BI TP

XC SUT & THNG K

n bn th hai

NGUYN VN THN

Trng i Hc Khoa Hc T Nhin Tp.HCM

LU HNH NI B

BI TP XC SUT V THNG K

c 2011, 2012 Nguyn Vn ThnBn quyn thuc tc gi.

Mi t chc, c nhn mun s dng tc phm di mi hnh thc phi c s ng

ca ch s hu quyn tc gi.

n bn u tin 9/2011

n bn th hai 9/2012

Li ni u

Ngy nay xc sut v thng k ton tr thnh mt khoa hc c nhiu ng dng

trong nhiu lnh vc khoa hc v k thut khc nhau nh: vt l, thin vn hc, ha

hc, sinh hc, y hc, tm l hc, kinh t hc . . .

V th m mn hc xc sut v thng k ton tr thnh mn bt buc c s

c ging dy hu ht cc trng i hc, cao ng cho cc sinh vin ngay t nm

nht hoc nm hai.

Mc ch ca ti liu ny l nhm gip bn c thng qua vic gii cc bi tp

(c trnh by di nhiu ng cnh, tnh hung v trong nhiu lnh vc khc nhau)

c th hiu ng bn cht ca nhng khi nim v phng php c bn nht ca xc

sut v thng k, v qua c th p dng c chng, i su tm hiu c phng

php thch hp cho nhng tnh hung c th trong chuyn nghnh m bn theo ui.

Ti liu gm c hai phn chnh cng thm phn ph lc:

Phn I l nhng bi tp v l thuyt xc sut gm khong 200 bi c su tm

v bin son gm bn chng:

Chng 1 ni v cc khi nim ti thiu ca l thuyt tp hp v gii tch t hp,nhm chun b cc kin thc bn c c th lnh hi v gii cc bi tp v sau

c d dng.

Chng 2 dnh cho cc bi tp v cc khi nim c bn ca l thuyt xc sutchng hn nh khng gian cc bin c, xc sut c in, xc sut hnh hc, xc

sut c iu kin, cng thc xc sut y ,. . .

Chng 3 trnh by cc bi tp v bin ngu nhin v hm phn phi cng ccc trng ca cc bin ngu nhin nh k vng, phng sai, trung v,. . .

Chng 4 trnh by cc bi tp v cc bin ngu nhin thng dng nh bin ngunhin c phn phi Bernulli, phn phi Poison, phn phi u, phn phi chun.

Phn II l nhng bi tp thng k ton hc gm khong 70 bi c su tm v

bin son bao gm ba chng:

Chng 5 dnh cho cc bi tp v l thuyt mu, tnh ton cc c trng camu nh trung bnh mu, phng sai mu,. . .

Chng 6 trnh by cc bi tp v l thuyt c lng, ch yu l c lngkhong cho trung bnh, t l ca tng th.

i

ii

Chng 7 ni n cc bi tp v l thuyt kim nh cc gi thuyt thng k.

Trong mi chng ti c chia ra thnh cc mc nh theo tng chuyn cc

bn c th rn luyn chuyn su v tp trung hn.

Ti liu gm nhng bi tp rn luyn k nng tnh ton, rn luyn t duy v

phng php chng minh cng nh gip bn c nm vng v vn dng cc khi nim

c bn v xc sut v thng k. Mt s bi tp c nh du (*) l cc bi tp kh,

th thch dnh cho cc sinh vin kh gii nm vng v vn dng sng to cc kin

thc hc trn lp.

Trong ti liu ny i km cc bi tp l cc ch thch, hng dn, p n ty theo

mc kh d ca chng.

V kh nng c hn, chc chn ti liu cn c nhiu thiu st, ti mong nhn c

s ng gp kin ca cc bn c n bn tip theo c hon thin hn.

Tp. H Ch Minh, Nguyn Vn Thn

Ma h, 2012

Mc lc

Li ni u i

1 Tp hp - Gii tch t hp 1

1.1 Tp hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2 Gii tch t hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

2 Bin c v xc sut 10

2.1 Bin c . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

2.2 Xc sut c in . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2.3 Xc sut hnh hc . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14

2.4 Cc cng thc tnh xc sut c bn . . . . . . . . . . . . . . . . . . . . 14

2.5 Cng thc xc sut y , cng thc Bayes . . . . . . . . . . . . . . . 19

3 Bin ngu nhin v hm phn phi 24

4 Mt s phn phi xc sut thng dng 36

4.1 Phn phi Bernoulli, nh thc . . . . . . . . . . . . . . . . . . . . . . . 36

4.2 Phn phi Poisson . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

4.3 Phn phi chun . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

5 L thuyt mu 46

6 c lng tham s thng k 49

6.1 c lng trung bnh tng th . . . . . . . . . . . . . . . . . . . . . . . 49

6.2 c lng t l tng th . . . . . . . . . . . . . . . . . . . . . . . . . . 52

6.3 Tng hp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

7 Kim nh gi thuyt thng k 54

7.1 So snh k vng vi mt s cho trc . . . . . . . . . . . . . . . . . . . 54

iii

MC LC iv

7.2 So snh hai k vng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57

7.3 So snh t l vi mt s cho trc . . . . . . . . . . . . . . . . . . . . . 60

7.4 So snh hai t l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61

A Cc bng phn phi 63

Chng 1

Tp hp - Gii tch t hp

1.1 Tp hp

Bi 1.1 (*). Cho dy tp hp A1, A2, . . . , An, . . .. Chng minh rng lun lun tn ti dytp hp B1, B2, . . . , Bn, . . ., sao cho:

(a) Cc Bi tng i mt ri nhau;

(b)i=1Ai =

k=1Bk.

Hng dn. Hy bt u vi hai trng hp d nht n = 2 v n = 3.

Ch thch.i=1Ai = {x|n, x An}.

Bi tp ny ch ra cch xy dng mt h cc tp ri nhau t mt h cc tp bt k.

Bi 1.2. Chng minh rng cc h thc sau y tng ng nu A v B l tp hp con ca:

A B = , A B,B A.Hng dn. Hy chng minh A B = A B B A A B = .

Bi 1.3. Khng nh sau c ng hay khng: "nu A,B,C l cc tp con ca tp sao cho

A B C v B A C

th B = " ?

