 1) + P(X= 2) + P(X= 3) = 0.425 + 0.14 + 0.015 = 0.58P(0 < X 4) = P(1 X< 4) = F(4) F(1) = 1 0.42 = 0.58Giibi3.23.(a) XcsutlyctransistorloiA,Blnltlp1=100150=23vp2= 1 p1=13.GiXlstransistorcrtra.X {1, 2, . . .}.TacntmP(X {9, 10}) = P(X= 9) + P(X= 10)= p81p2 + p91p2= 0.0217(b) Ginlstransistortnhtcrtra.Theogithit,P(X= n) = pn1=_23_nlog(1/3)log(2/3)= 2.7095Vyn = 3.Giibi3.25.Gi Ai:Tunglnthic6nt (i = 1, 2, 3)X:Stinthucsau1lnchiTac,P(X= 0) = P(A1A2A3)= P(A1)P(A2)P(A3)=565656= 0.579P(X= 2000) = P(A1A2A3 + A1A2A3 + A1A2A3)= P(A1)P(A2)P(A3) + P(A1)P(A2)P(A3) + P(A1)P(A2)P(A3)= 3.165656= 0.347P(X= 4000) = P(A1A2A3 + A1A2A3 + A1A2A3)= P(A1)P(A2)P(A3) + P(A1)P(A2)P(A3) + P(A1)P(A2)P(A3)= 3.161656= 0.069P(X= 6000) = P(A1A2A3)= P(A1)P(A2)P(A3)=161616= 0.005T,taccbngphnphixcsutcaXnhsau:X 0 2000 4000 6000P 0.579 0.347 0.069 0.005T,EX=4

i=1xiP(X= xi) = 100076(a) ngichivluvdihuvnthtacnA = EX= 1000(b) trungbnhmilnngichimt1ngnthA = EX + 1000 = 2000.Giibi3.27.(a) TacFX(x) = P(X< x) = P(X x) =___cx2nu0 x < 251 nux 25(i) VFXlintctrinnlimx25 FX(x) = F(25),tclc252= 1.Vyc =1252.(ii) Vi0 x < 25thfX(x) =2252x.Vix > 25thfX(x) = 0Vix = 25,taclimt0+FX(25 + t) FX(25)t= limt0+1 1t= 0limt0FX(25 + t) FX(25)t= limt01252(25 + t)21t= 0Suyralimt0+FX(25 + t) FX(25)t=limt0FX(25 + t) FX(25)t= 0HayfX(25) = F

