Phn 1 : L THUYTL thuyt xc sutCu 1: Phn bit cc khi nim: Hon v, t hp, chnh hp (lp v khng lp) ca mt tp con t tp n phn t. Nu cc cng thc xc nh cc s hon v, t hp, chnh hp. Th d minh ha.Ni dungHon v T hpChnh hpKhng lp LpKhi nimHon v ca n phn t l mt nhmc th t gm mt n phn t choT hp chp k ca n phnt( k n ) lmt nhmkhng phn bit th t, gmkphn t khc nhau chn t n phn t choChnh hp(khng lp) chpkcanphnt(k n ) l mt nhm (b) c th t gm k phn t khc nhau chn t n phn t choChnhhplpchpkcan phn t l mt nhm c th t gm k phn t chn t n phn t cho, trong mi phn t c th c mt 1,2,,k ln trong nhmCng thcShonvcan phntck hiu l nP!nP n Sthpchpkcan phn t k hiu l knC!!( )!k n kn nnC Ck n k S chnh hp chp k ca n phn t k hiu l knA!( ) !knnAn kS chnh hp lp chp k ca n phn t k hiu l knBk knB n Th dMt bn c 4hc sinh. Mi cch xp ch 4 hc sinhvo1bnl 1hon v ca 4 phn t. Do s cch xp l 44! 24 P Chn ngu nhin 2 quyn sch t trn gi sch c 3 quyn sch. Mi cch chnl1thp chp 2 ca 3 phn t. Do c 233 C cch chnCho 3 ch s 1,2,3. Mi s t nhin gm 2 ch s khc nhau c lp t 2 trong 3 ch s trn l mt chnh hp khng lp chp 2 ca 3 phn t. Do c th lp c 23A = 3.2 = 6 s.Cho 5 hn bi vo 3 hp. Mi cch xp 5 hn bi vo 3 hp l mtchnh hp lp chp5 ca 3. Do c 5 533 243 B cch xpCu 2 : Th no l mt php th ? Mt bin c (s kin) ?- Quan h gia cc bin c. php tnh ca cc bin c.- Bin c chc chn; khng th; xung khc; i lp biu din qua s Ven Euler ?Tr li:- Php th: l s th hin mt nhm cc iu kin xc nh (G) c tnh lp li.K hiu: TS kin (bin c) : l kt qu ca php th. K hiu: E, A, B, C, .C 2 loi s kin l:+ S kin ngu nhin (Random Effect) : c th xut hin hoc khng xut hin khi thc hin php th, khng ph thuc vo ch quan. S kin ngu nhin l i tng nghin cu ca khoa hc ngu nhin.+ S kin tt nh (Definity Events) : lun xut hin ( hoc) hoc lun khng xut hin khi thc hin php th ()* Quan h (relation) gia cc bin c:+ Quan h bao hm:A B + quan h tng ng:A B * Php tnh (Calculus):+ Hp (tng):A B C + Giao (tch):A B D + Tr (hiu):\ A B E + Hiu i xng: F = A B = (A\B) (B\A) + i lp (b): A i lp vi A ( c ci ny th ko c ci kia)+ Xung khc:A B (khng cng xy ra)+ Nhm y : 1 2...ni jA A AA A ' 1* Biu din qua s Ven Euler cc bin c:Bin c chc chn xy ra:Bin c khng th xy ra: (l mt tp rng) Bin c xung khc:Bin c b (i lp):Cu 3 : A, B,C l 3 bin c gn vi php th G. Biu din qua A, B, C cc bin c sau y :- Ch c A xy ra : ABC ( tng t cho B, C)- t nht mt trong 3 bin c xy ra: ABC- Nhiu nht mt bin c xy ra: (A B C) (A B C) (A B C) (A B C )- Khng c bin c no xy ra: ABCCu 4 : Cc nh ngha xc sut mt bin c. ngha ca xc sut l g?