Bai Giảng Xstk-libre

BI GING MNL THUYT XC SUT V THNG K TON

Mai Cm T B mn Ton kinh t

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PHN TH NHTL THUYT XC SUT

CHNG 1BIN C NGU NHIN V XC SUT

1. M U - Ni dung chng 1 Cc khi nim nn tng ca xc sut Cc nh ngha xc sut Hai nguyn l c bn ca xc sut Cc nh l XS dng tm XS ca bin c phc hp.

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Chng 1 2. Php th, bin c

2. PHP TH V CC LOI BIN C2.1. Php th v bin c+ Vic thc hin mt nhm cc iu kin c bn quan

st hin tng no c xy ra hay khng c gi l thc hin php th.

+ Hin tng c th xy ra (hoc khng xy ra) trong kt qu ca php th gi l bin c.

V d 1.1. Tung 1 con xc xc cn i, ng cht trn mt phng cng.

+ Php th tung 1 con xc xc+ iu kin c bn + Bin c A6 = xut hin 6 chm;

B = xut hin l chm 3

Chng 1 2. Php th, bin c

2. PHP TH V CC LOI BIN C2.2. Cc loi bin c+ Bin c chc chn (U)+ Bin c khng th c (V)+ Bin c ngu nhin (A, B, A1, A2,)

V d 1.1. Tung 1 con xc xcU = xut hin s chm nh hn 7V = xut hin 7 chmA6 = xut hin 6 chm (b/c ngu nhin)B = xut hin l chm (-----------------)

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Chng 1 3.Xc sut ca bin c

3. XC SUT CA BIN CXc sut ca mt bin c l mt con s c trng cho kh

nng khch quan xut hin bin c khi thc hin php th.

K hiu xc sut ca bin c A l P(A)V d 1.1. Tung 1 con xc xcA6 = xut hin 6 chm P(A6) = 1/6 B = xut hin l chm P(B) = 3/6 = 1/2 = 0,5

Theo kh nng tm XS c th chia bin c thnh 2 loi+ Bin c n gin+ Bin c phc hp

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Chng 1 4. nh ngha c in v XS

4. NH NGHA C IN V XC SUT4.1. Th d4.2. nh ngha c in v xc sutXc sut xut hin bin c A trong 1 php th l t s gia

s kt cc thun li cho A (k hiu m) v tng s cc kt cc duy nht ng kh nng c th xy ra khi thc hin php th (k hiu n).

( )m

P An

4.3. Cc tnh cht ca xc sut

0 P(A) 1; P(U) = 1; P(V) = 0

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4. NH NGHA C IN V XC SUT4.4. Cc phng php tnh xc sut bng nh ngha c

in.a. Phng php suy lun trc tipV d 1.2. Hp c 10 qu cu gm 3 qu trng v 7 qu

en. Ly ngu nhin t hp 1 qu cu. Tm xc sut ly c cu trng.

b. Phng php dng s Dng s cyV d 1.3. Tung 1 ng xu i xng, ng cht trn mt

phng cng 3 ln. Tm xc sut c ng 1 ln xut hin mt sp.

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b. Phng php dng s Dng s dng bngV d 1.4. Tung 1con xc xc cn i, ng cht trn mt

phng cng 2 ln. Tm xc suta. Tng s chm xut hin l 8.b. Tng s chm xut hin l 8 bit rng c (t nht 1) ln

xut hin mt 6 chm Dng s VennV d 1.5. Mt lp c 50 hc sinh, trong c 30 hc sinh

gii Ton, 20 hc sinh gii Vn, 15 hc sinh gii c Ton v Vn. Chn ngu nhin 1 hc sinh. Tm xc sut chn c hc sinh khng gii Ton v khng gii Vn. (khng gii mn no)

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c. Phng php dng cng thc ca gii tch t hp T hp chp k ca n phn t Cnk (0 k n) Chnh hp chp k ca n phn t Ank (0 k n) Hon v ca k phn t Pk Chnh hp lp k ca n phn t (0 k n)Quy c 0! = 1

V d 1.6. Mt hp c 10 sn phm (6 chnh phm v 4 ph phm). Ly ng thi 2 sn phm. Tm xc sut

a. Ly c 2 chnh phm.b. Ly c ng 1 chnh phm

knA

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4.5. u im v hn ch ca nh ngha c in u im: Khng cn thc hin php th Hn ch: + i hi s kt cc hu hn.

+ Cc kt cc phi tha mn tnh duy nht, ng kh nng.

c. Phng php dng cng thc ca gii tch t hpV d 1.7. Mt hp c 10 sn phm (2 sn phm xanh, 3

mu , 5 mu vng). Ly ng thi 4 sn phm. Tm xc sut

a. Ly c 4 sn phm cng mu.b. Ly c 3 mu.c. Ly ng thi 5 sn phm. Tm XS ly c 3 mu.

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Chng 1 5. nh ngha thng k v XS

5. NH NGHA THNG K V XC SUT5.1. nh ngha tn sutTn sut xut hin bin c A trong n php th l t s gia

s php th trong bin c xut hin (k) v tng s php th c thc hin

5.2. nh ngha thng k v xc sutXc sut xut hin bin c A trong 1 php th l mt s p

khng i m tn sut f xut hin bin c trong n php th s dao ng rt t xung quanh n khi s php th tng ln v hn.

( )k

f An

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Chng 1 6. nh ngha khc v XS

6. MT S NH NGHA KHC V XC SUT

6.1. nh ngha hnh hc v xc sut

6.2. nh ngha ch quan v xc sut

6.3. nh ngha tin v xc sut

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Chng 1 7. Nguyn l xc sut

7. NGUYN L XC SUT7.1. Nguyn l xc sut nh+ Nu mt bin c c xc sut rt nh thc t c th cho

rng trong mt php th bin c s khng xy ra.+ Mc xc sut c coi l nh ty thuc vo tng bi

ton v gi l mc ngha.+ Nguyn l XS nh l c s ca phng php kim nh.7.2. Nguyn l xc sut ln+ Nu mt bin c c xc sut rt ln thc t c th cho

rng trong mt php th bin c s xy ra.+ Mc xc sut ln gi l tin cy.+ Nguyn l XS ln l c s ca phng php c lng

bng khong tin cy.13

Chng 1 8. Lin h gia cc bin c

8. MI LIN H GIA CC BIN Cnh ngha 1. Bin c A gi l thun li cho bin c B, k

hiu , nu A xy ra th B cng xy ra. nh ngha 2. Bin c A gi bng bin c B, k hiu A = B,

nu A xy ra th B cng xy ra v ngc li. V d 1.8. Tung 1 con xc xcA i = xut hin i chm (i = 1,2,, 6)A = xut hin 1, 3 hoc 5 chmB = xut hin l chm A1 B

A = B

A B

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8.1. Tng cc bin cnh ngha 3. Bin c C c gi l tng ca 2 bin c A

v B, k hiu C = A + B, nu C ch xy ra khi v ch khi c t nht mt trong hai bin c A v B xy ra.

V d 1.. Mua ln lt 2 sn phm cng loiA i = ln th i mua c chnh phm (i=1,2)A = mua c t nht 1 chnh phm

= c mua c chnh phm A = A1 + A2

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nh ngha 4. Bin c A c gi l tng ca cc bin c A1, A2,, An nu A xy ra khi v ch khi c t nht mt trong n bin c thnh phn xy ra.

K hiu

1

n

ii

A

V d 1.10. X th bn 4 vin n c lp.A i = x th bn trng vin th i (1 = 1, 2, 3, 4)A = bia b trng n A = A1 + A2 + A3 + A4

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8.2. Tch cc bin cnh ngha 5. Bin c C c gi l tch ca 2 bin c A

v B, k hiu C = A.B, nu C xy ra khi v ch khi c 2 bin c A v B cng xy ra.

V d 1.11. Mua ln lt 2 sn phm cng loiA i = ln th i mua c chnh phm (i=1,2)B = mua c 2 chnh phm B = A1.A2

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nh ngha 6. Bin c A c gi l tch ca cc bin c A1, A2,, An nu A xy ra khi v ch khi tt c n bin c thnh phn xy ra.

K hiu

1

n

ii

A

V d 1.12. X th bn 4 vin n c lp.A i = x th bn trng vin th i (1 = 1, 2, 3, 4)B = x th bn trng c 4 vin n B = A1.A2.A3.A4

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8.3. Tnh xung khc ca cc bin cnh ngha 7. Hai bin c A v B gi l xung khc vi

nhau nu chng khng th ng thi xy ra trong kt qu ca mt php th.

Trng hp ngc li gi l khng xung khc.V d 1.13. Hp c 7 chnh phm v 3 ph phm. Ly 1

sn phmA = ly c chnh phm; B = ly c ph phm A v B xung khc.

V d 1.14. Hp c 7 chnh phm v 3 ph phm. Ly ln lt 2 sn phm. Ai = ln th i ly c chnh phm

A1 v A2 khng xung khc. 19


nh ngha 8. Nhm n bin c A1, A2,, An c gi l xung khc tng i nu bt k 2 bin c no trong nhm ny cng xung khc vi nhau.

V d 1.15. X th bn 4 vin n c lpA i = x th bn trng i vin (i=0,1,,4)A = bia b trng n

A0,A1,,A4 xung khc tng iA, A0, A1 khng xung khc tng i

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8.4. Nhm y cc bin cnh ngha . Nhm n bin c A1, A2,, An c gi l

nhm y cc bin c nu trong kt qu ca php th s xy ra 1 v ch 1 trong cc bin c .

A1, A2,, An l nhm y cc bin c

xung khc tng i1 2

1 2

...

, ,...,n

n

A A A U

A A A

V d 1.16. Tung 1 con xc xcA i = xut hin i chm ; B = xut hin l chmA1, A2, , A6 nhm y A2, A4, A6, B nhm y

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nh ngha 10. Hai bin c A v gi l i lp nu chng to thnh nhm y cc bin c.

V d 1.17. Tung 1 con xc xcA = xut hin chn chm = xut hin l chm

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V d 1.18. thi c 3 cu hi. A i = hc sinh tr li ng cu th i (i = 1, 2, 3)Biu din cc bin c sau qua A1, A2, A3

A = hc sinh tr li ng c 3 cuB = hc sinh tr li sai c 3 cuC = hc sinh ch tr li ng cu 3D = hc sinh tr li ng 1 cuE = hc sinh c tr li ng

Nu tr li ng cu 1 c 4 im, cc cu khc c 3 im, sai c 0 im.G = hc sinh t 7 im tr ln

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8.5. Tnh c lp ca cc bin cnh ngha 11. Hai bin c A v B gi l c lp vi nhau

nu vic xy ra hay khng xy ra ca bin c ny khng lm thay i xc sut xy ra ca bin c kia v ngc li.

Hai bin c khng c lp gi ph thuc.V d 1.1. Tung mt con xc xc 2 lnA i = ln th i xut hin 6 chm A1 v A2 c lp.V d 1.20. Hp c 7 chnh phm v 3 ph phm. Ly ln

lt 2 sn phm theo phng thc khng hon li.A i = ln th i ly c chnh phm A1 v A2 ph thuc. 24


nh ngha 12. Nhm n bin c A1, A2,, An gi l c lp tng i vi nhau nu mi cp 2 trong n bin c l c lp vi nhau.

nh ngha 13. Nhm n bin c A1, A2,, An gi l c lp ton phn vi nhau nu mi bin c c lp vi mi t hp ca cc bin c cn li.