X(25) = 0.VyfX(x) =___2252x nu0 x < 250 nux 25(iii) TacEX=_xfX(x)dx =_2502252x2dx =503(iv) TacP(X 10|X 5) =P(5 X 10)P(X 5)=F(10) F(5)1 F(5)=18(b) GiY lstinngichitc.TacEY = P(r < X 2r) + 10P(X r) 1= F(2r) F(r) + 10F(r) 1= F(2r) + 9F(r) 177TheogithitEY= 0.25tasuyraF(2r) + 9F(r) 1 =13252r21 = 0.25Vyr = 7.7522cm.Giibi3.29.(a) A = 2vF(x) =___x2khi0 x 10 khix < 01 khix > 1EX=23, V ar(X) = 0.055(b) A = 0.5vF(x) =___12(1 cos x) khi0 x 0 khix < 01 khix > EX=2, V ar(X) =242(c) A = vF(x) =___sin (x) khi0 x 120 khix < 01 khix >12EX=12 1, V ar(X) = 32(d) A = 3vF(x) =___1 1x3khi0 10 khix < 1EX=32, V ar(X) =3478Giibi3.31. Tac,X20 1 4P 0.3 0.5 0.2X + Y 2 1 0 1 2 3P 0.06 0.17 0.27 0.27 0.17 0.06T,EX2= 0.13EX4= 3.7V ar(X2) = EX4(EX2)2= 3.683E(X + Y ) = 0.5E[(X + Y )2] = 1.9V ar(X + Y ) = E[(X + Y )2] [E(X + Y )]2= 1.65Giibi3.33. TacY=___1 nuX 01 nuX< 0Do,FY (y) = P(Y< y) =___0 nu y 1P(X< 0) nu 1 < y 11 nu y> 1VyFY (y) =___0 nu y 1FX(0) nu 1 < y 11 nu y> 179Giibi3.35. Tronggiitchcstabit,nhngha7.1. ChoAlmttpconkhctrngca R.TaniAlmtkhongtrong Rnuvchnu[x, y] Avimix, y Asaochox y.nh l7.2.Cho fl mt hm s thc lin tc trn mt khong A. Lc , f(A) l mt khongtrong R.Trlibiton,tacntmFY (y) = P(Y< y) = P(F(X) < y)Viy> 1,tacFY (y) = 1.Viy 0,tacFY (y) = 0.Vi 0 1Nhnxt7.3. TathyrngY U(0, 1)vkhngphthucvophnphicaX.MtsbinngunhinthngdngGii bi 4.1. Gi Xlssnphmkhngttiuchuntrong10snphmlyra. Tac,X B(10, 20008000) = B(10, 0.25).Khi,xcsuttrong10snphmlyrac2snphmkhngttiuchunlP(X= 2) = C210(0.25)2(0.75)8= 0.282Giibi4.3. GiXlscontraitrongmtgianhc4conthX B(4; 0.5)(a) XcsutchaitraivhaigitrongbnaconlP(X= 2) = C24(0.5)2(0.5)2= 0.375(b) XcsutcmtcontraitrongsbnaconlP(X= 1) = C14(0.5)1(0.5)3= 0.25(c) XcsutcbnultrailP(X= 4) = C44(0.5)4(0.5)0= 0.0625Gii bi 4.5. Gi Xlstrnghpcnchmsccbittrong20casinh. Tac, X B(20; 0.01).(a) XcsutkhngctrnghpnocnchmsccbitlP(X= 0) = C020(0.01)0(0.99)20= 0.818(b) XcsutcngmttrnghpcnchmsccbitlP(X= 1) = C120(0.01)1(0.99)19= 0.16581(c) XcsutcnhiuhnmttrnghpcnchmsccbitlP(X> 1) = 1 [P(X= 0) + P(X= 1)] = 1 (0.818 + 0.165) = 0.017Khi xp x phn phi nh thc bng phn phi Poisson, ngha l X P(20 0.01) = P(0.2), tanhncP(X= 0) = e0.2= 0.819P(X= 1) = e0.20.211!= 0.164P(X> 1) = 1 [P(X= 0) + P(X= 1)]= 1 (0.819 + 0.164) = 0.017Ktlun:Vicmu20vtlbnhp = 0.01thktqucahailoiphnphinyxpxnhnhau.Giibi4.7. GiXlsngimcbnhAtrongnhm400ngi.Khi,X B(400, 0.1)(a) Cngthctnhxcsuttrongnhmcnhiunht50ngimcbnhAlP(X 50) =50