- Gi P(A) l tn sut c xut hin bin c A trong nh ngha xc sut theo tn sut. C th vit nh sau c khng?( ) lim ( )nP A P A- Hai bin c c xc sut bng nhau th c tng ng hay khng?- Mt bin c c xc sut 0, c th xy ra hay khng?Tr li : nh ngha xc sut mt bin c:-Gi s php th c n bin c ng kh nng c th xy ra, trong c m bin c ng kh nng thun li cho bin c A (A l tng kh nng ca m bin c s cp ny). Khi xc sut ca bin c A, k hiu P(A) c nh ngha bi cng thc sau:( )mP An , trong m l s trng hp thun li cho A, n l s trng hp ng kh nng.- C th vit ( ) lim ( )nP A P A v ta c ( )mP An%% ( ) limnmP An Khi cho s php th tng ln v hn th tn sut xut hin bin c A dn v mt s xc nh gi l xc sut ca bin c A.Hai bin c c xc sut bng nhau khng tng ng v 2 bin c A v B tng ng nhau A BB A ' C th xy ra mt bin c c xc sut bng khng. l trng hp khng xut hin kh nng no thun li cho A.2Cu 5: Hai bin c A v B l c lp khi v ch khi P(A B) = P(A).P(B).trong , A B l giao ca A v B, ngha l, n l bin c rng c hai bin c A v B u xy ra.Tng qut hn, mt tp hp bin c bt k (c th gm nhiu hn hai bin c) l c lp ln nhau khi v ch khi vi mi tp con hu hn A1, ..., An ca tp hp trn, ta c Mi quan h gia khi nim c lp v xung khcCngthc nhn xc sut:1. P(AB)=P(A). P(B)( Vi A, B c lp )P(AB)=P(B/A). P(A) = P( A/B). P(B)2. P(ABC) =P(A) . P(B) . P(C)( Vi A,B,C c lp)=P(A) . P(B/A) . P(C/AB)3. P( AB)= P(A) + P(B) ( Vi A B = )P( AB)= P( A) + P(B) - P(AB) ( Vi A, B bt k)4. P(ABC)= P(A) + P(B) + P(C)= P(A) + P(B) + P(C) - P(AB) - P(BC) - P(AC) + P(ABC) Chng minh nu A, B c lp th , B v, cng c lpHai bin c c lp l hai bin c xy ra nhng khng lin quan g n nhau cho nn bin c i ca A tcv B cng c lp vi nhau tc l xy ra khng h lin quan n nhau. Tng t nh vy th v cng c lp vi nhau Chng minh nu P(A/B)= P(A/) th A B c lpta c P(A/B)= P(A/)ngha l xc sut ca bin c A di iu kin B xy ra ging nh xc sut ca bin c A di iu kin B khng xy ra=> cho nn hai bin c AB vic xy ra hay khng xy ra bin c ny th khng nh hng g n vic xy ra ca bin c kia tc lhai bin c A, B c lp.Cu 6: Ti sao li gi l mt h y cc bin c? ngha ca cng thc xc sut y , cng thc Bayes.Tr li:-Gi l 1 h y cc bin c v khi ta thc hin php th ngu nhin s c nhiu kh nng xy ra, c th l i lp, xung khc tng i.Trong cc kh nng xy ra li xy ra nhiu giai on khc nhau na, nhng hot ng tip theo ph thuc vo hot ng xy ra trc =>Cn phi tnh xc sut thc hin c cng vic qua nhiu giai on nhng tng ca chng lun l bin c chc chn =>H y cc bin c.