Ch c lp ton phn c lp tng i

V d 1.21. Hp c 7 chnh phm v 3 ph phm. Ly ln lt c hon li 4 sn phm.

A i = ln th I ly c chnh phm (i = 1, 2, 3, 4)A1,,A4 c lp ton phn.

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Mt s ch 1, A v B xung khc AB = V2, Nu P(A) > 0; P(B) > 0

+ A, B c lp A, B khng xung khc+ A, B xung khc A, B khng c lp (ph thuc)

3, A, B c lp A v , v B, v c lp4, Php cng cc bin c c tnh cht ging hp cc tp

hp, nhn cc bin c ging giao cc tp hp.5, Quy tc i ngu De Morgan

B B

. ;A B AB AB A B 26


8.6. Xc sut c iu kinnh ngha 14. Xc sut ca bin c A c tnh vi iu

kin bin c B xy ra gi l xc sut c iu kin ca A.

K hiu P(A/B)

V d 1.22. Hp c 7 chnh phm v 3 ph phm. Ly ln lt 2 sn phm theo phng thc khng hon li.

A i = ln th i ly c chnh phm P(A2 / A1) = 6 / 9

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Chng 1 . nh l nhn xc sut

. NH L NHN XC SUTnh l 1. Nu A v B c lp vi nhau th

P(AB) = P(A).P(B)

1 1

( )n n

i ii i

P A P A

H qu. Nu A1, A2,, An c lp ton phn vi nhau th

V d 1.23. X th bn 4 vin n c lp vi XS trng u l 0,8. Nu trng 2 vin lin tip hoc ht n th dng bn. Tm XS

a, X th dng 2 vin nb, X th dng 3 vin n

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nh l 2. Nu A v B ph thuc nhau thP(AB) = P(A).P(B / A) = P(B).P(A / B)

( )( / )

( )

P ABP A B

P B

H qu 1. Nu P(B) > 0 th

Nu P(B) = 0 th P(A/B) khng xc nh

H qu 2. Nu P(A1A2 An-1) > 0 thP(A1A2 An) = P(A1)P(A2 /A1)P(An / A1A2 An-1)

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V d 1.25. Mt ngi mua 3 sn phm cng loi trn th trng. XS mua c chnh phm ln u l 0,8. Nu ln trc mua c chnh phm th XS mua c chnh phm ln tip theo l 0,5. Nu ln trc mua c ph phm th XS mua c chnh phm ln tip theo l 0,. Tm xc sut

a, Mua c 3 chnh phmb, Mua c 3 ph phm

V d 1.24. Mt hp c 7 chnh phm v 3 ph phm. Ly ln lt 2 sn phm theo phng thc khng hon li. Tm XS ly c 2 chnh phm.

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Ch . Nu bi ton cho bit s sn phm (v d 7 cp v 3 pp)

th php ly ln lt khng hon li tng t nh ly cng mt lc.

Nu bi ton khng cho bit s sn phm m cho t l (v d 70% cp v 30% pp) th XS mi ln ly c chnh phm hay ph phm l khng thay i, khng ph thuc vo phng thc ly (c hon li hay khng hon li)

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Chng 1 10. nh l cng xc sut

10. NH L CNG XC SUTnh l 3. Nu A v B xung khc vi nhau th

P(A+B) = P(A) + P(B)

1 1

( )n n

i ii i

P A P A

H qu 1. Nu A1, A2,, An xung khc tng i th

H qu 2. A1, A2,, An l nhm y cc bin c th

H qu 3. P(A) + P() = 1 P(A) = 1 P()1

( ) 1n

ii

P A

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V d 1.26. C 2 hp sn phmHp 1 7 chnh phm v 3 ph phmHp 2 5 chnh phm v 3 ph phmLy mi hp 1 sn phm. Tm xc sut ly c 2 sn

phm cng loi

V d 1.27. Trong hp c 10 chi tit, trong c 3 chi tit hng. Ly ngu nhin 5 chi tit. Tm xc sut ly c khng qu 2 chi tit hng.

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nh l 4. Nu A v B khng xung khc vi nhau th

P(A+B) = P(A) + P(B) P(AB)

V d 1.28 Mch c 2 bng in mc ni tip hot ng c lp vi nhau, vi xc sut hng u l 0,2. Tm xc sut mch b mt in do bng in hng.

? Lm VD trn cho trng hp 5 bng in mc ni tip.

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V d 1.30. Trong mt cuc thi th sinh phi thi 3 vng vi quy nh qua vng trc mi c d thi vng sau. Ti vng 1, 2, 3 s loi tng ng 50%; 40%; 30% s th sinh tham gia vng . Tm

a, T l th sinh b loi.b, T l th sinh b loi vng 1 trong s nhng th sinh b

loi.

V d 1.2. X th bn 3 vin n c lp vi XS trng ln lt l 0,6; 0,7; 0,8.

a, Tm XS x th bn trng ng 1 vinb, Tm XS x th bn trng vin th nht bit rng x th

bn trng ng 1 vin.

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Chng 1 11. Cc h qu

11. CC H QU CA L CNG V L NHN XC SUT

11.1. Cng thc Bernoullia. Lc BernoulliBi ton c gi l tha mn lc Bernoulli nu tha

mn cc iu kin sau + C n php th c lp+ Trong mi php th xc sut bin c A xy ra khng

thay i l p.

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Chng 1 11. Cc h qu

b. Cng thc BernoulliNu bi ton tha mn lc Bernoulli th xc sut sau n

php th bin c A xut hin ng k ln lvi k = 0, 1, 2, , n

q = 1 - p( ) k k n kn nP k C p q

V d 1.31. Mt thi trc nghim c 10 cu hi c lp,

mi cu hi c 4 phng n tr li trong ch c 1 phng n ng. Mt hc sinh tr li bng cch chn ngu nhin 1 trong 4 phng n. Tm xc sut hc sinh c 5 im.

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Chng 1 11. Cc h qu

11.2. Cng thc xc sut y H1, H,, Hn l nhm y cc bin c.A l bin c c th xy ra ng thi vi 1 trong n bin c

trn th

H1, H,, Hn gi l cc gi thuyt1

( ) ( ) ( / )n

i ii

P A P H P A H

V d 1.32. Mt l hng c 40% sn phm ca nh my A,

35% sn phm ca nh my B, cn li l ca nh my C. T l ph phm ca cc nh my ln lt l 3%; 4%; 5%. Tm t l ph phm ca l hng. ( Ly ngu nhin 1 sn phm, tm XS ly c ph phm)

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Chng 1 11. Cc h qu

11.3. Cng thc BayesH1, H,, Hn l nhm y cc bin c.A l bin c xy ra (P(A) > 0) th

P(Hi) l xc sut tin nghimP(Hi / A) l xc sut hu nghim.

1

( ) ( / )( / )

( ) ( / )

i ii n

j jj

P H P A HP H A

P H P A H

Ch : Sau khi tnh c cc XS P(Hi / A) th c th s dng cng thc y ln th 2, 3,

y ch yu cu s dng cng thc y 1 ln39

Chng 1 11. Cc h qu

V d 1.33. Mt l hng c 40% sn phm ca nh my A, 35% sn phm ca nh my B, cn li l ca nh my C. T l ph phm ca cc nh my ln lt l 3%; 4%; 5%. Ly ngu nhin 1 sn phm th c ph phm. Kh nng ph phm ca nh my no l cao nht?

V d 1.34. Mt nh my c t l ph phm l 10%. Sn phm trc khi a ra th trng phi c kim tra qua 1 my t ng. My kim tra c chnh xc 0% i vi chnh phm v 5% i vi ph phm. Sn phm c kt lun l chnh phm th c a ra th trng.

a, Kim tra 6 sp. Tm XS c 5 sp c a ra th trng .b, Tm t l ph phm trn th trng.c, Mua 6 sp trn th trng. Tm XS mua c 4 cp.

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Chng 1 11. Cc h qu

V d 1.35. C 2 hp sn phm vi b ngoi ging nhauHp I 6 chnh phm, 4 ph phmHp II 3 chnh phm, 7 ph phmLy ngu nhin 1 hp v t ly 1 sn phm.a, Tm XS ly c chnh phm.b, Bit rng ly c chnh phm. Tm XS sp ca hp I

? Lm bi ton trn cho trng hp l hng c 2 hp loi I v 8 hp loi II.

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Chng 1 Bi tp

Bi tp dng cng thc c in8, 10, 15, 25, 28, 44, 51, 91, 99

Bi tp dng L cng, nhn (trc tip)3, 41, 46 50, 54, 55, 56, 6, 7, 112

Bi tp dng CT Bernoulli58 61

Bi tp dng cng thc y , Bayes63, 65, 70, 71, 3, 101 104

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CHNG 2BIN NGU NHIN V QUY LUT

PHN PHI XC SUT

1. M U - Ni dung chng 2 nh ngha v phn loi bin ngu nhin (bnn) Quy lut phn phi XS ca bin ngu nhin Cc tham s c trng ca bin ngu nhin.

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Chng 2 2. nh ngha v phn loi bnn

2. NH NGHA V PHN LOI BIN NGU NHIN

2.1. nh nghaMt bin s c gi l bin ngu nhin nu trong kt qu

ca php th n ch nhn 1 v ch 1 trong cc gi tr c th c ca n ty thuc vo s tc ng ca cc nhn t ngu nhin.

K hiu bnn l X, Y, X1, X2, Gi tr c th c ca bnn l x, y, x1, x2, (X = x1), (X = x2), l cc bin c Nu X ch nhn cc gi tr x1,, xn th nhm bin c

(X = x1), , (X = xn) l nhm y cc bin c.44


V d 2.1. Tung 1 con xc xcGi X l s chm xut hin X l bnnX c th nhn 1, 2, 3, 4, 5, 6

V d 2.2. Gi Y l s ngi n xng ti 1 trm xng du trong 1 ngy Y l bnn

Y c th nhn 0, 1, 2,

V d 2.3. Gi Z l khong cch t im vin n chm bia n tm bia Z l bnn

Z c th nhn gi tr bt k trong on [0,R]

? Phn bit khi nim bin c v bin ngu nhin45


2.2. Phn loi bin ngu nhin+ Bnn ri rc Nu cc gi tr ca n lp nn mt tp hp

hu hn hoc m c.V d bnn X, Y trong v d 2.1, 2.2

+ Bnn lin tc nu cc gi tr c th c ca n lp y mt khong trn trc s.

V d bnn Z trong v d 2.3

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Chng 2 3. Quy lut PPXS ca bnn

3. QUY LUT PHN PHI XC SUT CA BIN NGU NHIN

3.1. nh nghaQuy lut phn phi xc sut ca bnn l s tng ng gia

cc gi tr c th c ca n v cc xc sut tng ng vi cc gi tr .

3.2. Bng PPXS ch dng cho bnn ri rcX x1 x2 xk xnP p1 p2 pk pn

1

0 1

1

i

n

ii

p

p

47


V d 2.4. Hp c 6 chnh phm v 4 ph phm. Ly ng thi 2 sn phm. Lp bng ppxs ca s chnh phm ly c.

V d 2.5. X th bn 4 vin n c lp vi xc sut trng mi vin u l 0,8. Nu trng 2 vin lin tip hoc ht n th dng bn. Lp bng ppxs ca s vin n c s dng

V d 2.6. thi c 2 cu hi c lp, mi cu 5 im. Xc sut hc sinh tr li ng cu 1, 2 tng ng l 0,6 v 0,7. Lp bng ppxs ca s im t c.