Bi 1.4. Chng minh rng nu A,B,C l cc tp hp con ca tp hp , sao cho

A B C v A C B, th A C =

Bi 1.5. Tm biu thc n gin ca cc biu thc sau:

(a) (A B)(A C)

(b) (A B)(A B)

1

1.1. TP HP 2

(c) (A B)(A B)(A B)

(d) (A B)(A B)(A B)

(e) (A B)(B C)

Bi 1.6. H thc no trong cc h thc sau y ng. Nu ng hy chng minh, nu saihy cho v d minh ha.

(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))Ch thch. i khi v s n gin v tin li ngi ta vit AB thay cho A B, A+B thay cho A B v Ahoc Ac thay cho A. Ch c nh trong Ac l vit tt ca t "complement" (phn b) trong ting Anh.

Bi 1.7. Chng minh rng:

(a) A B A B = A

(b) (A B)AB = AB BA

Bi 1.8. Chng minh

(a) Nu A B = AB th A = B

(b) A BC (A B)C

(c) Nu A1 A,B1 B v A B = th A1 B1 =

Bi 1.9. H thc no trong cc h thc sau y ng? i vi cc h thc sai, hy ch raiu kin h thc ng.

(a) A (B C) = (A B) (A C)

(b) A (B C) = (A B) C

(c) (A B) C = A (B C)

1.1. TP HP 3

(d) A \ (B C) = (A \B) (A \ C)

Bi 1.10. Cho A, B, C l cc tp con ca . t A4B = (A \B) (B \A). Chng minh:

(a) B 4A = A4B

(b) A4 = A

(c) A4A =

(d) A4 = A

(e) A4B = (A B) (A B)

(f) (A4B)4 C = A4 (B 4 C)Ch thch. Php ton 4 trn gi l hiu i xng. Hiu i xng ca hai tp hp A v B, k hiu l A4B,l tp hp gm cc phn t ch thuc A hoc ch thuc B, khng ng thi thuc c A v B.

Bi 1.11. Cho A, B, C l cc tp con ca . Chng minh:

(a) ((A B) (C D)) = (A B) (C D)

(b) (A B) (A B) (A B) (A B) =

(c) A \B = A (A4B)

(d) A B = (A4B)4 (A B)

(e) (A B)4 (B A) = A4B

(f) A4B = C 4D A4 C = B 4D

(g) A (B 4 C) = (A B)4 (A C)

(h) A4B = (A4 C)4 (C 4B)

Bi 1.12 (*). Cho A . nh ngha, IA, l hm ch (the indicator function, hay ngi tacn gi l hm c trng - the characteristic function) ca A nh sau:

IA : [0, 1]

vi

IA(x) :=

{1 nu x A0 nu x A

(a) Cho A, B l cc tp con ca . Chng minh rng

A = B nu v ch nu IA = IB

(b) Chng minh cc h thc sau

i. I = 1; I = 0

ii. IAB = IAIB

1.2. GII TCH T HP 4

iii. IAB = IA + IB IAIBiv. IA = 1 IAv. IA4B IA + IB (mod 2)vi. IA\B = IA(1 IB)

(c) Bng cch s dng cc h thc lin quan n hm ch trong cu b, chng minh cc hthc lin quan n tp hp trong bi 1.11

Ch thch. Cho s nguyn dng n, hai s nguyn a, b c gi l ng d theo m-un n nu chng c cng

s d khi chia cho n (tc l a b chia ht cho n). K hiu l a b (mod n). V d: 1 3 (mod 2).

Bi 1.13 (*). Cho l mt tp hp v gi s rng R l mt tp khc rng cc tp con ca. Ta ni rng R l mt vnh cc tp con ca nu

(A R v B R) (A B R v A \B R).

(a) Gi s R l mt vnh cc tp con ca . Chng minh rng R.

(b) Cho mt v d mt vnh R cc tp con ca sao cho / R.

(c) Gi R l mt tp cc tp con ca . Chng minh rng R l mt vnh nu v ch nu

(A R v B R) (A B R v A4B R).

(d) Cho S l mt tp cc tp con ca . Gi s rng

(A S v B S) (A B S v A \B S).

Chng minh rng S khng nht thit l mt vnh cc tp con ca .

(e) Chng minh rng giao ca hai vnh cc tp con ca l mt vnh cc tp con ca .

Hng dn. Trong cu (c), s dng kt qu cu (c) v (d) trong bi 1.11

1.2 Gii tch t hp

Bi 1.14. Nu mt ngi c 6 i v khc nhau v 4 i giy khc nhau. C bao nhiu cchkt hp gia v v giy?

p n. 24.

Bi 1.15. Mt lp c 40 hc sinh. Gio vin ch nhim mun chn ra mt ban cn s lpgm 3 ngi: 1 lp trng, 1 lp ph, 1 th qu. Hi gio vin ch nhim c bao nhiu cchchn ban cn s lp?

p n. 59280.

Bi 1.16. Mt l hng c 50 sn phm.

1.2. GII TCH T HP 5

(a) C bao nhiu cch chn ngu nhin cng lc 5 sn phm kim tra?

(b) C bao nhiu cch chn ngu nhin ln lt 5 sn phm?

p n. (a) 2118760 (b) 254251200.

Bi 1.17. Trong mt h thng in thoi ni b 3 s

(a) c bao nhiu my c cc ch s khc nhau?

(b) C bao nhiu my c s 9 cui cn cc ch s cn li u khc nhau?

p n. (a) 720 (b) 90.

Bi 1.18. M vng in thoi ca mt quc gia c dng mt dy gm 3 s. S u tin lmt s nguyn nm gia 2 v 9., s th hai l 0 hoc 1, v s th ba l mt s nguyn btk t 1 n 9.

(a) C th c ti a bao nhiu m vng?

(b) C bao nhiu m vng bt u vi s 4?

p n. (a) 144 (b) 18

Bi 1.19. Mt hp c 8 bi , 6 bi trng, 4 bi vng. Ngi ta chn ra 6 bi t hp . Hic bao nhiu cch chn nu:

(a) Khng yu cu g thm.

(b) Phi c 2 bi , 2 bi trng, 2 bi vng.