i=0Ci4000.1i0.9400i(b) Do n = 400 20 v p = 0.1 khng qu gn 0 hoc 1, nn ta c th p dng cng thc xpxphnphichun.P(X 50) = P_X 400.(0.1)_400.(0.1).(0.9)50 400.(0.1)_400.(0.1).(0.9)_ (1.667)= 0.953Gii bi 4.9. Vmigisnxutc20snphmnnmikhongthigian30phtmysnxutc10snphm. Gi Xlsthphmmysnxutratrong30pht. TheogithittacX B(10, 0.85).TacntmP(X {8, 9}) = P(X= 8) + P(X= 9)= C810(0.85)8(0.15)2+ C910(0.85)9(0.15)1= 0.6233Giibi4.11.82Gi Ai:"Xcscthixuthinngichit"(i = 1, 2, 3)B:"Ngichithng"TacP(B) = P(A1A2A3) = P(A1)P(A2)P(A3) =161616= 0.00463Vysaunhiulnchingilmcisthng.GiXlstinngilmcithuc.TacbngphnphicaXnhsau:X -5000 1000P 0.00463 0.99537KhitrungbnhmivnngilmcisthngEX= 972.2222.Giibi4.13.(a) TrchttalpbngphnphixcsutcaXvY nhsau:X 0 1P 0.2 0.8Y 0 1 2P 0.64 0.32 0.04Tac,P(X + Y= 0) = P(X= 0, Y= 0)= P(X= 0)P(Y= 0)= 0.512P(X + Y= 1) = P(X= 0, Y= 1) + P(X= 1, Y= 0)= P(X= 0)P(Y= 1) + P(X= 1)P(Y= 0)= 0.384P(X + Y= 2) = P(X= 0, Y= 2) + P(X= 1, Y= 1)= P(X= 0)P(Y= 2) + P(X= 1)P(Y= 1)= 0.096P(X + Y= 3) = P(X= 1, Y= 2)= P(X= 1)P(Y= 2)= 0.008TsuyrabngphnphixcsutcaX + Y lX + Y 0 1 2 3P 0.512 0.384 0.096 0.00883KimtraX + Y B(3,15).Tac,P(X + Y= 0) = C03_15_0_45_3P(X + Y= 1) = C13_15_1_45_2P(X + Y= 2) = C23_15_2_45_1P(X + Y= 3) = C33_15_3_45_0VyX + Y B(3,15).(b) TalpbngphnphixcsutcaXvY nhsau:X 0 1P 0.5 0.5Y 0 1 2P 0.64 0.32 0.04Tac,P(X + Y= 0) = P(X= 0, Y= 0)= P(X= 0)P(Y= 0)= 0.32P(X + Y= 1) = P(X= 0, Y= 1) + P(X= 1, Y= 0)= P(X= 0)P(Y= 1) + P(X= 1)P(Y= 0)= 0.48P(X + Y= 2) = P(X= 0, Y= 2) + P(X= 1, Y= 1)= P(X= 0)P(Y= 2) + P(X= 1)P(Y= 1)= 0.18P(X + Y= 3) = P(X= 1, Y= 2)= P(X= 1)P(Y= 2)= 0.02TsuyrabngphnphixcsutcaX + Y lX + Y 0 1 2 3P 0.32 0.48 0.18 0.0284GisX + Y cphnbnhthcdngB(3, p).Khi,P(X + Y= 0) = C03p0(1 p)3= 0.32Tasuyra,p = 0.316Mtkhc,vigitrpny,P(X + Y= 3) = C33p3(1 p)0= p3= 0.0316 > 0.02(vl)VyX + Y khngcphnbnhthc.Giibi4.15. GiXlsbthmycsaitrongmipht.Tac,X B(5000, 0.0004)(a) SbthtrungbnhmiphtmycsailEX= 5000.(0.0004) = 2(b) Sbthtinchcnhttrongmiphtmycsailsbthmxcsutmycsaillnnht,tclMod(X).Tac,5000.(0.0004) 0.9996 Mod(X) 5000.(0.0004) + 0.9996tcl1.0004 Mod(X) 2.9996SuyraMod(X) = 2.(c) Xcsuttrongmtphtmycsaitnht3bthlP(X 3) = 1 P(X< 3)= 1 [P(X= 0) + P(X= 1) + P(X= 2)]Don = 5000 100,p = 0.0004 0.01,np = 2 20.NntacthtnhccxcsutxpxtheolutPoissonnhsau,P(X 3) = 1 [P(X= 0) + P(X= 1) + P(X= 2)]= 1 e2_1 +211!+222!_= 0.323Giibi4.17.85(a) GiXlsthphmtrong10snphmckimtra.X {0, 1, 2}.TacP(X= 0) =C1023C1025= 0.35P(X= 1) =C12C923C1025= 0.5XcsutqutrnhsnxutcbocotyuculP(X {0, 1}) = P(X= 0) + P(X= 1) = 0.85GiY lslnqutrnhsnxutcbocotyucutrong8lnkimtra.TacY B(8, 0.85).TacntnhP(X 7) = P(X= 7) + P(X= 8)= C78(0.85)7(0.15)1+ C88(0.85)8(0.15)0= 0.6572(b) GiZlslnqutrnhsnxutcbocokhngtyucutrong8lnkimtra.TacZ B(8, 0.15) P(8 0.15 = 1.2).Do,P(Y 7) = P(Z 1) = e1.2_1.200!+1.211!_= 0.6626(c) Theogithit,tacntnhP(X= 0|X 1) =P(X= 0, X 1)P(X 1)=P(X= 0)P(X 1)=0.350.85= 0.4118Giibi4.19. TacP(X 1|X 1) =P(X 1, X 1)P(X 1)=P(X= 1)P(X 1)=e5.5e5+ e5.5=56Giibi4.21.86(a) Xcsutkhngphittc4chictucthulP(X 3) = P(X= 0) + P(X= 1) + P(X= 2) + P(X= 3)= e2_1 + 2 +222!+233!_= 0.857(b) Xcsutttc4chictucthulP(X 4) = 1 P(X 3) = 1 0.857 = 0.143(c) XcsutcahngkhngpngcyuculP(X> 4) = P(X 4) P(X= 4)= 0.143 e2244!= 0.053(d) TrungbnhstcthulEX= 2(e) Tac,P(X> 5) = 1 P(X 5)= 1 e2_1 + 2 +222!+233!+244!+255!_= 0.017 < 0.02Nhvystcnthitxcsutkhngpngcnhucuthubhn2%l5.Giibi4.23.(a) Theogithit,tacX P()vP(X= 1) = 3P(X= 0)e11!= 3e00! = 3Do,P(2 X 4) = P(X= 2) + P(X= 3) + P(X= 4)= e3_322!+333!+344!_= 0.616187(b) TacP(X= 0) = e3= 0.0498.Theogithit,U B(100, 0.0498).Do,P(U 1) = P(U= 0) + P(U= 1)= C0100(0.0498)0(0.9502)100+ C1100(0.0498)1(0.9502)99= 0.0377Giibi4.25. Gi Xil s khch hng xut hin trong pht th i (i = 1, 2, 3). Theo gi thitccXiclpnhauvXi P(5).Gi Y lskhchhngxuthintrongkhongthi gian3pht. TacY =X1 + X2 + X3.Theobi(4.20),thY P(15).TacntmP(Y 10) = 1 P(Y 9) = 1 9