48


3.3. Hm phn b xc sut dng cho c bnn ri rc v lin tc.

a. nh nghaHm phn b XS ca bnn X, k hiu F(x), l XS bnn X

nhn gi tr nh hn x, vi x l s thc bt kF(x) = P(X < x)

Nu X l bnn ri rc th

:

( )i

ii x x

F x p

49


V d 2.7. Cho bnn ri rc X sau

Tm hm phn b XS ca X v v th ca hm .

X 1 3 5

P 0,2 0,5 0,3

Gii.0 ; 1

0,2 ; 1 3( )

0,7 ; 3 5

1 ; 5

x

xF x

x

x

50

th hm F(x)


b. Tnh chtTnh cht 1: 0 F(x) 1Tnh cht 2: F(x) l hm khng gim

H qu 1: P(a X < b) = F(b) F(a)H qu 2: Nu X lin tc th P(X = x) = 0H qu 3: Nu X lin tc th

P(a < X < b) = P(a X < b) = P(a < X b) = P(a X b)Tnh cht 3: F(-) = 0; F(+) = 1

H qu 4: Nu X ch nhn gi tr trn [a, b]Vi x a th F(x) = 0Vi x > b th F(x) = 1

51


V d 2.8. Cho bnn X c hm phn b XS nh sau

Tm P(X < 3), P(X 2,4), P(2,4 X


3.4. Hm mt xc sut ch dng cho bnn lin tc.a. nh nghaHm mt XS ca bnn lin tc X, k hiu f(x), l o

hm bc nht ca hm phn b XS ca bnn .f(x) = F(x)

b. Tnh chtTnh cht 1: f(x) 0 vi mi x

Tnh cht 2:

Ch nu hm s f(x) c t/c 1 v 2 th n l hm mt XS ca mt bnn no .

( ) 1f x dx

53


Tnh cht 3:

Tnh cht 4:

( ) ( )x

F x f x dx

( ) ( )

b

a

P a X b f x dx

V d 2.. Cho bnn lin tc X c hm mt XS

a, Tm k.

0 ; (0,10)( )

; (0,10)

xf x

k x

54


V d 2.. Cho bnn lin tc X c hm mt XS

b, Xc nh F(x)c, Tm P(5 < X < 6); P(6 < X < 7) ;

0 ; (0,10)( )

; (0,10)

xf x

k x

c. ngha: Hm mt XS ca bnn X ti mi im x cho bit mc tp trung XS ti im .

55

Chng 2 4. Tham s c trng ca bnn

4. CC THAM S C TRNG CA BIN NGU NHIN

4.1. K vng tona. nh nghaK vng ton ca bnn X, k hiu E(X), c xc nh nh

sau:

+ Nu X l bnn ri rc th

+ Nu X l bnn lin tc th

Ch n v ca E(X) trng vi n v ca X.

1

( )n

i ii

E X x p

( ) ( )E X xf x dx

56


V d 2.10. Tm k vng ton ca bnn X

X 1 3 6 10

P 0,1 0,2 0,4 0,3

V d 2.11. Tm k vng ton ca bnn X

0 ; (0,1)( )

2 ; (0,1)

xf x

x x

57


b. Tnh chtTnh cht 1 E(C) = C ; C l hng sTnh cht 2: E(CX) = C.E(X)Tnh cht 3: E(X + Y) = E(X) + E(Y)

H qu 1:

Tnh cht 4 Nu X v Y c lp th E(XY) = E(X).E(Y)H qu 2:

1 1

( )n n

i ii i

E X E X

1 1

( )n n

i ii i

E X E X

58


c, Bn cht , ngha v ng dng ca k vng tonV d 2.12. Nhu cu hng ngy v mt loi thc phm ti

sng mt khu vc l bnn ri rc X c bng PPXS

Gi s khu vc ny ch c 1 ca hng v mi ngy ca hng nhp 100 kg thc phm.

Gi nhp l 40 nghn ng/kg, gi bn l 50 nghn ng/kg, nu thc phm khng bn c trong ngy th phi bn vi gi 20 nghn ng/ kg th mi ht hng.

Mun c li trung bnh cao hn th ca hng c nn nhp thm 20kg mi ngy hay khng?

X (kg) 80 100 120 150

P 0,2 0,4 0,3 0,1

59


4.2. Trung v md Trung v l gi tr nm chnh gia tp hp cc gi tr c

th c ca bnn.

4.3. Mt m0Mt l gi tr ca bnn tng ng vi+ XS ln nht nu l bnn ri rc+ Cc i ca hm mt XS nu l bnn lin tc.

60


4.4. Phng saia. nh nghaPhng sai ca bnn X, k hiu V(X), l k vng ton ca

bnh phng sai lch ca bnn so vi k vng ton ca nV(X) = E[X E(X)]2

Bin i V(X) = E(X2) [E(X)]2

Ch V(X) 0n v ca V(X) trng vi n v ca X2.

2 2 2 2

1

( ) ; ( ) ( )n

i ii

E X x p E X x f x dx

61


V d 2.13. Tm phng sai ca bnn X

X 1 3 6 10

P 0,1 0,2 0,4 0,3

V d 2.14. Tm phng sai ca bnn X

0 ; (0,1)( )

2 ; (0,1)

xf x

x x

62


b. Tnh chtTnh cht 1 V(C) = 0; C l hng sTnh cht 2: V(CX) = C2 V(X)Tnh cht 3 Nu X v Y c lp th

V(X + Y) = V(X) + V(Y)

H qu : X1, X2,,Xn l cc bnn c lp th

1 1

( )n n

i ii i

V X V X

63


c, Bn cht , ngha v ng dng ca phng saiPhng sai c trng cho mc ri ro ca cc quyt

nh.V d 2.15. Tin li sau 1 thng u t 1 t ng vo cc

ngnh A, B l cc bnn c lp X, Y

a, Mun li trung bnh cao hn th u t vo ngnh no?b, Mun ri ro thp hn th u t vo ngnh no?c, Mun ri ro thp nht th chia vn u t theo t l no?

X 0 15 30

P 0,3 0,5 0,2

Y -2 15 35

P 0,3 0,5 0,2

64


V d 2.15.

d, u t a t vo ngnh A v b t v ngnh B trong 1 thng. Tm trung bnh v phng sai ca tng tin li trong 1 thng?

e, u t 2 t vo ngnh A trong 1 thng, tm trung bnh v phng sai ca tin li thu c?

g, Mi thng u t 1 t vo ngnh A, c lp, tm trung bnh v phng sai ca tng tin li trong 2 thng? Tm XS tng tin li khng di 50 triu.

h, Tm XS u t vo A c li cao hn u t vo B.

X 0 15 30

P 0,3 0,5 0,2

Y -2 15 35

P 0,3 0,5 0,2

65


4.5. lch chun lch chun ca bnn X, k hiu (X), l cn bc hai ca

phng sai ca bnn

C n v trng vi n v ca X.

ng dng khi cn nh gi mc phn tn ca bnn theo n v o ca n th dng lch chun.

( ) ( )X X V X

66


4.6. H s bin thin CV

4.7. Gi tr ti hn xP(X > x ) =

4.8. H s bt i xng 3

4.. H s nhn 4

.100%( )

XCVE X

67

Chng 2 Bi tp

Bi tp v bnn ri rc1, 4, 5, 1 22, 30, 36, 37, 40, 41, 65, 67, 72, 77,82, 86

Bi tp bnn lin tc9, 24, 83, 84

68

CHNG 3MT S QUY LUT PHN PHI XC

SUT THNG DNG

1. M U - Ni dung chng 3 Vi bnn ri rc QL khng mt, nh thc, Poisson,

siu bi. Vi bnn lin tc QL u, ly tha, chun, Student, Khi

bnh phng, Fisher Snedercor.

69

Chng 3 2. QL khng mt A(p)

2. QUY LUT KHNG MT A(p)2.1. nh nghaBnn ri rc X nhn 1 trong 2 gi tr c th c l 0, 1, vi

cc XS tng ng c tnh theo cng thc

gi l phn phi theo quy lut khng mt vi tham s p.K hiu X ~ A(p)

Bng ppxs

1( ) ; 0,1; 1x xxP P P x p q x q p

X 0 1

P q p

2.2. Tham s c trng: E(X) = p; V(X) = pq2.3. ng dng

70

Chng 3 3. QL nh thc B(n,p)

3. QUY LUT NH THC B(n,p)3.1. nh nghaBnn ri rc X nhn 1 trong cc gi tr c th c l

0, 1,2,,n vi cc XS tng ng c tnh theo cng thc

gi l phn phi theo quy lut nh thc vi tham s n v p.K hiu X ~ B(n, p)

Bng ppxs

( ) ; 0,1,2,..., ; 1x x n xx nP P P x C p q x n q p

X 0 x n

P qn pnx x n xnC p q

71


Ch Nu bi ton tha mn lc Bernoulli vi 2 tham s l n, p v X l s ln xut hin bin c A sau n php th th X ~ B(n, p).

Cng thc XSP(a X a + h) = Pa + Pa+1 + + Pa+h

V d 3.1. Mt phn xng c 10 my hot ng c lp vi nhau. Xc sut trong mt ngy mi my b hng u l 0,1. Gi X l s my hng trong ngy.

a, X phn phi theo quy lut g?b, Tm XS trong 1 ngy c t 2 n 4 my hng.

72


3.2. Cc tham s c trng ca quy lut nh thcNu X ~ B(n, p) th E(X) = np; V(X) = npq.Mt m0 l gi tr ca X tha mn

np + p 1 = np q m0 np + p

V d 3.2. Mt thng nhn vin cho hng i cho hng 20 ngy vi XS bn c hng mi ngy u l 0,7.

a, Trung bnh trong 1 nm c bao nhiu ngy ngi bn c hng? Tm phng sai tng ng.

b, Tm s ngy bn c hng c kh nng cao nht trong 1 thng v XS ca gi tr .

73

Chng 3 3. QL nh thc B(n,p) Ch + Nu X1,,Xn c lp v Xi ~ A(p) th

X1 + + Xn ~ B(n, p)+ Nu X1 ~ B(n1, p); X2 ~ B(n2, p) v c lp vi nhau th

X1 + X2 ~ B(n1 + n2, p)

3.3. Quy lut phn phi xc sut ca tn sutGi X l s ln xut hin bin c A sau n php th. Tn

sut xut hin bin c A l Xf

n

f 0 x / n 1

P qn pnx x n xnC p q

74

Chng 3 4. QL Poisson P()

4. QUY LUT POISSON P()4.1. nh nghaBnn ri rc X nhn 1 trong cc gi tr c th c l 0, 1,2,

vi cc XS tng ng c tnh theo cng thc

gi l phn phi theo quy lut Poisson vi tham s .K hiu X ~ P()

4.2. Cc tham s c trngNu X ~ P() th E(X) = ; V(X) = ; -1 m0

( ) ; 0,1,2,...!

x

xP P P x e xx

Ch Nu X ~ B(n, p) vi n kh ln, p kh nh v np npq th coi nh X ~ P().