(c) C ng 2 bi vng.

p n. (a) 18564 (b) 2520 (c) 6006.

Bi 1.20. 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 phncng?

p n. 1260.

Bi 1.21. Mt t sn xut c 12 ngi, trong c 4 n, cn chia thnh 4 nhm u nhau.Hy tm s cch phn chia sao cho mi nhm c 1 n?

p n. 10080.

Bi 1.22. Xp 12 hnh khch ln 4 toa tu. Tm s cch sp xp:

(a) Mi toa c 3 hnh khch.

(b) Mt toa c 6 hnh khch, mt toa c 4 hnh khch, 2 toa cn li mi toa c 1 hnhkhch.

1.2. GII TCH T HP 6

p n. (a) 369600 (b) 665280.

Bi 1.23. (a) C bao nhiu cch xp 3 nam v 3 n ngi thnh mt hng?

(b) C bao nhiu cch xp 3 nam v 3 n ngi thnh mt hng nu mi nam v mi n ngicnh nhau?

(c) C bao nhiu cch xp nu 3 nam phi ngi cnh nhau?

(d) C bao nhiu cch xp nu khng c hai nam hoc hai n no c ngi cnh nhau?

p n. (a) 720 (b) 72 (c) 144 (d) 72.

Bi 1.24. C 6 hc sinh c sp xp ngi vo 6 ch ghi s th t trn mt bn di.Tm s cch xp

(a) 6 hc sinh vo bn.

(b) 6 hc sinh ny vo bn sao cho 2 hc sinh A, B ngi cnh nhau.

(c) 6 hc sinh ny ngi vo bn sao cho 2 hc sinh A, B khng ngi cnh nhau.

p n. (a) 720 (b) 240 (c) 480.

Bi 1.25. Nm ngi A, B, C, D, E s pht biu trong mt hi ngh. C bao nhiu cch spxp :

(a) B pht biu sau A.

(b) A pht biu xong th n lt B.

p n. (a) 120 (b) 24.

Bi 1.26. T 8 sinh vin n v 6 sinh vin nam, mt nhm lm vic gm 3 nam v 3 nphi c lp ra. C bao nhiu cch lp nhm nu

(a) 2 trong s cc sinh vin nam khng chu lm vic cng nhau?

(b) 2 trong s cc sinh vin n khng chu lm vic cng nhau?

(c) 1 nam v 1 n khng chu lm vic cng nhau?

p n. (a) 896 (b) 1000 (c) 910.

Bi 1.27. Mt ngi c 8 ngi bn. Ngi ny d nh mi 5 trong s 8 ngi bn thamd mt ba tic lin hoan. C bao nhiu cch chn nu

(a) 2 trong s cc ngi bn gin nhau v s khng tham d cng nhau?

(b) 2 trong s cc ngi bn s ch tham d cng nhau?

p n. (a) 36 (b) 26.

1.2. GII TCH T HP 7

Bi 1.28. Xt mt li cc im c cho nh hnh bn di. Gi s bt u ti im A,ta c th i mt bc ln trn hoc mt bc ngang sang phi v tip tc nh vy cho nkhi n c im B. Hi c bao nhiu ng i t A n B?

A

B

s s s s ss s s s ss s s s ss s s s s

p n. 35.

Hng dn. Ch rng n B, ta cn i 4 bc ngang sang phi v 3 bc ln trn.

Bi 1.29. Trong bi 1.28, c bao nhiu ng i t A n B qua C nh trong hnh bndi?

A

B

C

s s s s ss s s s ss s s s ss s s s s

p n. 18

Bi 1.30. Cc i biu n t 10 nc trong c Nga, Php, Anh, v M c xp ngivo mt hng gh. C bao nhiu cch xp ch sao cho i biu Anh v Php ngi k nhauv i biu Nga v M khng ngi k nhau?

p n. 564480.

Bi 1.31 (*). 8 mn qu ging nhau c chia cho 4 bn.

(a) C bao nhiu cch chia?

(b) C bao nhiu cch chia nu mi bn nhn t nht mt mn qu?

p n. (a) 165 (b) 35

Bi 1.32 (*). Ta c 20 triu ng u t vo 4 hng mc. Mi khon u t phi l bis ca 1 triu ng v mi hng mc u yu cu mt khon u t ti thiu nu ta u tvo . Cc khong u t ti thiu ny tng ng l 2, 2, 3 v 4 triu ng. C bao nhiucch u t nu ta u t

1.2. GII TCH T HP 8

(a) c 4 hng mc?

(b) t nht 3 trong 4 hng mc?

p n. (a) 220 (b) 572

Bi 1.33. Cc ch ci ca cc t sau y c th c sp xp theo bao nhiu cch?

(a) HANOI

(b) NGHEAN

(c) NHATRANG

p n. (a) 120 (b) 360 (c) 10080

Bi 1.34 (*). Ngi ta c th sp xp cc ch ci ca t

MUHAMMADAN

theo bao nhiu cch sao cho 3 ch ci ging nhau khng gn nhau?

Ch thch. Muhammadan, trong ting Anh, l mt tnh t v c ngha l (thuc/lin quan n) o Hi.

p n. 88080.

Bi 1.35. Cho s nguyn n 2, chng minh rng

(a) 1 + C1n + C2n + + Cnn = 2n

(b) 1 C1n + C2n + + (1)nCnn = 0(c) C1n + 2C

2n + + nCnn = n2n1

(d) C1n 2C2n + 3C3n + + (1)n1nCnn = 0(e) 2.1.C2n + 3.2.C

3n + + n(n 1)Cnn = n(n 1)2n2

Hng dn. S dng cng thc nh thc Newton.

Bi 1.36. Cho m, n l cc s nguyn dng. Chng minh rng

Cnn + Cnn+1 + C

nn+2 + + Cnn+m = Cn+1n+m+1

Hng dn. S dng h thc Ck+1n+1 = Ckn + C

k+1n .

Bi 1.37. Cho m,n, r l cc s nguyn dng. Chng minh rng

(a)mk=0

Crnk = Cr+1n+1 Cr+1nm

(b)mk=0

(1)kCkn = (1)mCmn1

1.2. GII TCH T HP 9

Hng dn. (a) S dng h thc Ck+1n+1 = Ckn + C

k+1n . (b) Quy np.