k=0e1515kk!= 0.9301Giibi4.27. TacX P()vp = P(X= 0) =e00!= e.GiY lsbnhquykhngcnhokhtrong20bnhtronghp.KhiY B(20, p).TaciukinP(Y 2) = 0.925 P(Y= 0) + P(Y= 1) + P(Y= 2) = 0.925 C020p0q20+ C120p1q19+ C220p2q18= 0.925 vi q= 1 p 171q20360q19+ 190q180.925 = 0Giiphngtrnhnytacq= 0.9501tcp = 0.0499.T = ln(0.0499) = 2.9977 3.Chrng,tacthkimtravi = 3,tacP(Y 2) 0.950220+ 20(0.9502)19(0.0498)+ 190(0.9502)18(0.0498)2 0.925Tuynhinkhngnhtthitphilsnguynvnlstrungbnhnhokhctrongmtcibnh.Giibi4.29.(a) GiAi:"snphmthi,csnxutbimyutin,lchnhphm"(i = 1, 2, . . .)88Xlssnphmcsnxutbimyutintrckhintorathphmutin.TacP(Ai) = 0.98vimii = 1, 2, . . .vP(X= 0) = P(A1) = 0.02P(X= 1) = P(A1A2) = 0(.98)(0.02). . .P(X= n) = P(A1A2. . . AnAn + 1) = (0.98)n(0.02). . .Do,EX =

n=0nP(X= n)= 0.02

n=0n(0.98)n(7.4)tp = 0.98 < 1,tacntm

n=0npn.Tacn

k=0pk=1 pn+11 p(7.5)Lyohmtheophaivca(7.5),n

k=0kpk1=npn+1(n + 1)pn+ 1(1 p)2(7.6)Nhnhaivca(7.6)chop,n

k=0kpk=npn+2(n + 1)pn+1+ p(1 p)2=npn+1(p 1) pn+1+ p(1 p)2=npn+1p + 1 pn+1(1 p)2+p(1 p)2(7.7)Tiptheotaschngminhlimnnpn= 0.Thtvy,dop < 1nn1p> 1hay1p= 1 + rvir > 0no.Tacnpn=n(1n)n=n(1 + r)nM(1 + r)n C2nr2=n(n1)2r2(dokhaitrinnhthcNewton).Nnnpn2(n 1)r2 0 khi n 89Vylimnnpn= 0.T,kthpvi(7.7)tasuyra