- ngha ca cng thc Bayes: Cng thc Bayes cn c gi l cng thc xc sut hu nghim khi v ch khi s kin xy ra, tm nguyn nhn gy ra s kin vi xc sut ln nht. cng thc Bayes c ng dng trong Khoa hc - K thut, c bit trong khoa hc lp rp.Cu 7: - Bin ngu nhin l 1 bin s nhn gi tr ty thuc s kin ngu nhin.- C 2 loi bin ngu nhin:+ Bin ngu nhin ri rc (discrete): Tp gi tr nhn l hu hn hoc m c.+ Bin ngu nhin lin tc (continious): Tp gi tr nhn lp y 1 khong hu hn hon v hn.)- Hm phn phi ca bin ngu nhin X c xc nh nh sau:F (x) = P {X < x}Trong x l bin ca hm F,x R K hiu X=F(x) ngha l bin X c hm phn phi l F(x).- Tnh cht ca hm phn phi:+ Hm phn phi xc nh vi mi x (- , +)+ Hm phn phi l hm khng gim: nu x1 < x2 th F(x1) F(x2)F(-) = 0, F(+) = 1, 0F(x)1 vi mi x (- , +)P {ax C l s kin ca xc sutC(p < 1 p=0) =>C= A0P(C) = 1 - P( C ) = 1 - kn mknCChoc:P(C) = 1.p k pkm n mkpnC CCd) Gi D l s kin c nhiu nht 1 ph phm => p 1 => C 2 trng hp l c 1 ph phm v khng c ph phm noGi D v D ln lt l s kin c 1 ph phm v khng c ph phm no6mD = 1 1.km n mC C mD = kn mCmD = mD + mD = 1 1.km n mC C + kn mCP(D) = 1 1.k km n m n mknC C CC +2/ M hnh i tuBi ton: 3 n sinh L, H, C i tu ha vi 10 toa tu. Tnh xc sut trong cc trng hp sau y:a) Mi toa tu khng cha qu 1 n sinhb) 3 n sinh ngi 3 toa khc nhauc) 3 n sinh ngi 3 toa lin k nhaud) 3 n sinh ngi cng 1 toae) L lun ngi toa uf) H, C ngi cc toa u cuig) Toa 5 khng c ai ngiGii:S trng hp ng kh nng: n = 10.10.10 = 1000 103a) Gi A l s kin cn tnh xc sut mA = 10.9.8 = 7201 cch chn l 1 chnh hp chp 3 ca 10 nn s cch chn l mA = 310A= 8.9.10 Xc sut mi toa tu khng cha qu 1 n sinh l: P(A) = Amn= 1%b) Gi B l s kin 3 n sinh ngi 3 toa khc nhau => BAc) Gi B l s kin 3 n sinh ngi 3 toa lin k3 c c th i ch cho nhau => c 3! = 6 cch sp xpC 8 v tr m cc c c th ngi lin k nhaumC = 3! . 8 = 48 => P(C) = 481000= 4,8%d) Gi D l s kin 3 n sinh ngi cng 1 toamD = 10 ; P(D) = 101000= 1%e) Gi E l s kin L ngi toa umE = 1.10.10 = 100P(E) = 1001000= 10%f) Gi F l s kin H, C ngi cc toa u cui => mF = 2.10 = 20 P(F) = 201000= 2%g) Gi G l s kin toa 5 khng c ai ngimG = 9.9.9 = 729P(G) = 7291000= 72,9%3/ M hnh xp ch ngiBi ton 1: 3 ngi L, H, C ngi trn 1 bng gh trng 10 ch. Tm xc sut:a) 3 nsinh ngi 3 v tr lin k7b) L lun ngi v tr uc) H, C lun ngi v tr u cuiGii:S trng hp ng kh nng : n = 10.9.8 = 720a) Gi A l s kin 3 n sinh ngi 3 v tr lin kC 3 ! = 6 cch xp ch cho 3 ngi C 8 ch m 3 ngi c th ngi lin k mA = 6.8 = 48P(A) = 487206,6%b) Gi B l s kin L lun ngi v tr umB = 1.9.8 = 72 P(B) = 72720= 10%c) Gi C l s kin C, H lun ngi v tr u cuimC = 2.8 = 16 P(C) = 167202,2%Bi ton 2 : Mt t gm 10 ngi t chc lin hoan ngi quanh bn trn. Mi ngi ngi vo ch mt cch ngu nhin. Tm kh nng cho A v B ngi cnh nhauGii:S trng hp ng kh nng : n = 10! A c th ngi 1 trong 10 ch, B c th ngi 2 ch bn cnh A => m = 10.2.8! Xc sut cn tm l : P = 10.2.8!10!