Bi 1.38. Chng minh rng vi cc s nguyn dng n, k

C0nCkn C1nCk1n1 + C2nCk2n2 + (1)kCknC0nk = 0

Tng qut hn,ki=0

CinCkinit

i = Ckn(1 + t)k

Bi 1.39 (H thc Vandermonde). Gi s m,n, r l cc s nguyn dng. Chng minhrng

C0mCrnm + C

1mC

r1nm + + CrmC0nm = Crn

Ch thch. H thc ny c tm ra bi nh ton hc Alexandre-Thophile Vandermonde vo th k 18.

Hng dn. So snh cc h s tr trong hai v ca h thc (1 + t)m(1 + t)nm = (1 + t)n.

Bi 1.40. Chng minh rng(C0n)2

+(C1n)2

+ + (Cnn )2 = Cn2nHng dn. p dng bi 1.39.

Bi 1.41. Chng minh rng

nk=0

2n!

(k!)2[(n k)!]2 = (Cn2n)

2

Hng dn. p dng bi 1.40.

Bi 1.42 (*). Cho r n l cc s nguyn dng. Chng minh rngnk=0

(1)nkCknkr ={

0 nu r < nn! nu r = n

Hng dn. Xt o hm cp r ca (1 et)n ti t = 0.

Bi 1.43. Chng minh rng

C1n1

1 C2n

1

2+ + (1)n1Cnn

1

n= 1 +

1

2+

1

3+ + 1

n

Hng dn. Tch phn trn [0, 1] h thc

n1k=0

(1 t)k = [1 (1 t)n]t1.

Chng 2

Bin c v xc sut

2.1 Bin c

Bi 2.1. Mt hp bt c 3 cy bt xanh, , tm. Xt php th ly ra mt cy bt t hp,sau tr li hp v rt ra cy bt th hai.

(a) Hy m t khng gian mu.

(b) Trong trng hp cy bt th nht khng c tr li hp, hy m t khng gian mu.

Bi 2.2. Khi no th c cc ng thc sau:

(a) A+B = A

(b) AB = A

(c) A+B = AB

Hai s kin A v A+B c xung khc khng?

p n. (a) A = , B = (b) A = , B = (c) A = B; C.

Bi 2.3. Mt chic tu thy gm mt bnh li, 4 ni hi, 2 tuc bin. 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 hotng tt, tuc bin th j hot ng tt. Bit rng tu hot ng tt khi v ch khi bnh li,t nht 1 ni hi v t nht mt tuc bin u hot ng tt. Gi D l s kin tu hot ngtt. Hy biu din D v D qua A,Bi, Cj .

Bi 2.4. C 4 sinh vin lm bi thi. K hiu Bi(i = 1, . . . , 4) l bin c sinh vin th i lmbi thi t yu cu. Hy biu din cc bin c sau y:

(a) C ng mt sinh vin t yu cu.

(b) C ng ba sinh vin t yu cu.

(c) C t nht mt sinh vin t yu cu.

10

2.1. BIN C 11

(d) Khng c sinh vin no t yu cu.

Bi 2.5. Tung hai con xc sc. Gi E l bin c tng s nt l l, F l bin c xut hinmt mt nt, v G l bin c tng s nt l 5. Hy m t cc bin c sau EF , E F , FG,EF c, v EFG.

p n. EF = {(1, 2), (1, 4), (1, 6), (2, 1), (4, 1), (6, 1)}; FG = {(1, 4), (4, 1)}; EFG = {(1, 4), (4, 1)}.

Hng dn. Trc ht hy vit ra khng gian mu v cc bin c E, F v G.

Bi 2.6. Xt php th: Gieo mt xc xc 2 ln. M t khng gian bin c s cp ng viphp th trn?

Gi A: "Tng s nt chia ht cho 3, B: "Tr tuyt i ca hiu s nt l s chn. Biudin A,B?

Bi 2.7. A, B v C thay phin nhau ln lt tung mt ng xu. Ngi u tin tung cmt nga l ngi thng cuc. Khng gian mu ca th nghim ny c nh ngha nh sau

S = {1, 01, 001, 0001, . . . , 0000 }

(a) Hy gii thch khng gian mu trn.

(b) Hy m t cc bin c sau theo cch biu din ca S:

(i) A = A thng.

(ii) B = B thng.

(iii) (A B)c.

Gi s rng A tung u tin, sau n B, n C, ri quay li A, tip tc nh vy.

Bi 2.8. Mt h thng my c nm b phn. Mi b phn c th hot ng hoc b h.Xt mt php th quan st tnh trng ca cc b phn ny, v kt qu ca php th cghi li trong mt vector (x1, x2, x3, x4, x5), vi xi bng 1 nu b phn i hot ng v bng 0nu b h.

(a) C bao nhiu bin c s cp trong khng gian mu ca th ngim ny?

(b) Gi s rng h thng hot ng nu b phn 1 v 2 u hot ng, hoc nu b phn3 v 4 u hot ng, hoc nu b phn 1, 3 v 5 u hot ng. Gi W l bin c hthng hot ng. Hy biu din W .

(c) Gi A l bin c cc b phn 4 v 5 u b h. A c bao nhiu bin c s cp?

(d) Hy biu din bin c AW .

p n. (d) AW = {(1, 1, 1, 0, 0), (1, 1, 0, 0, 0)}.

Bi 2.9. Xt mt php th bao gm xc nh loi cng vic lao nghoc lao ng tr choc lao ng chn tayv ni sinhmin Bc, min Trung, hoc min Namca 15 thnhvin thuc mt i bng nghip d. Hi c bao nhiu bin c s cp

2.2. XC SUT C IN 12

(a) trong khng gian mu?

(b) trong bin c c t nht mt trong cc thnh vin l lao ng tr c?

p n. (a) 615 (b) 615 315.

Bi 2.10. Cho A,B l hai bin c ngu nhin bit. Tm bin c X t h thc:

X +A+X +A = B

p n. X = B.