k=0kpk=limnn

k=0kpk=p(1 p)2=0.980.022= 2450Thayvo(7.4)tacEX= 0.02(2450) = 49(b) GiY lsthphmtrong10snphmcchn.TacY B(10, 0.02).TacntmP(Y 2) = P(Y= 0) + P(Y= 1) + P(Y= 2)= C010(0.02)0(0.98)10+ C110(0.02)1(0.98)9+ C210(0.02)2(0.98)8= 0.9991(c) TaxpxY biphnphiPoissonsau:Y P(10(0.02)) = P(0.2).Do,P(Y 2) = P(Y= 0) + P(Y= 1) + P(Y= 2)= e0.2_0.200!+0.211!+0.222!_= 0.9989(d) Ginlssnphmtnhtphiclyrathayucu.Zlsthphmtrongnsnphm.TacZ B(n, 0.02)vP(Z 1) 12 1 P(Z= 0) 12 P(Z= 0) 12 C0n(0.02)0(0.98)n12 0.98n12 n ln(1/2)ln(0.98)= 34.3096Vyn = 35.Giibi4.31. VX N(, 0.012)nnY=X 0.1 N(0, 1).90(a) TacntnhP(X 1.15) = P_X 0.11.15 0.1_= P(Y 1.5)= 1 (1.5)= 1 0.9332 = 0.0668(b) Gi chiu di xe cn tm l x0.1. Ta cn phi c P(X x0.1) = 0.9 hay P(X x0.1) = 0.1,tclP_X 40.4x0.140.4_= _x0.140.4_= 0.1.Suyrax0.1= 4 + 0.4z0.1= 4 0.4z0.9= 4 0.4(1.282) 3.49Gii bi 4.33. Gi Xltrnglngtri cyth XN(500, 42). Vi Y =X 5004thY N(0, 1).Do,(a) Tltricyloi1lP(X> 505) = P_X 5004>505 5004_= P(Y> 1.25)= 1 (1.25)= 0.106(b) Tltricyloi2lP(495 X 505) = P_495 5004X 5004505 5004_= P(1.25 Y 1.25)= 0.788(c) Tlcaloi3lP(X< 495) = P_X 5004 180) = 1 P(X 180)= 1 P_X 1754180 1754_= 1 P(Y 1.25)= 1 (1.25)= 1 0.894 = 0.106(b) tlngitrngthnhcchiucaot166cmn177cmlP(166 X 177) = P_166 1754X 1754177 1754_= P(2.25 Y 0.5)= (0.5) (2.25)= (0.5) + (2.25) 1= 0.692 + 0.988 1 = 0.68(c) Tac,P(X< h0) = P_X 1754h01754_= P_Y h01754_= _h01754_= 0.33Tathy0.33 < 0.5nnh01754< 0.Do,_175 h04_= 1 0.33 = 0.67.Suyra,175 h04= 0.44,tclh0= 173.24(d) Tatmatiukinsau:P(175 a X 175 + a) = 0.992Tac,P(175 a X 175 + a) = P_175 a 1754X 1754175 + a 1754_= P_a4 Y a4_= 2_a4_1Tasuyra_a4_= 0.95,tcla = 6.6Giibi4.37. TacfX(x) =122exp_x22(2)_= ln [fX(X)] = ln_12_X24HnnaE(X2) = V ar(X) + (EX)2= 2 + 02= 2.DoH = E_ln_12_+X24_= ln(2) +14E(X2)= ln(2) +24 1.766LthuytmuGiibi5.1.(a) Tacbngtnssau:xi61 62 63 64 65 66 67 68 69 70ni1 2 2 4 8 4 8 5 2 1Ttatnhc,x =1nk

i=1nixi= 65.811s2=1n 1k

i=1ni(xix)2= 4.436s = 2.106(b) Tacbngtnsutsau:xi70 69 68 67 66 65 64 63 62 61fi0.027 0.055 0.135 0.216 0.108 0.216 0.108 0.054 0.054 0.027Tbngtrn,tathyrngP(X 66) = 0.541 12P(X 66) = 0.568 12Do,trungvcachiucaosinhvinlpnyl66.94Giibi5.3. Tacbngtnssau:xi43 44 45 46 47 48 49 50 51 52ni1 2 2 3 1 2 4 4 3 2Trungbnhmu:x =1nk

i=1nixi= 48.125Phngsaimu:s2=1n 1k

i=1ni(xix)2= 7.245lchtiuchun:s =s2= 2.692Giibi5.5. Tac,n

i=1yi=n

i=1(a + bxi)= na + bn

i=1xi1nn

i=1yi= a + b1nn

i=1xiy = a + bxvs2y=1n 1n

i=1(yiy)2=1n 1n

i=1(a + bxi(a + bx))2=1n 1n

i=1b2(xix)2= b2s2xsy= |b|sx95Giibi5.7. Tacgitrtrungtmcatngkhongl5.5, 15.5, 25.5, 35.5, 45.5, 55.5, 65.5, 75.5, 85.5, 95.5Tasdngphpbinisau:yi=xi55.510Khi,gitrcayisl 5, 4, 3, 2, 1, 0, 1, 2, 3, 4.Ttatnhc,y =1nk