75

Chng 3 5. QL siu bi M(N, n)

5. QUY LUT SIU BI M(N, n)5.1. nh nghaBnn ri rc X nhn 1 trong cc gi tr c th c l 0,

1,2,,n vi cc XS tng ng c tnh theo cng thc

gi l phn phi theo quy lut siu bi vi tham s N v n.K hiu X ~ M(N, n)

5.2. Cc tham s c trng

( )x n xM N M

x nN

C CP P P x

C

76

Chng 3 6. QLPP u U(a, b)

6. QUY LUT PHN PHI U U(a, b)6.1. nh nghaBnn lin tc X gi l phn phi theo quy lut u trn

khong (a, b) nu hm mt XS ca n c dng

K hiu X ~ U(a,b)? V th ca f(x)6.2. Tham s c trng

1; ( , )

( )0 ; ( , )

x a bf x b a

x a b

2( )( ) ; ( )

2 12

a b b aE X V X

77

Chng 3 7. QL ly tha E()

7. QUY LUT LY THA E()7.1. nh nghaBnn lin tc X gi l phn phi theo quy lut ly tha

(quy lut m) nu hm mt XS ca n c dng

Trong l hng s dngK hiu X ~ E()

? V th ca f(x)

; 0( )

0 ; 0

xe xf x

x

78

Chng 3 7. QL ly tha E()

th hm f(x) (v d vi = 0,8)

7.2. Tham s c trng2

1 1( ) ; ( )E X V X

79

0

0,1

0,2

0,3

0,4

0,5

0,6

0,7

0,8

0,9

Chng 3 8. QL Chun N(, 2)

8. QUY LUT CHUN N(, 2)8.1. nh nghaBnn lin tc X nhn gi tr trong khong (-,+) gi l

phn phi theo quy lutChun vi cc tham s v 2nu hm mt XS ca n c dng

K hiu X ~ N(, 2)

? V th ca f(x)

2

2

( )

21( ) ;2

x

f x e x R

80


81

0

0,1

0,2

0,3

0,4

0,5

0,6

th hm f(x) (v d vi = 2; = 0,8)

8.2. Tham s c trngE(X) = ; V(X) = 2


8.3. Quy lut phn phi Chun haa. nh nghaBnn lin tc U nhn gi tr trong khong (-,+) gi l

phn phi theo quy lut Chun ha nu hm mt XS ca n c dng

K hiu U ~ N(0, 1)

? V th ca (u)

2

21

( ) ;2

u

u e u R

Ch Nu X ~ N(, 2) th (0, 1)X

U N

82


b. Cc tham s c trng E(U) = 0; V(U) = 1

Trong

83

c. Hm phn b XS (u) v hm 0(u)0

0

0

( ) ( ) ( ) ( ) 0,5 ( )u u

u t dt t dt t dt u

0

0

( ) ( )u

u t dt Tnh cht + 0(-u) = -0(u)

+ Vi u 5 th 0(u) = 0,5Gi tr hm ny ph lc 5 hoc bng s TKT


8.4. Gi tr ti hn chun uGi tr ti hn chun mc , k hiu l u l gi tr ca

bnn U tha mn iu kinP( U > u) =

P( U < u ) = 1

Bng gi tr ti hn chun ph lc 6 trang 52Bng s TKT tra u ti dng ca bng gi tr ti hn

Student.

84


8.5. Cng thc xc sutNu X ~ N, 2) th

85

0 0

0

0

0

( )

( ) 0,5

( ) 0,5

(| | ) 2

b aP a X b

bP X b

aP a X

P X


Quy tc hai sigma

86

0 0

2(| | 2 ) 2 2 (2) 0,9544P X

Quy tc ba sigma

0 0

3(| | 3 ) 2 2 (3) 0,9973P X

ng dng Nu QLPPXS ca bnn tha mn quy tc hai sigma v quy tc ba sigma th xem nh bn phn phi chun.


87

V d 3.3. Trng lng sn phm l bnn phn phi chun vi trung bnh l 400g v phng sai l 400 g2.

a, Tm t l sn phm c trng lng t 360g n 440gb, Tm t l sn phm nh hn 350gc, Tm t l sn phm nng hn 360gd, Sn phm t tiu chun nu c trong lng sai lch so

vi mc trung bnh khng qu 30g. Tm t l sn phm t tiu chun.


88

V d 3.4. Tui th sn phm l bnn phn phi chun vi trung bnh l 3,6 nm v lch chun l 1,2 nm. Khi bn c 1 sp th ca hng li 70 nghn ng, nu sp b hng trong thi gian bao hnh th ca hng phi chi 100 nghn ng cho vic bo hnh (chu l 30 nghn ng). Quy nh thi gian bo hnh l 1 nm.

a, Tm t l sn phm phi bo hnh.b, Nu mun bo hnh cho 2,5% s sp th quy nh thi

gian bo hnh bao lu?c, Tm tin li trung bnh trn mi sp bn ra.d, Mun tin li trn mi sp bn ra l 65 nghn ng th

quy nh thi gian bo hnh bao lu?


89

V d 3.5. Chiu di sp c sn xut t ng l bnn phn phi chun vi trung bnh l 60 cm. Bit rng c 10% s sp c chiu di trn 63cm. Sn phm c chiu di t 55 cm tr ln th t tiu chun.

a, Tm phng sai v chiu di sn phm.b, Tm t l sn phm t tiu chun.c, Tm xs trong 10 sp c ng 8 sp t tiu chun.


8.6. Phn phi xc sut ca tng cc bnn c lp cng tun theo mt quy lut

90

Nu X1 ~ N(1, 12) v X1 ~ N(2, 22) v c lp thX1 + X2 ~ N(1 + 2, 12 + 22)

V d 3.6. u t 100 triu vo ngnh A thu c li l X, u t 100 triu vo B thu c li l Y, X v Y c lp v X ~ N(10; 16); Y ~ N(;). (thi gian 1 qu)

a, Nu u t 40 triu vo A v 60 triu vo B trong 1 qu th xs thu c li trn 10 triu bng bao nhiu?

b, Tm xs u t vo A c li cao hn u t vo B.


8.7. S hi t ca quy lut nh thc v quy lut Poisson v quy lut chun

91

Nu X ~ B(n, p) vi n 5 v

Th coi nh X ~ N( = np, 2 = npq)

1 10,3

1

p p

p p n

V d 3.7. Ngi ta kim tra cht lng 00 chi tit. XS chi tit t tiu chun l 0,. Gi X l s chi tit t tiu chun trong s 00 chi tit c kim tra.

a, C th coi nh X phn phi theo ql Chun c khng?b, Tm xs trong c t 800 n 850 chi tit t t/c.


8.8. ng dng ca quy lut chunQL Chun c p dng rng ri trong thc t.

92

L do Nu bnn X l tng ca mt s ln cc bnn c lp v gi tr ca mi bin ch chim mt v tr nh trong tng th X s c pp xp x chun.

V d + Kch thc ca chi tit do 1 my t ng sn xut pp

Chun.+ Chiu cao, cn nng, ch s thng minh, ca con

ngi pp Chun.+ Nng sut lao ng pp Chun.

Chng 3 . QL Khi bnh phng 2(n)

. QUY LUT KHI BNH PHNG 2(n).1. nh ngha.2. Cc tham s c trngQuan tm n gi tr ti hn khi bnh phng n bc t do

mc ngha , k hiu lTra bng ph lc 7 hoc bng ca TKT

S dng Gi s c cc bnn Xi (i = 1, , n) c lp cng pp chun ha th

93

2( )n

2 2 2

1

( )n

ii

X n

Chng 3 10. QL Student T(n)

10. QUY LUT STUDENT T(n)10.1. nh ngha10.2. Cc tham s c trngQuan tm n gi tr ti hn student n bc t do mc

ngha , k hiu lTra bng ph lc 8 hoc bng ca TKT

S dng Gi s c cc bnn c lp U v V vi U ~ N(0,1) v V ~ 2(n) th

94

( )nt

( )/

UT T n

V n

Chng 3 11. QL Fisher F(n1, n2)

11. QUY LUT FISHER - SNEDECOR F(n1, n2)11.1. nh ngha11.2. Cc tham s c trngQuan tm n gi tr ti hn Fisher (n1, n2) bc t do mc

ngha , k hiu lTra bng ph lc hoc bng ca TKTTnh cht

S dng Gi s c cc bnn c lp U v V vi U ~ 2(n1) v V ~ 2(n2) th

95

1 2( , )n nf

11 2

2

/( , )

/

U nF F n n

V n

1 2

2 1

( , )1 ( , )

1n nn n

ff

Chng 3 Bi tp

Bi tp v bnn pp Nh thc5, 11, 18, 77

Bi tp bnn pp Chun41, 42, 44, 47, 78, 80, 81, 87,88, 89, 91, 93

96

Bi tp khc22, 23, 25, 34, 36

CHNG 4

BIN NGU NHIN HAI CHIUHM CC BIN NGU NHIN

1. M U - Ni dung chng 4 Khi nim bnn nhiu chiu. Bnn 2 chiu ri rc, bng ppxs, cc tham s c trng. Bnn 2 chiu lin tc, hm mt xs, cc tham s. Hm cc bnn.

97

Chng 4 2. Khi nim

2. KHI NIM V BNN HAI CHIU2.1. nh nghaHai bnn 1 chiu c xt mt cch ng thi to nn bnn

2 chiu.K hiu (X, Y)V d thu nhp v tiu dng ca 1 ngi.

chiu di v chiu rng ca 1 sn phm.

2.2. Phn loi+ Bnn 2 chiu ri rc nu X, Y u ri rc+ Bnn 2 chiu lin tc nu X, Y u lin tc

98

Chng 4 3. Bng ppxs

3. BNG PHN PHI XC SUT3.1. Bng phn phi xc sut ng thi

Vi 0 P(xi, yj) 1 v 99

X Y y1 yj ymx1 P(x1, y1) P(x1, yj) P(x1, ym)xi P(xi, y1) P(xi, yj) P(xi, ym)xn P(xn, y1) P(xn, yj) P(xn, ym)

1 1

( , ) 1n m

i ji j

P x y

Chng 4 3. Bng ppxs

V d 4.1. Mt thi trc nghim c 2 cu hi c lp vi xs hc sinh lm ng u l 0,2. ng cu 1 c 4 im, cu 2 c 6 im, sai c 0 im. Gi X l s cu tr li ng, Y l s im t c ca hc sinh.

Lp bng ppxs ng thi ca bnn 2 chiu (X, Y).

Ta c bng sau

100

X Y 0 4 6 10

0 ? 0 0 0

1 0 ? ? 0

2 0 0 0 ?

Nhng c xs bng 0 ng vi cc bin c khng th c.

Chng 4 3. Bng ppxs

3.2. Bng phn phi xc sut bina. Bng pps bin ca X

T bng trn ta c bng ppxs bin ca X nh sau

101

X Y y1 yj ym P(x)x1 P(x1, y1) P(x1, yj) P(x1, ym) = P(x1) xi P(xi, y1) P(xi, yj) P(xi, ym) = P(xi) xn P(xn, y1) P(xn, yj) P(xn, ym) = P(xn)

X x1 xi xnP P(x1) P(xi) P(xn)

Chng 4 3. Bng ppxs

b. Bng ppxs bin ca Y

Tng t nh trn ta c bng ppxs bin ca Y nh sau

102

Y y1 yj ymP P(y1) P(yj) P(ym)

T cc bng ppxs bin ca X v Y ta c th tm c cc tham s c trng ca X, Y tng ng

+ E(X), V(X), + E(Y), V(Y),

Chng 4 3. Bng ppxs

3.3. Bng phn phi xc sut c iu kina. Bng ppxs c iu kin ca Y khi ca X = xj

Ly tng xs ti dng xi chia cho tng ca dng ta c

103

X Y y1 yj ymx1 P(x1, y1) P(x1, yj) P(x1, ym)xi P(xi, y1) P(xi, yj) P(xi, ym) = P(xi)xn P(xn, y1) P(xn, yj) P(xn, ym)

( , )( / ) : ( / )

( )i j

j i j ii

P x yP Y y X x P y x

P x

Chng 4 3. Bng ppxs

Bng ppxs c iu kin ca Y khi X = xi nh sau

104

Y/X=x i y1 yj ymP P(y1/xi) P(yj/xi) P(ym/xi)

b. Tng t ta c bng ppxs c iu kin ca X khi Y = yj.