= 294/ M hnh bn sng:Bi ton : 2 x th bn vo bia 1 cch c lp. Xc sut trng tng ng l 0,7 v 0,8. Tm cc xc sut:a) C ng 1 x th trng chb) C t nht 1 x th trng chGii: a) P(A)=0,7 => P()= 1 - 0,7 = 0,3 P(B)=0,8 => P(B)= 1 - 0,8 = 0,2Gi A1 l bin c c ng 1 x th trng ch.X th A trng v x th B trt hoc ngc liA1 = A.B i lp .B (A,B) v (, B) l cc s kin i mt xung khc=> P(A1)=P(A).P(B) + P().P(B)= 0,7 . 0,2 + 0,3 .0,8 = 0,38b) C t nht 1 x th trng ch =>( ) ( ) ( ) ( ) P A B P A P B P A B + = 0,7 + 0,8 - 0,56 = 0,94L thuyt thng k tonCu 1: i tng v ni dung nghin cu ca Thng k ton l g?Thng k ton hc l mt ngnh khoa hc nghin cu vic thu thp thng tin qua cc d liu ca cc i tng cn nghin cu t rt ra nhng kt lun bng s v bn cht i tng ty theo yu cu nghin cu. i tng nghin cu ca thng k ton: thng tin qua cc d liu.Ni dung ca nghin cu thng k ton: t thng tin qua cc d liu rt ra kt lun bng s v bn cht i tng.Cu 2: Cc khi nim c s ca thng k ton1.Khi nimTrong nhng vn thc t, ta thng phi nghin cu mt hay nhiu du hiu nh tnh hoc nh lng ca cc phn t thuc mt tp hp no , chng hn nh: chiu cao ca thanh nin Vit Nam, tnh hnh thu nhp v chi tiu ca cc h gia nh8 nghin cu tp hp cc phn t theo du hiu nghin cu, ta c th s dng phng php nghin cu ton b, ngha l kho st du hiu nghin cu trn tng phn t ca tp hp. Nhng phng php ny gp nhiu kh khn.V vy, trong thc t, ngi ta thng p dng phng php nghin cu chn mu. Ni dung ca phng php ny l t tp hp nghin cu, c gi l tng th, chn ra mt s cc phn t, c gi l mu, kho st du hiu nghin cu trn mu, da vo m phn tch, rt ra kt lun cho tng th.C s khoa hc ca phng php ny l l thuyt xc sut v thng k ton.2. Mt s phng php chn mu ch yu:a)Chn mu ngu nhin, n gin: l loi mu c chn trc tip t danh sch c nh s ca tng th. Cc phn t ca mu c chn ra t tng th bng cch rt thm theo mt bng s ngu nhin.Phng php ny c u im l cho php thu c mt mu c tnh i din cao nu gia cc phn t ca tng th khng c g l khc bit nhiu. Nu kt cu ca tng th phc tp th chn theo phng php ny s kh m bo tnh i din. Mt nhc im na l trong trng hp quy m ca tng th kh ln th vic nh s tt c cc phn t s rt kh khn.b)Mu h thng: l loi mu m ch c phn t u tin c chn ngu nhin, sau da trn mt quy tc hay mt th tc no chn ra cc phn t tip theo. Chng hn, trn mt danh sch gm N sinh vin, cn chn ra mt mu kch thc n, ta chia danh sch thnh n phn bng nhau, phn T1 gm N/n phn t, chn ngu nhin ra 1 phn t, sau c cch N/n