i=1niyi= 0.653s2y=1n 1k

i=1ni(yiy)2= 8.349Do,x = 10y + 55.5= 48.97s2x= 102s2y= 834.9Giibi5.9.n = 8, x = 3.0775, s2= 0.096Giibi5.11.n = 97, x = 19.133, s2= 0.054clngthamsthngkGiibi6.1. Ta cn = 100 > 30,2cha bit.Khong clng 95% cho gi trtrung bnhthtcdng_x z12sn, x + z12sn_Trong, = 0.05,z12= z0.975= 1.96,x = 0.1,s = 0.014.Do,khongclng95%chogitrtrungbnhthtl(0.0973, 0.1027)Gii bi 6.3. Tacn=8 30, = 100. Do , khong tin cy cho tui th trung bnh ca bng ncdng_x z12n, x + z12n_Trong,x = 1000, = 0.05,z12= z0.975= 1.96.Vykhongtincy95%chotuithtrungbnhcabngnl(980.4, 1019.6).(b) Tacdungsaicaclnglz12n= z1210010= 15.Tsuyraz12= 1.5Trabngtatmc = 0.13362vdotincyl1 = 0.8664 = 86.64%(c) Ta bitrngkhincngln thdungsaicngnh.Hnna,tathy rngkhin = 29 < 30tht280.97510029= 38.0304 > 25.Dogitrnphilnhnhocbng30.Tiukin1.96.100n 25Suyran 61.4656.Vytacnthnghimtnht62bngGiibi6.9.(a) n = 53,trungbnhmux = 12.21,lchchuns = 0.103.(b) Tac,kchthcmun=53 30,2chabit.Do,vitincy0.95,khongtincycal_x z0.975sn, x + z0.975sn_Thayx,svz0.975= 1.96vobiuthctrn,tatmckhongtincycal(12.18, 12.24)(c) Tabitrngkhi ncnglnth dungsai cngnh. Hnna, vi n=53trongcub)tathy = 0.028 > 0.02.Do,gitrnphilnhn53.Tiukinz0.975sn 0.02Suyran 101.89.Vytaphiquansttnht102trnghp.98Giibi6.11.(a) Tbngsliucamu,tacn = 128,x = 1391.41vs = 234.45.(b) Tacn = 128 30,2chabit.Vitincy0.95,khongtincychol_x z0.975sn, x + z0.975sn_Thayx,svz0.975= 1.96,tac(1350.79, 1432.03)(c) Tabitrngkhincnglnthdungsaicngnh.Hnna,vin=128trongcub)tathy = 40.6 > 30.Do,gitrnphilnhn128.Tiukinz0.975sn 30Suyran 234.63.Vytaphiquansttnht235bngn.Gii bi 6.13. Tacn = 2000,f=13802000= 0.69,nf= 1380 > 5,n(1 f) = 620 > 5.Do,khongtincy95%chotlphiubucdng_f z12_f(1 f)n, f+ z12_f(1 f)n_tcltlphiubutithiulf z12_f(1 f)nThayf =0.69, n=2000, =0.05, z0.975=1.96vobiuthctrntac0.6697, tcl66.97%.Giibi6.15. saiskhngvtqu2%thz12_f(1 f)n 0.02Trong,f= 0.9, = 0.1,z12= z0.95= 1.65.Do,n 612.5625Vytacnphikhmtnht613ngi.Giibi6.17.99(a) Khongtincychotlpvitincy1 l_f z12_f(1 f)n, f+ z12_f(1 f)n_Theogithittactnsutkhibnhlf=4050= 0.8.Vi tin cy 0.95 ta c = 0.05 v z0.975= 1.96. Do , khong tin cy cho t l p vi tincy0.95l(0.69, 0.91).Vi tin cy 0.99 ta c = 0.01 v z0.995= 2.58. Do , khong tin cy cho t l p vi tincy0.99l(0.65, 0.946).(b) Tacsaiscaclngl = z12_f(1 f)nVi tin cy 0.95 ta c = 0.05v z0.975= 1.96. Do sais khng vt qu 0.02 tacniukin1.96 _0.8 0.2n 0.02Suyran 1536.64.Vytacnquansttnht1537trnghp.Giibi6.19. Tac f=20500= 0.04,nf= 20 > 5,n(1 f) = 480 > 5.Do ,khong tin cy95%chotlccnhdutronghcdng_f z1+2_f(1 f)n, f+ z1+2_f(1 f)n_Trong, = 0.05,z1+2= z0.975= 1.96.Do,khongtincy95%chotlccnhdutronghl(0.0228, 0.0572).T,khongtincy95%choscctronghl(34965.03, 877719.3)Giibi6.21.(a) Tacx =1n