T cc bng ppxs c iu kin ca Xv Y ta c th tm c cc tham s c trng tng ng

+ E(Y/X=xi), E(X/Y=yj) cc k vng ton c iu kin+ V(Y/X=x i), V(X/Y=y j) cc phng sai c iu kin

Chng 4 4. Tham s

4. CC THAM S C TRNG4.1. K vng tonT cc bng ppxs bin ta tm c cc k vng ton sau

Cc k vng ton trn phn nh gi tr trung bnh ca mi thnh phn.

105

1 1 1

1 1 1

( ) ( ) ( , )

( ) ( ) ( , )

n n m

i i i i ji i j

m m n

j j j i jj j i

E X x P x x P x y

E Y y P y y P x y

Chng 4 4. Tham s

4.2. Phng saiT cc bng ppxs bin ta tm c cc phng sai sau

Cc k vng ton trn phn nh mc phn tn ca cc gi tr ca mi thnh phn so vi gi tr trung bnh ca thnh phn .

106

2 2

1 1

2 2

1 1

( ) ( , ) [ ( )]

( ) ( , ) [ ( )]

n m

i i ji j

m n

j i jj i

V X x P x y E X

V Y y P x y E Y

Chng 4 4. Tham s

4.3. Hip phng saiHip phng sai ca cc bnn X v Y, k hiu cov(X,Y), l

k vng ton ca tch cc sai lch ca cc bnn vi k vng ton ca chng.

i vi cc bnn ri rc th tnh nh sau

107

( , ) {[ ( )][ ( )]}

( ) ( ) ( )

Cov X Y E X E X Y E Y

E XY E X E Y

1 1

( , ) ( , ) ( ) ( )n m

i j i ji j

Cov X Y x y P x y E X E Y

Hip phng sai o mc cht ch ca s ph thuc

gia X v Y. n v ca cov(X,Y) l tch cc n v ca X v Y.

Chng 4 4. Tham s

4.4. H s tng quanH s tng quan ca cc bnn X v Y, k hiu XY, l t s

gia hip phng sai v tch cc lch chun ca cc bnn .

Ch + H s tng quan khng c n v o+ XY = YX+ -1 XY 1+ Nu X v Y c lp th XY = 0+ Nu XY = 1 th X v Y ph thuc hm s vi nhau.

108

cov( , ) cov( , )

( ) ( )XY

X Y

X Y X Y

V X V Y

Chng 4 4. Tham s

ngha H s tng quan ca X v Y dng c trng cho mc cht ch ca s ph thuc gia hai bnn X v Y.

Nhn xt v XY+ XY 0 th X v Y ph thuc.+ XY > 0 th X v Y ng bin (cng chiu)

XY < 0 th X v Y nghch bin (ngc chiu)+ |XY | cng gn 1 th X v X ph thuc cng cht ch.

109

Chng 4 4. Tham s

Khi nim+ X v Y l (c) tng quan vi nhau nu cov(X, Y) 0

(hoc XY 0).+ X v Y l khng tng quan vi nhau nu cov(X,Y) = 0

(hoc XY = 0).+X v Y tng quan X v Y ph thuc

110

Cng thcNu X v Y ph thuc th

V(aX + bY) = a2V(X) + b2V(Y) + 2ab.cov(X,Y)p dng chia vn nh th no ri ro l thp nhp

Chng 4 4. Tham s

4.5. K vng c iu kinK vng c iu kin ca bnn Y vi iu kin X = xi tnh

theo cng thc sau

C th lp bng ppxs c iu kin ri tnh k vng ton.

111

Tng t c cng thc tnh E(X/yj)

1

( / ) ( / ) ( / )m

i i j j ij

E Y x E Y X x y P y x

Chng 4 4. Tham s

4.6. Hm hi quyHm hi quy ca Y i vi X l k vng c iu kin ca

bnn Y i vi X

Tng t c hm hi quy ca X i vi Y

112

Cc hm hi quy cho bit trung bnh ca bnn ny ph thuc vo bnn kia nh th no.

1( ) ( / )g x E Y x

2( ) ( / )g y E X y

Chng 4 4. Tham s c trng

V d 4.2. Cho bit X l thu nhp ca v, Y l thu nhp ca chng trong 1 gia nh (n v nghn USD). Bnn 2 chiu (X, Y) c bng ppxs nh sau

113

X Y 10 20 30 50

0 0,1 0,1 0,05 0,05

15 0,05 0,15 0,15 a

30 0 0,05 0,1 b

a. Tm a, b bit thu nhp trung bnh ca v l 14,25 nghn USD

b. Thu nhp ca v v ca chng c c lp vi nhau khng?

Chng 4 4. Tham s c trng

114

c. Tm thu nhp trung bnh ca chng.d. Tm thu nhp trung bnh ca chng khi v thu nhp 20

nghn USD.e. So snh thu nhp trung bnh ca v ng vi cc mc

thu nhp khc nhau ca v. V th.g. Thu nhp ca v v ca chng bin i cng chiu

hay ngc chiu, mc ph thuc cht ch th no?h. Tnh trung bnh v phng sai ca tng thu nhp ca

v v chng. (lm bng 2 cch)

Bi tp46, 47, 49, 50, 51, 52, 53, 55, 56, 58, 59, 60, 64, 65

CHNG 5

CC NH L GII HN

1. M U - Ni dung chng 5 Bt ng thc Trbsp. nh l Trbsp. nh l Bernoulli. nh l gii hn trung tm Linderberg - Lewi.

115

Chng 5 2. Bt Trbsp

2. BT NG THC TRBSPNu X l bin ngu nhin c k vng ton v phng sai

hu hn th vi mi s dng ty ta u c

116

Trng hp 2 V(X) th bt ng thc trn v ngha.

2

2

( )(| ( ) | ) 1

( )(| ( ) | )

V XP X E X

V XP X E X

Chng 5 3. L Trbsp

3. NH L TRBSPCc bin ngu nhin X1, X2, , Xn, c lp tng i,

c k vng ton hu hn v cc phng sai u b chn bi hng s c (ngh l V(Xi) c vi mi i) th vi mi dng b ty ta c

117

1 1... ( ) ... ( )lim 1n nn

X X E X E XP

n n

nh l Trbsp l c s ca php o lng trong thc t. N cng l c s ca phng php mu trong thng k.

Chng 5 3. L Trbsp

Bt ng thc Trbsp (trng hp ring ca nh l Trbsp)

Cc bin ngu nhin X1, X2, , Xn, c lp tng i, c cc k vng ton E(Xi) = m (i = 1, 2, ) v cc phng sai u b chn bi hng s c (ngha l V(Xi) c vi mi i) th vi mi dng b ty ta c

118

1 ...lim 1nn

X XP m

n

Chng 5 4. L Bernoulli

4. NH L BERNOULLIf l tn sut xut hin bin c A trong n php th c lp

v p l xc sut xut hin bin c trong mi php th th vi mi dng b ty ta c

L ny cn gi l lut s ln Bernoulli

119

lim 1n

P f p

nh l Bernoulli l c s l thuyt cho nh ngh thng k v xc sut, do n cng l c s cho mi ng dng ca nh l thng k v xc sut trong thc t.

Chng 5 5. L gii hn trung tm

5. NH LGII HN TRUNG TMNu X1, X2, , Xn, l cc bin ngu nhin c lp

cng tun theo mt quy lut ppxs no vi k vng ton v phng sai hu hn

E(Xk) =a; V(Xk) = 2 vi mi kth qlppxs ca bnn

vi

120

( )

( )c n nn

n

U E UU

V U

s hi t khi n ti qlpp chun ha N(0,1).

1

n

n ii

U X

PHN TH HAITHNG K TON

CHNG 6C S L THUYT MU

1. M U - Ni dung chng 6 Tng th Mu ngu nhin v mu c th Thng k Quy lut PPXS ca mt s thng k v ng dng.

121

Chng 6 2. Khi nim phng php mu

2. KHI NIM V PHNG PHP MU nghin cu mt tp hp c th s dng cc phng

php nghin cu sau2.1. Nghin cu ton b thng k ton b tp hp v

phn tch tng phn t ca n theo du hiu nghin cu.

122

Kh khn+ Quy m ca tp hp qu ln th kh iu tra ton b.+ C nhiu trng hp cc n v iu tra b ph hy ngay

trong qu trnh iu tra.+Trong nhiu trng hp khng th c c danh sch

ca tng th

Chng 6 2. Khi nim phng php mu

2.2. Nghin cu mu+ T tng th rt ra mt mu c kch thc n.+ Xc nh cc tham s c trng ca mu.+ Xc nh quy lut phn phi xc sut ca cc tham s

c trng mu.+ T cc tham s c trng mu rt ra kt lun v tng

th.

123

Chng 6 3. Tng th nghin cu

3. TNG TH NGHIN CU3.1. nh nghaTon b tp hp cc phn t ng nht theo mt du hiu

nghin cu nh tnh hay nh lng no c gi l tng th nghin cu hay tng th.

S lng phn t ca tng th gi l kch thc tng th, k hiu N.

Du hiu nghin cu, k hiu , c th l nh tnh hay nh lng.

124

Bin ngu nhin gc X l bin ngu nhin i din v lng ha cho du hiu nghin cu ca tng th


3.2. Cc phng php m t tng tha. Bng phn phi tn sGi s trong tng th, du hiu nghin cu nhn cc gi

tr x1, x2,, xk vi cc tn s tng ng N1, N2,, NkKhi ta c bng phn phi tn s nh sau

125

iu kin

Gi tr ca x1 xi xkTn s Ni N1 Ni Nk

1

0 ;ik

ii

N N i

N N


b. Bng phn phi tn sutK hiu pi l tn sut ca gi tr xi trong tng th.

Khi ta c bng phn phi tn sut nh sau

126

iu kin

Gi tr ca x1 xi xkTn sut pi p1 pi pk

1

0 1;

1

i

k

ii

p i

p

ii

Np

N


3.3. Cc tham s c trng ca tng tha. Trung bnh tng thTrung bnh tng th, k hiu m, l trung bnh s hc ca

cc gi tr ca du hiu nghin cu trong tng th.

Vi bng phn phi tn s nh trn th

Nu X l bin ngu nhin gc th m = E(X).

127

Thc t ngi ta cn tnh cc loi trung bnh sau+ Trung bnh nhn+ Trung bnh iu ha

1

1 ki i

i

m N xN


b. Phng sai tng thPhng sai tng th, k hiu 2, c tnh nh sau

Nu X l bin ngu nhin gc th 2 = V(X).Phng sai tng th phn nh mc phn tn ca cc gi

tr ca du hiu xung quanh trung bnh tng th.

128

lch chun ca tng th l

2 2

1

1( )

k

i ii

N x mN

2


c. Tn sut tng thNu tng th nghin cu c kch thc N, trong c M

phn t mang du hiu nghin cu th tn sut tng th, k hiu p, c xc nh bi cng thc

Thc cht tn sut tng th p l trng hp ring ca trung bnh tng th m v phn nh c cu tng th theo du hiu nghin cu .