Bi 2.11. Cho A, B l cc tp con ca . Tm iu kin cn v tn ti mt tp conX ca tha AX +BX = .p n. B A

Bi 2.12. Xt php th: Bn khng hn ch vo 1 bia cho n khi trng bia ln u tinth dng. Biu din khng gian bin c s cp ca bin c trn. Ch ra mt h y ccbin c.

Hng dn. C nhiu h y cc bin c cho khng gian mu ny. Hy tm mt h n gin nht.

Bi 2.13. Gieo hai con xc xc cn i v ng cht. Gi Ai l bin c xy ra khi s nt mt trn con xc xc th nht l i(i = 1, . . . , 6), Bk bin c xy ra khi s nt mt trncon xc xc th hai l k(k = 1, . . . , 6).

(a) Hy m t cc bin c A6B6, A3B5

(b) Vit bng k hiu cc bin c:

A: hiu gia s nt mt trn con xc xc th nht v th hai c tr s tuyt ibng ba.

B: s nt mt trn hai con xc xc bng nhau.(c) Hy ch ra mt nhm y cc bin c.

2.2 Xc sut c in

Bi 2.14. Mt nhm n ngi xp ngu nhin thnh mt hng di.

(a) Tm xc sut 2 ngi nh trc ng cnh nhau.

(b) Tm xc sut 2 ngi ng cch nhau 2 ngi.

(c) Tm xc sut 2 ngi ng cch nhau r ngi (0 < r < n 2).(d) Xt trng hp khi h xp thnh mt vng trn.

p n. (a) 2n(b) 2(n3)

(n1)n (c)2(nr1)(n1)n (d) Nu r =

n22

th P = 1n1 . Nu r 6= n22 th P = 2n1 .

2.2. XC SUT C IN 13

Bi 2.15. Thang my ca mt ta nh 7 tng, xut pht t tng mt vi 3 ngi khch.Tnh xc sut :

(a) Tt c cng ra tng bn.

(b) Tt c cng ra mt tng.

(c) Mi ngi ra mt tng khc nhau.

p n. (a)1

63(b)

6

63(c)

6 5 463

Bi 2.16. C n qu cu c phn ngu nhin ln lt vo n hp, mi hp c th chanhiu qu cu. Khi phn bit hp v cu, tm xc sut mi hp cha mt qu cu.

p n. n!nn

Bi 2.17. Cho mt l hng gm n sn phm trong c m sn phm xu. Ly ngu nhint l hng k sn phm. Tm xc sut sao cho trong s sn phm ly ra c ng s snphm xu (s < k).

p n.CsmC

ksnm

Ckn

Bi 2.18. Ta gieo lin tip 4 ln mt ng tin cn i ng cht. Tm xc sut ca ccbin c:

(a) A: C hai mt sp.

(b) B: C ba mt nga.

(c) C: C t nht mt mt sp.

p n. (a) 0.375 (b) 0.25 (c) 0.9375

Bi 2.19. Mi hai sn phm c sp ngu nhin vo ba hp. Tm xc sut hp thnht c cha ba sn phm.

p n. 0.212

Bi 2.20 (*). Gieo ng thi hai con xc xc ng cht cn i n ln lin tip. Tm xcsut xut hin t nht mt ln hai mt trn cng c 6 nt.

p n. 1 ( 3536

)n.

2.3. XC SUT HNH HC 14

2.3 Xc sut hnh hc

Bi 2.21. Mt thanh st thng c b thnh ba khc mt cch ngu nhin. Tm xc sut ba khc to c thnh mt tam gic. Bit rng thanh st di l (n v di.)

p n. 0.25

Bi 2.22 (* Bi ton Butffon). Trn mt phng c cc ng thng song song cch unhau 2a, gieo ngu nhin mt cy kim c di 2l (l < a). Tm xc sut cy kim ct mtng thng no .

p n. 2lapi

.

Bi 2.23. Trn ng trn bn knh R c mt im A c nh, chn ngu nhin mt imB. Tm xc sut cung AB khng qu R.

p n. 13.

Bi 2.24. Trn on thng OA ta gieo mt cch ngu nhin hai im B,C c ta tngng l OB = x,OC = y(y x). Tm xc sut sao cho di ca on BC b hn dica on OB.

p n. 0.25

2.4 Cc cng thc tnh xc sut c bn

Bi 2.25. Mt h thng c cu to bi 3 b phn c lp nhau. H thng s hot ngnu t nht 2 trong 3 b phn cn hot ng. Nu tin cy ca mi b phn l 0.95 th tin cy ca h thng l bao nhiu?

p n. 0.9928

Hng dn. Gi

Bi l bin c B phn th i hot ng tt (i = 1, 2, 3) H l bin c H thng hot ng tt

Biu din H theo Bi v tnh P (H).

Bi 2.26. Mt hp c 7 bi v 3 bi en.

(a) Ly ngu nhin 1 vin bi t hp ra kim tra. Tnh xc sut nhn c bi en.

(b) Ly ngu nhin ln lt c hon li 2 bi. Tnh xc sut ly c 2 bi en.

(c) Ly ngu nhin ra 2 vin bi t hp. Tnh xc sut ly c 2 bi en.

p n. (a) 0.3 (b) 0.09 (c) 0.067

2.4. CC CNG THC TNH XC SUT C BN 15

Bi 2.27. Cho P (A) = 13 , P (B) =12 v P (A+B) =

34 .

Tnh P (AB), P (A.B), P (A+B), P (AB), P (AB).

p n. 112, 14, 1112, 14, 512.

Bi 2.28. T l ngi b bnh tim trong mt vng dn c l 9%, b bnh huyt p l 12%,b c hai bnh l 7%. Chn ngu nhin mt ngi trong vng. Tnh xc sut ngi

(a) B bnh tim hay b bnh huyt p.

(b) Khng b bnh tim cng khng b bnh huyt p.

(c) Khng b bnh tim hay khng b bnh huyt p.

(d) B bnh tim nhng khng b bnh huyt p.

(e) Khng b bnh tim nhng b bnh huyt p.

p n. (a) 0.14 (b) 0.86 (c) 0.93 (d) 0.02 (e) 0.05

Hng dn. Gi

A l bin c nhn c ngi b bnh tim B l bin c nhn c ngi b bnh huyt p

Ta c: P (A) = 0.09;P (B) = 0.12;P (AB) = 0.07

Biu din cc bin c trong tng cu theo A v B v tnh xc sut cc bin c .