phn t cho vo mu cho n khi n phn t.Nhc im ca phng php ny l d mc sai s h thng khi cc phn t ca tng th khng c sp xp mt cch ngu nhin m theo mt trt t ch quan no . Tuy vy, do tnh n gin, mu h thng thng c dng cp chn mu cui cng v khi tng th tng i thun nht.c)Mu phn t chn mu phn t, trc ht ngi ta phn chia tng th thnh cc t c thun nht cao chn ra cc phn t i din cho tng t. Vic phn t c hiu qu khi tng th nghin cu khng thun nht theo du hiu nghin cu. Sau khi phn t th kch thc mu c phn b cho mi t theo 1 quy tc no , chng hn t l thun vi kch thc mi t.3. ngha ca muTrong ngnh chn mu, kho st khng nhiu cc n v nghin cu nn thng c tin hnh trong thi gian ngn. D liu c x l, phn tch nhanh chng nn thng tin thu c t iu tra chn mu c tnh thi s, cp nht.Chi ph cho cng tc t chc nghin cu gim. Do , nghin cu chn mu tit kim c nhn lc, vt lc, ti chnh.C th m rng ni dung nghin cu hoc i su tm hiu mt no ca i tng.C th tuyn chn nhng iu tra vin tt: C trnh , c kinh nghim, c iu kin tp hun th thng tin thu c c tnh chnh xc cao.4. Cc c trng muGi s (X1, X2, ,Xn) l mu ngu nhin sinh ra t X c EX=, DX = 2a. Thng kHm T = T(X1, X2,,Xn) c gi l mt thng k.V d : T = T(X1, X2,,Xn) = max {X1, X2,,Xn} l mt thng kb. Trung bnh mu (k vng mu)Thng k trung bnh mu, k hiu : X, xc nh bi :Trn mu c th (x1,x2,,xn), thng k X nhn gi tr :Nu mu sp xp theo bng phn phi tn s th :K vng mu s l mt bin ngu nhin c cc c trng :c. Phng sai muPhng sai mu, k hiu MS xc nh bi :Vi mu thc nghim c sp xp theo tn s, phng sai mu nhn gi tr tnh theo cng thc :9Dng cc php bin i, ta c:Thng k :Gi l phng sai mu iu chnhd. lch tiu chun mu- thng kc gi l lch tiu chun mu- thng k S = c gi l lch tiu chun mu iu chnhe. Phn b ca X v S2* Nu (X1, X2,, Xn) l mu ngu nhin c rt ra t bin ngu nhin chun N(, 2) th:n.X cng c phn phi chun c phn phi X c phn phi chunX v S2 c lp vi nhau v ngc li c phn phi student vi n-1 bc t do* Nu (X1, X2,, Xn) l mu t bin ngu nhin chun N(1, 12) , cn (Y1, Y2,, Yn) l mu t bin ngu nhin chun N(2, 2 2) c lp vi mu trn. Khi :C phn phi student vi n + m - 2 bc t do, trong Sx2, Sy2 l 2 phng sai mu tng ng vi mu X v mu Y.5. Phn phi thc nghim v hm phn phi thc nghimGi s mu thc nghim (x1, x2, , xn) sinh t X. Ta xy dng hm:c gi l hm phn phi thc nghim6.nh l Glivenco.Gi s F(x) l hm phn phi ca bin ngu nhin X m ta cn tm. Fn(x) l hm phn phi thc nghim nhn c t mu ngu nhin c n sinh ra t X. Khi :Nh vy, hm phn phi thc nghim l 1 xp x ca hm phn phi l thuyt. Vi n c nh, hm phn phi thc nghim cho ta hnh nh hnh hc v phn phi l thuyt cn tm. nh l Glivenco l cch