ki=1nixi= 35.89,s2=1n1

ki=1ni(xix)2= 3.21,s = 1.792.Tacn=100>30,2chabit.Do,khongtincy95%chokhilngtrungbnhccqucamcdng_x z12sn, x + z12sn_Trong, = 0.1,z12= z0.975= 1.96.T,khongtincy95%chokhilngtrungbnhccqucaml(35.539, 36.241)100(b) Tacf=5100= 0.05,nf= 5 5,n(1 f) = 95 5.Do,khongclngchotlloi2vitincy90%cdng_f z12_f(1 f)n, f+ z12_f(1 f)n_Trong, =0.1, z12=z0.95=1.65. Vykhongclngchotlloi 2vi tincy90%l(0.014, 0.086)KimnhgithuytthngkGiibi7.1. Tacnkimnhccgithuyt___H0: = 380H1: = 380yltrnghpn = 36 30v2chabit,nntadngz =n(x )s=36(350 380)40= 4.5Tathy |z| >z12=z0.975=1.96. DotabcbgithuytH0. Nghalli bococagimckhngngtincy.Giibi7.3. Tacnkimnhccgithuyt___H0: = 25H1: < 25yltrnghpn = 15 < 30v2chabit,Xcphnphichun,nntadngt =n(x )s=15(24 25)2= 1.9365Tathyt 30v2chabit,nntadngz =n(x )s=100(0.9856 1)0.0208= 6.9204Ta thy |z| z0.975= 1.96. Do ta bc b gi thuyt H0. Ngha l my hot ng khng bnhthng.Giibi7.9.103(a) Taavbnggitrsaux 155 165 175 185 195 205Sngi 3 9 11 3 2 1Tatnhck = 6n = 29x =1nk

i=1nixi= 173.2759s2=1n 1k

i=1ni(xix)2= 143.3498s = 11.9729(b) Tacn=29 tn112= t190.975= 2.093.DotabcbgithuytH0.Nghalchnkingctcdnglmthayitrnglng.Giibi7.15. Tacnkimnhccgithuyt___H0: 1= 2H1: 1 = 2Tatnhcz =x1x2_21n1+22n2=18 24_3220+3220= 6.3246106Tathy |z| > z1= z0.95= 1.65.Do,tabcbgithuytH0nghalhailoichtnlngnyctctchykhcnhau.Giibi7.17. Tacnkimnhccgithuyt___H0: X= YH1: X> YTatnhcs2=(n11)s2x + (n21)s2yn1 + n22=(50 1)72+ (40 1)9.2250 + 40 2= 64.795s = 8.0495vt =x ys_1n1+1n2=60 528.0495_150+140= 4.6851Tathyt > tn1+n221= t880.95 z0.95= 1.65.Do,tabcbgithuytH0nghalhmlngngtrongmusau5gilmvicgimi.Giibi7.19. GiX,Y lnltltrnglngtrssinhnngthnvthnhth.Tacnkimnhccgithuyt___H0: X= YH1: X< YTatnhcs2=(n11)s2x + (n21)s2yn1 + n22=(8000 1)0.32+ (2000 1)0.228000 + 2000 2= 0.08s = 0.2828107vt =x ys_1n1+1n2=3 3.20.2828_18000+12000= 28.2885Ta thy t < tn1+n221= t99980.95 z0.95= 1.65. Do , ta bc b gi thuyt H0ngha l trng lngtrungbnhcatrssinhthnhthcaohnnngthn.Giibi7.21. Tatnhcn1= 10n2= 10x =1n1n1