Bi 2.29. Bn qun mt s cui cng trong s in thoi cn gi (s in thoi gm 6 chs) v bn chn s cui cng ny mt cch ngu nhin. Tnh xc sut bn gi ng sin thoi ny m khng phi th qu 3 ln. Nu bit s cui cng l s l th xc sut nyl bao nhiu ?

p n. 0.3; 0.6

Hng dn. Gi Ai l bin c gi ng ln th i (i = 1, 2, 3)

Biu din cc bin c cn tm theo Ai v p dng cc cng thc tnh xc sut tm xc sut ca cc

bin c ny.

Bi 2.30 (*). (a) Cho A,B l hai bin c c lp. Chng minh rng A,B; A,B v A,Bu l cc cp bin c c lp.

(b) Cho A1, A2, . . . , An l n bin c c lp. Chng minh rng A1, A2, . . . , An cng l n binc c lp. T suy ra rng nu xt n bin c B1, B2, . . . , Bn vi Bi = Ai hoc Bi = Aith B1, B2, . . . , Bn cng l n bin c c lp.

Bi 2.31. Mt t x s pht hnh N v, trong c M v c thng. Mt ngi mua rv (r < N M). Tnh xc sut ngi c t nht mt v trng thng.p n. 1 C

rNMCrN

.

2.4. CC CNG THC TNH XC SUT C BN 16

Bi 2.32. Mt ngi c 3 con g mi, 2 con g trng nht chung mt lng. Mt ngi nmua, ngi bn bt ngu nhin ra mt con. Ngi mua chp nhn mua con .

(a) Tm xc sut ngi mua c con g mi.Ngi th hai n mua, ngi bn li bt ngu nhin ra mt con.

(b) Tm xc sut ngi th hai mua c g trng, bit rng ngi th nht mua c gmi.

(c) Xc sut trn bng bao nhiu nu ngi bn g qun mt rng con g bn cho ngi thnht l g trng hay g mi?

p n. (a) 0.6 (b) 0.5 (c) 0.4

Bi 2.33 (*). C mt nhm n sinh vin, mi ngi c mt o ma ging ht nhau. Mthm tri ma, c nhm cng n lp v treo o mc o. Lc ra v v vi vng mi ngily h ha mt ci o. Tnh xc sut c t nht mt sinh vin chn ng o ca mnh.

p n. 1 12!

+ 13 + (1)n1 1

n!.

Hng dn. Gi

Ai l bin c Sinh vin th i nhn ng o ca mnh (i = 1, . . . , n) A l bin c C t nht mt sinh vin nhn ng o ca mnh

Biu din A theo Ai v p dng cng thc cng xc sut.

Bi 2.34 (*). Mt ngi vit n l th v b n l th ny vo trong n phong b vit sna ch. Tm xc sut sao cho c t nht mt l th c b ng vo phong b ca n.

Hng dn. Tng t bi 2.33.

Bi 2.35. Ba x th, mi ngi bn mt vin n vo mc tiu vi xc sut trng ch cami ngi l 0.6; 0.7; 0.8. Tm xc sut

(a) ch c ngi th hai bn trng.

(b) c ng mt ngi bn trng.

(c) c t nht mt ngi bn trng.

(d) c ba ngi u bn trng.

(e) c ng hai ngi bn trng.

(f) c t nht hai ngi bn trng.

(g) c khng qu hai ngi bn trng.

p n. (a) 0.056 (b) 0.188 (c) 0.976 (d) 0.336 (e) 0.452 (f) 0.788 (g) 0.664

Hng dn. Gi Ai l bin c X th th i bn trng (i = 1, 2, 3)

Biu din cc bin c cn tm theo Ai v p dng cc cng thc tnh xc sut.

2.4. CC CNG THC TNH XC SUT C BN 17

Bi 2.36. Cho hai bin c xung khc A v B, sao cho P (A) 6= 0, P (B) 6= 0.Chng minh rng A v B ph thuc nhau.

Hng dn. Dng nh ngha.

Bi 2.37. Ba con nga a, b, c trong mt cuc ua nga. Nu xut hin bac c ngha l b nch trc, sau l a v v cui l c. Khi tp hp tt c cc kh nng xut hin l

= {abc, acb, bac, bca, cab, cba}.

Gi s rng P [{abc}] = P [{acb}] = 1/18 v bn kh nng cn li u c xc sut xy ra l2/9. Hn na, ta nh ngha cc bin c

A = "a n ch trc b" v B = "a n ch trc c"

(a) Hai bin c A v B c to thnh mt h y ca ?

(b) Hai bin c A v B c c lp nhau?

p n. (a) khng (b) c.

Bi 2.38. C tn ti hai bin c xung khc v c lp khng?

Hng dn. Hy vit ra cc nh ngha hai bin c xung khc v hai bin c c lp nhau.

Bi 2.39. Mt my tnh in t gm c n b phn. Xc sut hng trong khong thi gianT ca b phn th k bng pk(k = 1, 2, . . . , n). Nu d ch mt b phn b hng th my tnhngng lm vic. Tm xc sut my tnh ngng lm vic trong khong thi gian T .

p n. 1 (1 p1)(1 p2) (1 pn).

Bi 2.40. Chng minh rng nu

P (A|B) > P (A), th P (B|A) > P (B)

Bi 2.41. Gi s P (AB) = 1/4, P (A|B) = 1/8 v P (B) = 1/2. Tnh P (A).p n. 5/16.

Bi 2.42. Bit rng ta nhn c t nht mt mt nga trong 3 ln tung ng xu clp. Hi xc sut t c c 3 mt nga l bao nhiu?

p n. 1/7

Hng dn. p dng cng thc xc sut c iu kin.