tm dng ca hm phn phi thc nghim.10Cu 3 : Cc loi c lng tham s (c trng). Cc phng php c lng. Nu cc cng thc c bn.Tr li:+) nh ngha: l s dng 1 thng k nh gi i tng+) cc loi c lng tham s( c trng) l: c lng khng chch, c lng hiu qu, c lng vng, c lng t l m ng, c lng trung bnh m ng, c lng phng sai m ng+) Cc phng php c lng l:-c lng im cho k vng, phng sai v xc sut-c lng khong cho k vng v xc sut+) cc cng thc:Tnh trung bnh mu: =Tnh phng sai m ng: DX=VarX=Tnh phng sai mu:=; Cng thc T tnh khong tin cy i vi s trung bnh trong phn phi chun khi cha bit phng sai: nSx n t * ) 2 / )( 1 ( < < +nSx n t * ) 2 / )( 1 ( Cu 4 : Th no l kim nh gi thuyt thng k? Cch tin hnh kim nh gi thuyt thng k. Cc kt qu kim nh chnh.Tr li:Kim nh gi thuyt thng k (statistical hypothesis test) l phng phpra quyt nh s dng d liu, hoc t th nghim hoc t nghin cu quan st (observational study)(khng c kim sot). Trong thng k (statistics), mt kt qu c gi l tin cy mang tnh thng k (statistically significant) nu n t c kh nng din ra theo mt ngng xc sut cho trc (v d 5% hay 10%).-Cch tin hnh kim nh gi thuyt thng kBc 1, nh nghin cu cn phi nh ngha mt gi thuyt o (null hypothesis), tc l mt gi thuyt ngc li vi nhng g m nh nghin cu tin l s tht.Bc 2, nh nghin cu cn phi nh ngha mt gi thuyt ph (alternativehypothesis), tc l mt gi thuyt m nh nghin cu ngh l s tht, v iu cn c chng minh bng d kinBc 3, sau khi thu thp y nhng d kin lin quan, nh nghin cu dng mt hay nhiu phng php thng k kim tra xem trong hai gi thuyt trn, gi thuyt no c xem l kh d. Cch kim tra ny c tin hnh tr li cu hi: nu gi thuyt o ng, th xc sut m nhng d kin thu thp c ph hp vi gi thuyt o l bao nhiu. Gi tr ca xc sut ny thng c cp n trong cc bo co khoa hc bng k hiu P value. iu cn ch y l nh nghin cu khng th nghim gi thuyt khc, m ch th nghim gi thuyt o m thi.Bc 4, quyt nh chp nhn hay loi b gi thuyt o, bng cch da vo gi tr xc sut trong bc th ba.Bc 5, khi nh nghin cu bc b gi thuyt o, th gi thuyt ph c mc nhin cng nhn, nhng nh nghin cu khng th xc nh gi thuyt ph no l ng vi s tht.- Cc kt qu chnh ca gi thuyt thng kPhn 2 : BI TPBi tp v Bin ngu nhin Hm phn phi Hm mt xc sut. Cc phn phi xc sut thng gp Cc c trng ca bin ngu nhinCu 1: Bn 3 pht vo mc tiu. Xc sut trng ca mi pht l 0,7. C mi pht trng c 5 im. gi x l s im at c sau 3 pht. Tm hm phn phi xc sut ca x.Gii:Cng thc: k k n knC p qC: p = 0,7.q = 1 p = 1 0,7 = 0,3.Ta c lut phn phi:X 0 5 10 1511P = Pk03C . 0,70. 0,33= 0,02713C . 0,71. 0,32= 0,18923C . 0,72. 0,3 = 0,44133C . 