i=1xi= 16.015y =1n2n2

i=1yi= 16.005s2x=1n11n1

i=1(xix)2= 0.000917sx= 0.0303s2y=1n21n2

i=1(yiy)2= 0.00065sy= 0.0255Tacnkimnhccgithuyt___H0: 1= 2H1: 1 = 2Tatnhcs2=(n11)s2x + (n21)s2yn1 + n22=(10 1)0.000917 + (10 1)0.0006510 + 10 2= 0.000783s = 0.028108vt =x ys_1n1+1n2=16.015 16.0050.028_110+110= 0.7986Tathy |t| < tn1+n221= t180.95= 1.734.Do,tachpnhngithuytH0nghalhaimyrtncvobnhnhnhau.Giibi7.23.(a) TaavbnggitrsauTrnglng 3100 3300 3500 3700 3900Sbtrai 1 3 8 10 3Sbgi 2 10 10 5 1Tatnhck = 5N1= 25N2= 28x =1N1k

i=1nixi= 3588y =1N2k

i=1niyi= 3450s2x=1N11k

i=1ni(xix)2= 40266.67sx= 200.6656s2y=1N21k

i=1ni(yiy)2= 37407.41sy= 193.4099(b) Tacnkimnhccgithuyt___H0: X= YH1: X = Y109Tatnhcs2=(N11)s2x + (N21)s2yN1 + N22=(25 1)40266.67 + (28 1)37407.4125 + 28 2= 38752.94s = 196.8577vt =x ys_1N1+1N2=3588 3450196.8577_125+128= 2.5476Ta thy |t| > tn1+n221= t510.95 z0.95= 1.65. Do , ta bc b gi thuyt H0ngha l trnglngbtraivbgilcssinhkhcnhau.(c) Nhphaimuli,tacTrnglng 3100 3300 3500 3700 3900Strssinh 3 13 18 15 4Tatnhck = 5N = 53z =1Nk

i=1nizi= 3515.094s2z=1N 1k

i=1ni(ziz)2= 42844.7sz= 206.9896TacN= 53 30,2chabit.Do,khongtincychoscnngtrungbnhcatrssinhcdng_z z12szN, z + z12szN_Trong, = 0.05, z12= z0.975= 1.96.Thayvotatmckhongtincy95%choscnngtrungbnhcatrssinhl(3459.367, 3570.821).110Giibi7.25. Tacnkimnhccgithuyt___H0: p = 0.98H1: p < 0.98Tacn = 500,f=500 28500= 0.944,nf= 472 5vn(1 f) = 28 5.Do,tadngz =n(f p)pq=500(0.944 0.98)0.98 0.02= 5.7499Tathyz z12= z0.995= 2.58.DotabcbgithuytH0.Nghalbinphpkthutmilmthayitlphphm.(b) Tacnkimnhccgithuyt___H0: p = 0.02H1: p = 0.02Tacn = 800,f=24800= 0.03,nf= 24 5vn(1 f) = 776 5.Do,tadngz =n(f p)pq=800(0.03 0.02)0.02 0.98= 2.0203Tathy |z| z12= z0.995= 2.58.DotachpnhngithuytH0.Nghalnhmybocotlphphmlchpnhnc.Giibi7.31. Gip1:tlnxAp2:tlnxBTacnkimnhccgithuyt___H0: p1= p2H1: p1 = p2112Tacn1= 250 30n2= 160 30f1=140250= 0.56f2=80160= 0.5 p =n1f1 + n2f2n1 + n2=250 0.56 + 160 0.5250 + 160= 0.5366Tatnhcz =f1f2_ p q_1n1+1n2_=0.56 0.5_0.5366(1 0.5366)_1250+1160_= 1.1885Ta thy |z| < z12= z0.975= 1.96. Do ta chp nhn gi thuyt H0. Ngha l t l n hai xbngnhau.Giibi7.33. Gip1:tltrssinhctrnglngtrn3000gamthnhphp2:tltrssinhctrnglngtrn3000gamnngthnTacnkimnhccgithuyt___H0: p1= p2H1: p1 = p2113Tacn1= 150 30n2= 200 30f1=100150= 0.6667f2=98200= 0.49 p =n1f1 + n2f2n1 + n2=150 0.6667 + 200 0.49150 + 200= 0.5657Tatnhcz =f1f2_ p q_1n1+1n2_=0.6667 0.49_0.5657(1 0.5657)_1150+1200_= 3.3005Tathy |z| > z12= z0.975= 1.96.DotabcbgithuytH0.Nghaltltrssinhctrnglngtrn3000gamthnhphvnngthnkhcnhaumcngha5%.