Bi 2.43. Tung mt con xc sc hai ln c lp nhau. Bit rng ln tung th nht c snt chn. Tnh xc sut tng s nt hai ln tung bng 4.

p n. 1/18

2.4. CC CNG THC TNH XC SUT C BN 18

Bi 2.44. Gi s P (A) = P (B) = 1/4 v P (A|B) = P (B). Tnh P (AB).p n. 3/16

Bi 2.45. Bn lin tip vo mt mc tiu n khi trng mc tiu th ngng. Tm xc sutsao cho phi bn n vin th 6, bit rng xc sut trng ch ca mi vin n l 0.2 vcc ln bn l c lp.

p n. 0.0655

Hng dn. Gi

Ai l bin c Bn trng ln th i A l bin c Phi bn n vin th 6

Biu din A theo Ai v p dng cc cng thc tnh xc sut.

Bi 2.46. Gi s cc bin c A1, . . . , An c lp c xc sut tng ng P (Ak) = pk(k =1, . . . , n). Tm xc sut sao cho:

(a) khng mt bin c no trong cc bin c xut hin.

(b) c t nht mt bin c trong cc bin c xut hin.

T suy ra cng thc khai trin tchnk=1

(1 pk)

p n. (a)nk=1(1 pk) (b) 1

nk=1(1 pk).

Bi 2.47. C ba tiu ch ph bin cho vic chn mua mt chic xe hi mi no l A: hps t ng, B: ng c V6, v C: iu ha nhit . Da trn d liu bn hng trc y, tac th gi s rng P (A) = 0.7, P (B) = 0.75, P (C) = 0.80, P (A+B) = 0.80, P (A+C) = 0.85,P (B +C) = 0.90 v P (A+B +C) = 0.95, vi P (A) l xc sut ngi mua bt k chn tiuch A, v.v. . . . Tnh xc sut ca cc bin c sau:

(a) ngi mua chn t nht mt trong 3 tiu ch.

(b) ngi mua khng chn tiu ch no trong 3 tiu ch trn.

(c) ngi mua ch chn tiu ch iu ha nhit .

(d) ngi mua chn chnh xc mt trong 3 tiu ch.

p n. (a) 0.95 (b) 0.05 (c) 0.15 (d) 0.3

Bi 2.48. Gi s A, B l hai bin c bt k. Ta nh ngha khong cch d(A,B) gia A vB nh sau:

d(A,B) = P (A4B)Chng minh rng nu A, B, C l cc bin c th

d(A,C) d(A,B) + d(B,C)y l bt ng thc tam gic cho hm khong cch d.

2.5. CNG THC XC SUT Y , CNG THC BAYES 19

Hng dn. S dng bi 1.11(h).

Bi 2.49 (*). Gi s A, B l hai bin c bt k. Ta nh ngha khong cch d(A,B) gia Av B nh sau

d(A,B) =

{P (A4B)P (AB) nu P (A B) 6= 0

0 nu P (A B) = 0Chng minh rng nu A, B, C l cc bin c th

d(A,C) d(A,B) + d(B,C)

y l bt ng thc tam gic cho hm khong cch d.

2.5 Cng thc xc sut y , cng thc Bayes

Bi 2.50. Gi s P (B|A1) = 1/2, P (B|A2) = 1/4 vi A1 v A2 l hai bin c ng khnng v to thnh mt h y cc bin c. Tnh P (A1|B).p n. 2/3

Bi 2.51 (*). Mt hp ng 10 phiu trong c 2 phiu trng thng. C 10 ngi lnlt rt thm (khng hon li). Tnh xc sut nhn c phiu trng thng ca mi ngi.

p n. 0.2

Bi 2.52. C hai hp ng bi. Hp 1 ng 20 bi trong c 5 bi v 15 bi trng. Hp 2ng 15 bi trong c 6 bi v 9 bi trng. Ly mt bi hp 1 b vo hp 2, trn u rily ra mt bi. Tnh xc sut nhn c bi ? bi trng?

p n. 0.391; 0.609

Hng dn. Gi

A l bin c Bi nhn c t hp 2 l bi B l bin c Bi t hp 1 b sang hp 2 l bi

Bi 2.53. Trong mt vng dn c, c 100 ngi th c 30 ngi ht thuc l. Bit t lngi b vim hng trong s ngi ht thuc l l 60%, trong s ngi khng ht thuc l l30%. Khm ngu nhin mt ngi v thy ngi ny b vim hng.

(a) Tm xc sut ngi ny ht thuc l.

(b) Nu ngi ny khng b vim hng th xc sut ngi ny ht thuc l l bao nhiu.

p n. (a) 0.462 (b) 0.197

Hng dn. Gi

A l bin c Ngi ny ht thuc B l bin c Ngi ny b vim hng

2.5. CNG THC XC SUT Y , CNG THC BAYES 20

Bi 2.54. Mt chic taxi gy tai nn v b chy khi hin trng vo na m. Trong thnhph c hai hng taxi, gi l taxi en v taxi Trng. Ta bit rng 85% taxi trong thnh phl en v 15% l Trng. C mt nhn chng lc tai nn xy ra, theo nhn chng, taxi gyra tai nn l Trng. Mt kho st v tin cy ca nhn chng ch ra rng, di cc iukin tng t v thi gian, a im, nh sng,. . . nh lc xy ra tai nn, nhn chng c khnng xc nh chnh xc mu sc ca mt taxi trong 80% trng hp.

(a) Khng lm php ton, bn ngh rng taxi en hay Trng c kh nng gy ra tai nn lnhn?

(b) Tnh xc sut taxi gy tai nn l Trng.

(c) So snh p n hai cu hi trn.

(d) Hy kho st nhy cm ca p n trong cu (b) vi cc thng tin sau. Gi s rng0 p 1 v 100p% taxi l Trng v 100(1p)% taxi l en. tin cy ca nhn chngvn l 80%. Chng minh rng xc sut taxi Trng gy tai nn ln hn 0.5 nu v chnu p > 0.2. Bit rng nhn chng ni rng taxi gy tai nn l Trng.

(e) Hy kho st nhy cm ca p n trong cu (b) vi cc thng tin sau. Gi s rng0 p 1 v 100p% taxi l Trng v 100(1 p)% taxi l en. Gi s rng 0 q 1 v tin cy ca nhn chng l 100q%, tc l nhn chng c th xc nh chnh xc muca mt taxi trong 100q% trng hp. Xc nh min bn trong hnh vung

{(p, q) : 0 p 1, 0 q 1}

m xc sut taxi Trng gy tai nn ln hn 0.5. Bit rng nhn chng ni rng taxi gytai nn l Trng.