0,73 .0,30 = 0,343Hm phn phi xc sut ca X:FX (x) = Cu 2: Tin hnh bn vo mc tiu cho n lc trng th dng li. Xc sut trng mi pht l . Gi x l s ln bn. Tm hm phn phi xc sut ca x.Gii:Tin hnh bn vo mc tiu cho n lc trng th dng li. Xc sut trng mi pht l. Gi x l s ln bn. tm hm phn phi xc sut ca x.Ta c y l bin ngu nhin ri rc.Ta c bng phn phi xc sut.X 0 1 2 3 nP 0(1- )(1- )2 (1- )n-1Nu x 1 bin c (X f(x) = 0Vi => F(x) = x2 => f(x) = 2xVi x > 1 => F(x) = 1 => f(x)=0Ti x = 0 c F(0-) = = = = 0F (0+) = = == 0=> f(0) = F (0) = 012Ti x =1 cF (1-) = = = F (1+) = = = = 0Vy:f(x) = Xc sut bin ngu nhin X nhn gi tr trong khong l:P = F F = = Cu 5:{ }.!keP KK , trong M Suy ra( )( )22 2 2012 21 02! ( 1)( 1)! ! 1KKK KK KeD M M KKe eK KK K + + + + + Vy D Mt s bi tp thng k dng ton (dng tng hp)Cu 1:a) Ta c:n = 111 = 1- = 0,95 => = 0,05Tnh: =1114 * 8 6 * 5 , 6 10 * 75 , 5 16 * 25 , 5 24 * 75 , 4 30 * 25 , 4 8 * 75 , 3 8 * 25 , 3 5 * 5 , 2 + + + + + + + += 4,7Phng sai mu: = ( - ) =1101* ( 238,375 - 22,09 ) = 1,96 => S = = 1,402Tm t: = 0.05 => t110 ( 0,05/2 ) = 1,98Vy khong tin cy l:(nSx n t * ) 2 / )( 1 ( ;+nSx n t * ) 2 / )( 1 ( ) = (1,92 ; 7,48 ) ngha: c lng thu nhp trung bnh ca cc h trong thnh ph trn nm trong khong t 1,92 n 7,48 c kh nng ng 95%, sai 5%b) H c thu nhp 500 ngn ng/ngi/thng tr ln=> mc thu nhp/ngi/nm ca h l t 6 triu ng tr ln c gi l nhng h c thu nhp cao=> t l h thu nhp cao l == 0,09 = = 0,027 ( ) = 0,005vy khong tin cy l:p ( f - < p < f +) = ( 0,036 < p < 0,143 )Vy vi tin cy 95% th t l h thu nhp cao dao ng trong khong t 0,036 n 0,143c) Tnh c lng trung bnh ca ch tiu X nhng h c thu nhp caotng t phn a ta c:= = 7,1 = 0,508 => S = = 0,713Vy khong tin cy l: (7,1 -; 7,1 +) = ( 7 ; 7,21 )Vi tin cy l 95% th thu nhp trung bnh ca nhng h c thu nhp cao dao ng t 7 n 7,21d) Vi tin cy 5%=> = 0,05 = 1 - => = 0,9513Cu 2:S liu v doanh s ca 1 siu th trong 1 ngyDoanh s ( triu ng / ngy ) S ngy24 530 1236 2542 3518 2454 1560 1265 1070 6a. c lng doanh s bn trung bnh trong 1 ngy ca siu th vi tin cy 95%b. Nhng ngy c doanh s trn 60 triu l ngy t hng. c lng t l ngy bn t hng vi tin cy 90%c. c lng doanh s bn trung bnh ca 1 ngy bn t hng vi tin cy 95%( gi thit doanh s bn t hng l i lng ngu nhin c phn phi chun)d. Trc y doanh s bn trung bnh ca siu th l 35 triu/ngy s liu bng trn thu c khi c 1 phng php bn hng mi. Hy nhn xt v phng php bn hng vi mc ngha 5%Gii:a, Kch thc mu : n=144 >30Doanh s trung bnhx = 40.8 (triu ng/ngy) Theo gi thit ta c: =95% => U/2 =1.96 tnh c S=15.4 =U/2.(S /cn n)=(1.96).(15.4/12)=2.5=>P {40.8-2.5