129

Mp

N

Chng 6 4. Mu

4. MU 4.1. nh ngha+ Mu ngu nhin kch thc n l tp hp ca n bin

ngu nhin c lp X1, X2, , Xn c thnh lp t bin ngu nhin gc X trong tng th nghin cu v c cng quy lut phn phi xc sut vi X.

W = (X1, X2,, Xn) l mu ngu nhin th- X1, X2, , Xn c lp- X1, X2, , Xn c cng quy lut ppxs vi X, khi

E(Xi) = E(X) = m; V(Xi) = V(X) = 2

+ Mu c th: khi thc hin php th i vi mu ngu nhin th thu c mu c th l w = (x1, x2,,xn)

130

Chng 6 4. Mu

4.2. Cc phng php chn mua. Mu n ginb. Mu thng k.c. Mu chm.d. Mu phn t.e. Mu nhiu cp.

4.3. Thang o cc gi tr mua. Thang nh danh.b. Thang th bc.c. Thang o khong.d. Thang o t l

131

Chng 6 4. Mu

4.4. Cc phng php m t s liu muXt mu c th w = (x1, x2,, xn)

a. Bng phn phi tn sSp xp cc s liu mu theo th t tng dn.+ Nu mu c n gi tr khc nhau th tn sut mi gi tr

bng 1.+ Nu mu c k gi tr khc nhau vi tn s tng ng th

c bng phn phi tn s nh sau

Vi n1 + n2 + + nk = n132

Gi tr ca x1 xi xkTn s ni n1 ni nk

Chng 6 4. Mu

V d 6.1. Cn mt s sn phm cng loi th c bng sau

133

Trng lng (kg) 5 6 7 8S sn phm 2 12 7 4

V d 6.2. Theo di thi gian hon thnh ca mt s sn phm th c bng sau

Thi gian hon thnh (pht) 10 - 12 12 - 14 14 - 16 16 - 18

S sn phm 10 40 30 20

Chng 6 4. Mu

b. Bng phn phi tn sutK hiu fi = ni / n l tn sut xut hin gi tr xi ca mu.C bng phn phi tn sut nh sau

Vi f1 + f2 + + fk = 1

134

Gi tr ca x1 xi xkTn sut fi f1 f i fk

Chng 6 5. Thng k

5. THNG K 5.1. nh nghaTng th c bin ngu nhin gc XMu ngu nhin W = (X1, X2,, Xn)Thng k l hm ca cc bin ngu nhin X1, X2,, Xn,

k hiu l G G = f(X1, X2, , Xn)

Bn cht thng k G l bin ngu nhin.Khi mu ngu nhin W nhn w = (x1, x2,,xn) th G nhn

gi tr c th lg = f(x1, x2,,xn)

135

Chng 6 5. Thng k

5.2. Mt s thng k c trng muMu ngu nhin W = (X1, X2,, Xn)a. Trung bnh mu l mt thng k, k hiu , v l trung

bnh s hc ca cc gi tr mu

Tnh cht

BT chng minh cc tnh cht trn.

136

X

1 1

1 1;

n k

i i ii i

X X x n xn n

2

( ) ; ( ) ; ( )E X m V X Se Xn n

Chng 6 5. Thng k

b. Phng sai mu l mt thng k, k hiu S2, c tnh theo cng thc sau

Trong

Tnh cht E(S2) = 2

BT chng minh tnh cht trn.

lch chun mu:

137

2 22 2 2 2;1 1n nS X X s x xn n 2 2

1 1

1 1;

n k

i i ii i

X X x n xn n

2S S

Chng 6 5. Thng k

V d 6.3. Cn mt s sn phm cng loi th c bng sau

Tnh trung bnh mu v phng sai mu.

138

Trng lng (kg) 5 6 7 8S sn phm 2 12 7 4


Tnh trung bnh mu v phng sai mu.


S sn phm 10 40 30 20

Chng 6 5. Thng k

Lin quan n S2 cn c cc khi nim sau+ Tng bnh phng cc sai lch SS

+ lch bnh phng trung bnh MS

Lin h

139

2

1

( )n

ii

SS X X

22 2

1

1( )

n

ii

MS X X X Xn

2

1 1

SS nS MS

n n

Chng 6 5. Thng k

Phng sai S*2

Nu bit trung bnh tng th m th tnh c phng sai S*2

Tnh cht E(S*2) = 2

140

*2 2

1

1( )

n

ii

S X mn

Chng 6 5. Thng k

c. Tn sut mu l mt thng k, k hiu f, c tnh theo cng thc sau

Trong Y l (bin m) s phn t mang du hiu nghin cu trong mu.

Tnh cht vi p l tn sut tng th ca du hiu nghin cu th

141

Yf

n

(1 ) (1 )( ) ; ( ) ; ( )

p p p pE f p V f Se f

n n

Vi X l s phn t mang du hiu nghin cu trong 1 phn t ca

tng th th X l bnn gc v X ~ A(p)

Chng 6 5. Thng k

V d 6.5. Kim tra 100 sn phm ca l hng th c 10 ph phm. Khi t l ph phm trn mu l

142


Tm t l sn phm c thi gian hon thnh lu hn 14 pht.


S sn phm 10 40 30 20

100,1

100f

Chng 6 6. Mu hai chiu

6. MU NGU NHIN HAI CHIU6.1. nh ngha6.2. Phng php m t mu ngu nhin hai chiu

143

Chng 6 7. Quy lut PPXS

7. QUY LUT PHN PHI XC SUT CA MT S THNG K C TRNG MU

7.1. Trng hp tng th c bnn gc X ~ N(, 2)Vi W = (X1, X2, , Xn) l mu ngu nhin th+ X1, X2,, Xn l cc bnn c lp+ Xi c cng quy lut ppxs vi X, suy ra

E(Xi) = E(X) = ; V(X i) = V(X) = 2

Ta chng minh c cc kt lun sau

144


p dng + Tnh xs+ Tm a, b sao cho

V d 6.7. Trng lng sn phm l bnn phn phi chun vi trung bnh l 15 kg v lch chun l 3 kg.

a. Tm xc sut trng lng trung bnh ca 25 sp t 14 kg n 16 kg.

b. Vi mc xc sut 0, th trng lng trung bnh ca 25 sp ti a l bao nhiu?

145

2

1

1(1) ;

n

ii

X X Nn n

( ) ?P a X b

( ) 1P a X b


146

*22 2

2

22 2

2

( )(2) (0,1)

( )

( )(3) ( )

( 1)(4) ( 1)

( )(5) ( 1)

X X nU N

Se X

n Sn

n Sn

X nT T n

S


p dng + Tnh xs+ Tm a, b sao cho

V d 6.8. Trng lng sn phm l bnn phn phi chun vi trung bnh l 15 kg v lch chun l 3 kg.

a. Tm xc sut phng sai v trng lng ca 25 sp nh hn 5,2 kg2.

b. Vi mc xc sut 0, th lch chun v trng lng ca 25 sp ti a l bao nhiu kg?

147

22 2

2

( 1)(4) ( 1)

n Sn

2( ) ?P a S b

2( ) 1P a S b


7.2. Trng hp c 2 tng th vi cc bnn gc X1 ~ N(1, 12) v X2 ~ N(2, 22)

Rt ra 2 mu ngu nhin c lpW1 = (X11 , X12, , X1n1) kch thc n1W2 = (X21 , X22, , X2n2) kch thc n2Cc trung bnh mu s c lp vi nhauCc phng sai mu c lp vi nhauTa c cc kt lun sau

148

2 21 2

1 2 1 21 2

(6) ;X X Nn n


Nu n1 > 30 v n2 > 30 th T N(0,1)149

1 2 1 2

2 21 2

1 2

2 22 21 1 2 2

1 22 21 2

2 2 21 2

1 2 1 21 22 2

1 1 2 2

1 2 1 2

( ) ( )(7) (0,1)

( 1) ( 1)(8) ( 2)

(9)

( ) ( )( 2)

( 1) ( 1) 1 12

X XU N

n n

n S n Sn n

Khi

X XT T n n

n S n Sn n n n


Nu n1 > 30 v n2 > 30 th T N(0,1)

150

2 21 2

1 2 1 2

2 21 1 2 2

21 2 1 12 2 2 2

2 1 1 1 2 2

(10)

( ) ( )( )

/ /

( 1)( 1) /;

( 1) ( 1)(1 ) / /

Khi

X XT T k

S n S n

n n S nk C

n C n C S n S n

21

2 21 1 1

1 222 222 2

2

/ 1(11) ( 1, 1)

/1

S nF F n n

Sn


7.3. Trng hp tng th c bnn gc X ~ A(p)+ Vi p l tn sut tng th ca du hiu nghin cu

X l s phn t mang du hiu nghin cu trong 1 phn t ca tng th th X ~ A(p) l bnn gc.

+ f l tn sut mu ca du hiu nghin cuVi mu c kch thc n 100 ta c cc kt lun sau

151

( )(13) (0,1)

(1 )

(1 )(12) ,

f p nU

p pf N p

n

Np p


p dng (12) + Tnh xs P(a < f < b)+ Tm a, b sao cho P(a < f < b) = 1 -

V d 6.. Bit rng t l ph phm ca l hng l 10%.a. Tm xc sut khi kim tra (mu) 100 sn phm ca l hng

th t l ph phm trn mu ln hn 13%.b. Tm xc sut khi kim tra 100 sn phm ca l hng th c

ti thiu l 6 ph phm.c. Vi mc xc sut 0, nu kim tra 200 sn phm ca l

hng th t l ph phm ti a l bao nhiu? C ti a bao nhiu ph phm?

V d 6.10. Trng lng sn phm l bnn phn phi chun vi trung bnh l 15 kg v lch chun l 3 kg. Tm xc sut trong 100 sn phm c t hn 20 sn phm nh hn 12 kg.

152


7.4. Trng hp c 2 tng th vi cc bnn gc X1 ~ A(p1 ) v X2 ~ A(p2)

+ Vi p1, p2 l cc tn sut tng th ca du hiu nghin cu

+ f1, f2 l cc tn sut mu ca du hiu nghin cuVi cc mu c kch thc 100 ta c cc kt lun sau

153

1 1 2 21 2 1 2

1 2

1 2 1 2

1 1 2 2

1 2

(1 ) (1 )(14) ,

( ) ( )(15) (0,1)

(1 ) (1 )

p p p pf f N p p

n n

f f p pU N

p p p pn n

CHNG 7

C LNG CC THAM S CA BIN NGU NHIN

1. M U - Ni dung chng 7 Phng php c lng im. Phng php c lng bng khong tin cy.Bi ton c lng Cho bin ngu nhin gc X vi quy

lut phn phi xc sut bit xong cha bit tham s no ca n. Phi c lng (xc nh mt cch gn ng) gi tr ca .

154

Chng 7 2. PP c lng im

2. PHNG PHP C LNG IM2.1. Phng php hm c lnga. Khi nimGi s cn c lng tham s .Lp mu ngu nhin W = (X1, X2, , Xn)

Chn lp thng k (da vo chng 6).

Lp mu c th w v tnh c gi tr c th ca thng k

l chnh l c lng im ca .

l hm ca cc bnn Xi nn gi l pp hm c lng.155

1 2( , ,..., )nf X X X

1 2( , ,..., )nf x x x


V d 7.2. Cho tng th vi bnn gc X. Lp mu ngu nhin W = (X1, X2, X3) v cc c lng sau

Trong 3 c lng trn, c lng no l khng chch, c lng no l hiu qu nht ca trung bnh tng th

Ch + Trng hp tng qut trung bnh tng th l E(X) = m+ Nu bnn gc X pp chun trung bnh tng th l E(X) =+ Nu bnn gc X pp A(p) trung bnh tng th l E(X) = p

W l mnn nn cc Xi c lp, E(Xi) =E(X), V(Xi) = V(X)156

1 2 31 21 2 3 1 2 3

1 2 3 1 1; ;

2 6 2 3

X X XX XG G G X X X


b. Cc tiu chun la chn hm c lng c lng khng chch

Thng k ca mu l c lng khng chch ca tham s nu

? Cng thc tnh chch

V d 7.1.

(chng minh)

l cc c lng khng chch ca m, p, 2157

( )E

2 2

2

( )

( )

( )

, ,

E X m

E f p

E S

X f S


c lng hiu qu Thng k ca mu l c lng hiu qu (nht) ca

tham s nu n l c lng khng chch v c phng sai nh nht so vi cc c lng khng chch khc c xy dng trn cng mu .

Nu l cc c lng khng chch ca v

th l hiu qu hn

? Cng thc tnh hiu qu

158

1 2( ) ( )V V

1 2,

1 2


V d 7.3. Cho tng th vi bnn gc X. Lp 2 mu ngu nhin c lp vi kch thc n1, n2 v cc trung bnh mu tng ng. Xt h c lng

Chng minh G l c lng khng chch ca trung bnh tng th (m). Vi = ? th G l hiu qu nht ca m.

Nu X pp chun th m = .

V d 7.4. Cho tng th vi bnn gc X pp A(p). Lp 2 mu ngu nhin c lp vi kch thc n1, n2 v cc tn sut mu f1, f2. Xt h c lng f = f1 + (1-)f2. Vi = ? th f l c lng hiu qu nht ca p.

159

1 2(1 ) ; 0 1G X X


Bt ng thc Cramer Rao Nu bin ngu nhin gc X c hm mt xc sut f(x, ) tha mn mt s iu kin nht nh v * l mt c lng khng chch bt k ca

V d 7.5. Cho tng th vi bnn gc X pp chun. Chng minh rng trung bnh mu l c lng hiu qu nht ca trung bnh tng th.

V d 7.6. Chng minh rng tn sut mu f l c lng hiu qu nht ca tn sut tng th p.

160

2

1( *)

ln ( , )V

f xnE


c lng vngThng k ca mu lngvng ca tham s nu n hi t theo xc sut n khi n .

Ngha l vi mi > 0 ta c

V d 7.7. Cho tng th vi bnn gc X pp chun. Chng minh rng trung bnh mu l c lng vng ca trung bnh tng th.

V d 7.8. Chng minh rng tn sut mu l c lng vng ca tn sut tng th.

c lng 161

lim | | 1n

P


c. Mt s kt lun Trung bnh mu l c lng khng chch, hiu qu nht

v vng ca trung bnh tng th v ng thi l c lng tuyn tnh khng chch tt nht, do nu cha bit trung bnh tng th th c th dng trung bnh mu c lng n.

Tn sut mu l c lng khng chch, hiu qu nht v vng ca tn sut tng th v ng thi l c lng tuyn tnh khng chch tt nht, do nu cha bit tn sut tng th th c th dng tn sut mu c lng n.

C E(S2) = E(S*2) = 2 nn c th dng 1 trong 2 thng k ny c lng phng sai tng th.

162


2.2. Phng php c lng hp l ti aCn c lng tham s bng pp c lng hp l ti a.Bit hm mt xc sut f(x, ) ca bnn gc X.Lp mu ngu nhin W = (X1, X2, , Xn)Ta xy dng hm hp l L ca tham s nh sau

L() = L(x1, x2,,xn, ) = f(x1 , ).f(x2, ) f(xn, )trong w (x1 , x2 ,, xn ) l gi tr c th bt k ca mu.Gi tr c th ca thng k trn mu w gi l c lng

hp l ti a ca nu ng vi gi tr hm hp l t cc i.

Khi thay w bi W th ta c hm (L) hp l ti a ca . 163


Cc bc tm c lng hp l ti aV L v lnL t cc i ti cng mt gi tr ca nn ta s

tm lnL t cc i nh sau+ Tm L v lnL, rt gn.+ Tm o hm bc nht v bc hai ca lnL theo .+ Tm nghim ca o hm bc nht, k hiu l .+ Chng minh

Khi l c lng im hp l ti a ca

l hm c lng hp l ti a ca . 164

2

2

ln0

d L

d

1 2( , ,..., )ng x x x

1( ,..., )ng X X


V d 7.. Cho tng th vi bnn gc X pp chun. Tm c lng hp l ti a ca trung bnh tng th.

Bi tp. + Tng th c bnn gc X ~ A(p). Tm c lng hp l ti

a ca tham s p.+ Tng th c bnn gc X ~ P(). Tm c lng hp l ti

a ca tham s .+ Tng th c bnn gc X ~ E(). Tm c lng hp l ti

a ca tham s .

165

Chng 7 3. PP c lng khong

3. PHNG PHP C LNG BNG KHONG TIN CY

3.1. Khi nimKhong (G1, G2) c gi l khong tin cy ca tham s

vi tin cy (1 ) cho trc nu tha mnP(G1 < < G2 ) = 1

I = G2 G1 gi l di khong tin cy.C s ca phng php c lng bng khong tin cy l

nguyn l xc sut ln.

166


Cc bc tm khong G1, G2) Lp mu ngu nhin W = (X1, X2, , Xn) v xy dng

thng k G = f(X1, X2, , Xn, ) (da vo chng 6) sao cho quy lut phn phi xc sut ca G l n ton xc nh v khng ph thuc v cc i s ca n.

Vi tin cy (1 ) cho trc tm c cp gi tr khng m 1, 2 sao cho 1 + 2 = . T tm c cp gi tr ti hn g1-1 v g2 tha mn

Bin i tng ng ta c

167

1 21 1 2( ) 1 ( ) 1P g G g

1 2( ) 1P G G


Thc t thng yu cu tin cy (1 ) kh ln nn theo nguyn l xc sut ln, bin c (G1 < < G2) s xy ra khi thc hin mt php th c th.Thc hin 1 php th th c mu c th w v tm c cc gi tr c th ca Gi l gi

Kt lun Vi tin cy (1 ).100% , t mu c th cho ta c lng c tham s nm trong khong (g1, g2)

Cc khong tin cy thng dng+ KTC 2 pha (c bit i xng)+ KTC bn tri c lng gi tr ti a+ KTC bn phi c lng gi tr ti thiu.

168


3.2. c lng k vng ton ca bnn phn phi chun (c lng trung bnh tng th )

a. Nu bit phng sai 2

Chn thng k

Vi tin cy (1 ) ta tm c 2 gi tr 1 + 2 = v 2 gi tr ti hn chun l u1-1 v u2 tha mn

P(u1-1 < U < u2) = 1 Thay cng thc ca U ri bin i tng ng ta c

169

( )(0,1)

X nU N

2 1(1 )P X u X u

n n


b. Nu cha bit phng sai 2

Chn thng k

Tng t phn (a) ta c

Khong tin cy ngu nhin ca l

Vi mu c th ta c khong tin cy c th ca .170

( )( 1)

X nT T n

S

2 1

( 1) ( 1) (1 )n nS S

P X t X tn n

2 1

( 1) ( 1);n nS S

X t X tn n


Cc trng hp thng dng+ KTC i xng 1 = 2 = /2 (t.bnh thuc khong no)

+ KTC c lng gi tr ti a 1 = , 2 = 0

+ KTC c lng gi tr ti thiu 1 =0, 2 =

Vi mu c th ta c khong tin cy c th ca .171

( 1) ( 1)/2 /2n nS SX t X t

n n

( 1)nSX tn

( 1)nSX tn


+ Vi KTC i xng, ta c di KTC l

Sai s ca c lng l

+ Mun c KTC vi tin cy (1 ) m di KTC khng vt qu I0 cho trc th phi iu tra mu c kch thc n tha mn

Tng t, mun 0 th 172

( 1)/22nSI t

n

( 1)/22nI S t

n

2 2( 1)/220

4' n

Sn t

I

2 2( 1)/220

' nS

n t


V d 7.10. Bit thi gian gia cng 1 chi tit my (pht) phn phi chun. Cho s liu sau

a. Vi tin cy 0,5 hy c lng thi gian trung bnh gia cng chi tit (bng KTC i xng)

b. Mun gi nguyn tin cy 0,5 m chnh xc ca c lng tng gp i th phi iu tra thm bao nhiu chi tit.

c. Cho bit thi gian trung bnh gia cng chi tit ti a l bao nhiu? Ly = 0,05.

173

Thi gian gia cng (pht) 8 9 10

S chi tit 4 10 6


V d 7.11. iu tra thu nhp ca 40 nhn vin cng ty A thy trung bnh (mu) l 5,5 triu ng/thng v lch chun mu l 0,8 triu ng/thng. Vi tin cy 5% hy cho bit thu nhp trung bnh ti thiu ca nhn vin cng ty ny.

174


3.3. c lng phng sai ca bnn phn phi chun (c lng phng sai tng th 2)

a. Nu bit trung bnh b. Nu cha bit trung bnh Khong tin cy ngu nhin ca 2 l

Cc trng hp thng dng+ Khong tin cy hai pha ca 2 l

175

2 1

2 22

2( 1) 2( 1)1

( 1) ( 1)n n

n S n S

2 22

2( 1) 2( 1)/2 1 /2

( 1) ( 1)n n

n S n S

Chng 7 3. PP c lng khong+ Khong tin cy c lng gi tr ti a ca 2 l

+ Khong tin cy c lng gi tr ti thiu ca 2 l

V d 7.12. iu tra thu nhp ca 40 nhn vin cng ty A thy trung bnh (mu) l 5,5 triu ng/thng v lch chun mu l 0,8 triu ng/thng. Vi tin cy 5% hy cho bit phng sai v thu nhp ca nhn vin cng ty ny nm trong khong no?

176

22

2( 1)1

( 1)0

n

n S

22

2( 1)

( 1)n

n S


V d 7.13. (tip 7.10) Bit thi gian gia cng 1 chi tit my (pht) phn phi chun. Cho s liu sau

a. Vi tin cy 0,5 hy c lng phng sai v thi gian gia cng chi tit .

b. Cho bit lch chun v thi gian gia cng chi tit ti a l bao nhiu? Ly = 0,05.

177


S chi tit 4 10 6


3.4. c lng xc sut p ca bnn phn phi khng mt (c lng tn sut tng th p)

Vi p l tn sut tng th ca du hiu nghin cu (p = M/N)f l tn sut mu ca du hiu nghin cu

a. Nu mu c kch thc nh (t c)b. Nu mu c kch thc ln (n 100)Khong tin cy ca p vi tin cy (1 ) l

? Hy vit khong tin cy c lng gi tr ti a, gi tr ti thiu ca p.

178

2 1

(1 ) (1 )f f f ff u p f u

n n


+ Khong tin cy i xng ca p l

di KTC v sai s ca c lng l

+ Mun c khong tin cy vi tin cy ( 1 ) m

(n l kch thc mu mi)

179

/2 /2

(1 ) (1 )f f f ff u p f u

n n

20 /22

0

20 /22

0

4 (1 )'

(1 )'

f fI I n u

I

f fn u

/2 /2

(1 ) (1 )2 ;

f f f fI u u

n n


V d 7.14. Kim tra 200 sn phm ca mt l hng th c 20 ph phm. Ly = 0,05 cho cc cu hi sau

a. Hy c lng t l ph phm ti a ca l hng.b. Mun c khong tin cy 5% m di khong tin cy

khng vt qu 2% th phi kim tra thm bao nhiu sn phm na?

V d 7.15. iu tra 500 h gia nh mt thnh ph th c 400 h dng bp ga, trong c 100 h dng bp ga ca hng A. Gi s mi h dng 1 bp ga. = 0,05.

a. c lng t l h dng bp ga TP ny.b. Bit hng A bn 15000 bp ga TP ny. Hy c

lng ti a s h gia nh dng bp ga TP ny.180

CHNG 8

KIM NH GI THUYT THNG K

1. M U - Ni dung chng 8 Khi nim.

Kim nh gi thuyt v tham s ca 1 bin ngu nhin.

Kim nh gi thuyt v tham s ca 1 bin ngu nhin.

181

Chng 8 2. Khi nim

2. KHI NIM2.1. Gi thuyt thng k Gi thuyt thng k l gi thuyt v

+ Cc tham s c trng ca bin ngu nhin,+ Dng phn phi XS ca bin ngu nhin,+ S c lp ca cc bin ngu nhin.

K hiu H0 l gi thuyt gc H1 l gi thuyt i

y ch xt H0 l gi thuyt n (dng = 0)H1 l gi thuyt kp

182

Chng 8 2. Khi nim

Nhng pht biu c cha du bng (=, , ) l H0Nhng pht biu khng cha du bng (, ) l H1

V d 8.1. Vit cp gi thuyt H0 , H1 cho cc cu hi saua) C th cho rng t l nam ca Vit Nam l 50% hay

khng?b) C th cho rng trng lng trung bnh ca sn phm

ln hn 6 kg hay khng?c) C th cho rng mc phn tn v tin lng khng

vt qu 1,5 triu ng/thng c khng?d) C th cho rng VN t l sinh vin nam bng t l

sinh vin n hay khng?

183

Chng 8 2. Khi nim

Phng php chung kim nh+ Gi s H0 ng, da vo thng tin ca mu rt ra mt

bin c A no sao cho P(A) = b n mc c th coi bin c A khng xy ra trong 1 php th c th (theo nguyn l XS b)

+ Thc hin 1 php th i vi mu ngu nhin (ngha l xt trn 1 mu c th).

Nu A xy ra th H0 sai v bc b nNu A khng xy ra th cha c c s bc b H0

184

Chng 8 2. Khi nim

2.2. Tiu chun kim nh gi thuyt thng k Vi mu ngu nhin W, chn thng k

G = f( X1, X2 , , Xn, 0)sao cho nu H0 ng th quy lut PPXS ca G l hon ton xc nh. (0 l tham s lin quan n gi thuyt cn kim nh)

Trn mu c th, G nhn gi tr c th lGqs = f(x1, x2,,xn, 0)

185

Chng 8 2. Khi nim

2.5. Sai lm khi kim nh a. Sai lm loi 1: bc b gi thuyt H0 trong khi H0 ngXS mc sai lm loi 1 l

b. Sai lm loi 2: tha nhn gi thuyt H0 trong khi H0 saiXS mc sai lm loi 2 l (1 ) l lc kim nhNu gim th tng v ngc li.

2.6. Th tc kim nh gi thuyt thng k

186

0( W / )P G H

1( W / )P G H

Chng 8 2. Khi nim

2.3. Min bc b gi thuyt H0 vi mc ngha l W

Min cn li trn trc s gi l min tha nhn H0im ngn cch gia min bc b v min tha nhn gi

l im ti hn.

2.4. Quy tc kim nh gi thuyt thng k+ Nu th bc b H0, tha nhn H1+ Nu th cha c c s bc b H0

187

0

W

( W / )

R

P G H

W

Wqs

qs

G

G

Chng 8 3. K mt tham s

3. KIM NH GI THUYT V MT TNG TH3.1. Kim nh gi thuyt v k vng ton ca bin

ngu nhin phn phi chun (K trung bnh)a. Khi bit phng sai (t c)b. Khi cha bit phng sai

188

H0 v tiu chun H1 Min bc b H0H0 : = 0 0

> 0

< 0

0( )X nTS

( 1)/2W :| | nT T t ( 1)W : nT T t ( 1)W : nT T t


V d 8.2. Bit thi gian gia cng 1 chi tit my (pht) phn phi chun. Cho s liu sau

Vi mc ngha 5% c th cho rng thi gian trung bnh gia cng chi tit l pht c khng?

V d 8.3. iu tra thu nhp ca 40 nhn vin cng ty A thy trung bnh l 5,5 triu ng/thng v lch chun l 0,8 triu ng/thng. Vi mc ngha 5% c th cho rng thu nhp trung bnh ca nhn vin cng ty ny khng di 6 triu/thng c khng?

189


S chi tit 5 12 8


3.2. Kim nh gi thuyt v phng sai ca bin ngu nhin phn phi chun (K phng sai)

Ch + n v o+ Phn bit chiu bin i ca phn tn ( dao

ng,) vi n nh ( ng u,.. .).190

H0 v tiu chun H1 Min bc b H0H0 : 2 = 20 2 20

2 > 20

2 < 20

22

20

( 1)n S

2 2( 1)/22

2 2( 1)1 /2

W :n

n

2 2 2( 1)W : n 2 2 2( 1)1W : n


V d 8.4. (tip 8.2) Bit thi gian gia cng 1 chi tit my (pht) phn phi chun. Cho s liu sau

Vi mc ngha 5% c th cho rng phng sai v thi gian gia cng chi tit l 2 pht2 c khng?

V d 8.5. (tip 8.3) iu tra thu nhp ca 40 nhn vin cng ty A thy trung bnh l 5,5 triu ng/thng v lch chun l 0,8 triu ng/thng. Vi mc ngha 5% c th cho rng phn tn v thu nhp ca nhn vin cng ty ny khng qu 0,5 (triu/thng) c khng? 191


S chi tit 5 12 8


3.3. Kim nh gi thuyt v xc sut p ca bin ngu nhin phn phi khng mt (K tn sut)

V d 8.6. Kim tra 200 sn phm ca mt nh my th c 20 ph phm. Vi mc ngha 5% c th cho rng t l ph phm ca nh my nh hn 15% hay khng?

192

H0 v tiu chun H1 Min bc b H0H0 : p = p0 p p0

p > p0

p < p0

0

0 0

( )

(1 )

f p nU

p p

/2W :| |U U u W :U U u W :U U u


V d 8.7. iu tra 200 cng nhn th c 110 nam v 0 n. Ly = 0,05 cho cc cu hi sau

a. C th cho rng t l cng nhn nam cao hn t l cng nhn n hay khng?

b. Cho bit t l cng nhn nam thuc khong no?

V d 8.8. iu tra 500 h gia nh mt thnh ph th c 400 h dng bp ga, trong c 100 h dng bp ga ca hng A. Gi s mi h dng 1 bp ga. = 0,05 .

a. C th cho rng t l h dng bp ga ca hng A TP ny khng vt qu 15% hay khng?

b. Bit hng A bn 20.000 bp ga TP ny. C th cho rng s h gia nh dng bp ga TP ny nh hn 100.000 h c khng?

193

Chng 8 4. K hai tham s

4 KIM NH GI THUYT V HAI TNG THCh kt lun l kim nh 2 tham s khi c 2 mu c lp4.1. Kim nh gi thuyt v hai k vng ton ca hai

bin ngu nhin phn phi chuna. Khi bit phng sai (t c)b. Khi cha bit c hai phng sai

194


1 > 2

1 < 2

1 2

2 21 2

1 2

X XT

S Sn n

/2W :| |T T u W :T T u W :T T u


4.2. Kim nh gi thuyt v hai phng sai ca hai bin ngu nhin phn phi chun

Ch Phn bit chiu bin i ca phn tn ( dao ng,) vi n nh ( ng u,.. .).

195


12 > 22

12 < 221 2

2 1

2122

( , )1 ( , )

1/ : n n

n n

SF

S

T c ff

1 2

1 2

( 1, 1)/2

( 1, 1)1 /2

W :n n

n n

F fF

F f

1 2( 1, 1)W : n nF F f 1 2( 1, 1)1W : n nF F f


V d 8.. Cn 25 sn phm loi A th thy trong lng trung bnh l 2,6 kg v lch chun l 0,4 kg. Cn 40 sn phm loi B th thy trung bnh l 2,3 kg v lch chun l 0,6 kg.

Ly = 0,05 cho cc cu hi saua. C th cho rng trng lng trung bnh ca sn phm

loi A v loi B l nh nhau hay khng?b. C th cho rng trng lng sn phm loi A n nh

hn loi B hay khng?

196


4.3. Kim nh gi thuyt v hai xc sut ca hai bin ngu nhin phn phi khng mt

197

H0 v tiu chun H1 Min bc b H0H0 : p1 = p2 p1 p2

p1 > p2

p1 < p2

1 2

1 2

1 1 2 2

1 2

1 1(1 )

f fU

f fn n

n f n ff

n n

/2W :| |U U u W :U U u W : |U U u

CHNG

KIM NH PHI THAM S

1. M U - Ni dung chng Kim nh gi thuyt v s c lp ca 2 du hiu nh

tnh.

Kim nh Jarque Bera v bin ngu nhin phn phi chun.

198

Chng 1. K s c lp

1. KGT v s c lp ca 2 du hiu nh tnhH0 du hiu A v du hiu B c lp vi nhauH1 du hiu A v du hiu B ph thuc nhauBng s liu

199

A B B1 B2 Bk Tng

A1 n11 n12 n1k n1A2 n21 n22 n2k n2

Ah nh1 nh2 nhk nhTng m1 m2 mk n

Chng 1. K s c lp

Tiu chun K

Min bc b

V d .1. C kin cho rng gii tnh v mc lng ca cng nhn l ph thuc nhau. Ngi ta iu tra mt s cng nhn v c s liu sau

Vi mc ngha 5% hy kt lun v kin trn. 200

2ij2

1 1

2 2 2[( 1)( 1)]

1

W :

h k

i j i j

h k

nn

n m

Gii tnh Lng Cao Trung bnh Thp

Nam 15 35 10

N 5 30 5

Chng 2. K phn phi chun

1. K Jarque Bera v bin ngu nhin phn phi chunH0 bin ngu nhin X phn phi chunH1 bin ngu nhin X khng phn phi chunVi a3 l h s bt i xng, a4 l h s nhn

Tiu chun K

Min bc b

V d .2. C kin cho rng im thi tt nghip mn Ton ca hc sinh lp 12 khng phn phi chun. iu tra im thi ca 40 hc sinh th thy h s bt i xng l 0,4; h s nhn l 2,5. Vi mc ngha 5% hy kt lun v kin trn.

201

2 23 4

2(2)

( 3)

6 24

W :

a aJB n

JB JB

Thng k Bi tp

Bi tp chng 658, 62, 63, 65, 67

Bi tp chng 777, 78, 80, 83, 84, 85, 90, 91

202

Bi tp chng 865, 66, 67, 69, 70, 71, 76, 77, 79, 80, 82, 83

Bi tp chng 1, 2, 3, 4, 5, 6

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