Ch thch. Bi ton ny minh ha rng phn on trc gic c th sai v mt s thng tin quan trng

khng c xt ti v b cho l khng cn thit hoc khng lin quan.

p n. (b) 0.41

Bi 2.55. Gi s c mt loi bnh m t l ngi mc bnh l 1/1000. Gi s c mt loixt nghim, m ai mc bnh khi xt cng ra phn ng dng tnh, nhng t l phn ngdng tnh nhm (false positive) l 5% (tc l trong s nhng ngi khng mc bnh c 5%s ngi th ra phn ng dng tnh). Hi khi mt ngi xt nghim b phn ng dngtnh, th kh nng mc bnh ca ngi l bao nhiu?

Ch thch. y l mt bi ton c 3 nh ton hc Cassels, Shoenberger v Grayboys em 60 sinh vin

v cn b y khoa ti Harvard Medical School nm 1978. Kt qu l ch c 18% ngi tr li ng.

p n. 2%. (iu c ngha l trong s nhng ngi xt nghim ra dng tnh, c khong 98% s ngil khng mc bnh!)

Bi 2.56 (*). Mt trung tm chn on bnh dng mt php kim nh K. Xc sut mt ngi n trung tm m c bnh l 0.8. Xc sut ngi khm c bnh khi php kimnh dng tnh l 0.9 v xc sut ngi khm khng c bnh khi php kim nh mtnh l 0.5. Tnh cc xc sut

(a) php kim nh l dng tnh.

2.5. CNG THC XC SUT Y , CNG THC BAYES 21

(b) php kim nh cho kt qu ng.

p n. (a) 0.75 (b) 0.8

Hng dn. Gi

A l bin c Ngi khm c bnh B l bin c Php kim nh dng tnh

Bi 2.57 (*). Mt cp tr sinh i c th do cng mt trng (sinh i tht) hay do haitrng khc nhau sinh ra (sinh i gi). Cc cp sinh i tht lun lun c cng gii tnh.Cc cp sinh i gi th gii tnh ca mi a c lp vi nhau v c xc sut l 0.5. Thngk cho thy 34% cp sinh i l trai; 30% cp sinh i l gi v 36% cp sinh i c gii tnhkhc nhau.

(a) Tnh t l cp sinh i tht.

(b) Tm t l cp sinh i tht trong s cc cp sinh i c cng gii tnh.

p n. (a) 0.28 (b) 0.4375

Hng dn. Gi

A l bin c Nhn c cp sinh i tht B l bin c Nhn c cp sinh i c cng gii tnh

Bi 2.58. C 10 hp bi, trong c 4 hp loi I, 3 hp loi II, cn li l hp loi III. Hploi I c 3 bi trng v 5 , hp loi II c 4 bi trng v 6 bi , hp loi III c 2 bi trng v2 bi .

(a) Chn ngu nhin mt hp v t ly h ha 1 bi. Tm xc sut c bi .

(b) Chn ngu nhin mt hp v t ly 1 bi th c bi trng. Tm xc sut bi ly rathuc loi II.

p n. (a) 0.58 (b) 0.286

Hng dn. Gi

Ai l bin c Ly c hp th i (i = 1, 2, 3) B l bin c Ly c bi

Bi 2.59. C hai l sn phm, l th nht c 10 sn phm loi I v 2 sn phm loi II. Lth hai c 16 sn phm loi I v 4 sn phm loi II. T mi l ta ly ngu nhin mt snphm. Sau , t 2 sn phm thu c ly h ha ra mt sn phm. Tm xc sut snphm ly ra sau cng l sn phm loi I.

p n. 0.79

Hng dn. Gi

2.5. CNG THC XC SUT Y , CNG THC BAYES 22

Ai l bin c Sn phm ly ra ln u l th i l loi I (i = 1, 2) B l bin c Sn phm ly ra ln sau l loi I

Bi 2.60. C 2 l g. L th nht gm 15 con, trong c 3 con g trng. L th hai gm20 con, trong c 4 g trng. Mt con t l th hai nhy sang l th nht. Sau t lth nht ta bt ngu nhin ra mt con. Tm xc sut con g bt ra l g trng.

p n. 0.2

Bi 2.61. Ba my t ng sn xut cng mt loi chi tit, trong my I sn xut 25%,my II sn xut 30% v my III sn xut 45% tng sn lng. T l ph phm ca cc myln lt l 0.1%; 0.2%; 0.4%. Tm xc sut khi chn ngu nhin ra 1 sn phm t kho th

(a) c chi tit ph phm.

(b) chi tit ph phm do my II sn xut.

p n. (a) 0.00265 (b) 0.226

Bi 2.62. Gi s 3 my M1, M2, M3 sn 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 lymt linh kin c sn xut bi mt trong 3 my trn mt cch ngu nhin, kt qu l cmt ph phm. Tm xc sut linh kin ny c sn xut bi my M3.

p n. 0.5526

Bi 2.63 (*). Ba khu pho cng bn vo mt mc tiu vi xc sut trng ch ca mikhu l 0.4; 0.7; 0.8. Bit rng xc sut mc tiu b tiu dit khi trng mt pht n l30%, khi trng 2 pht n l 70%, cn trng 3 pht n th chc chn mc tiu b tiu dit.Gi s mi khu pho bn 1 pht.

(a) Tnh xc sut mc tiu b tiu dit.

(b) Bit rng mc tiu b tiu dit. Tnh xc sut khu th 3 c ng gp vo thnhcng .

p n. (a) 0.6412 (b) 0.8883

Hng dn. Gi

Ai l bin c Khu pho th i bn trng (i = 1, 2, 3) Bk l bin c Mc tiu trng k pht n (k = 0, 1, 2, 3) B l bin c Mc tiu b tiu dit.

Bi 2.64 (*). Hp I c 10 linh kin trong c 3 b hng. Hp II c 15 linh kin trong c 4 b hng. Ly ngu nhin t mi hp ra mt linh kin.

(a) Tnh xc sut c 2 linh kin ly ra u hng.

2.5. CNG THC XC SUT Y , CNG THC BAYES 23

(b) S linh kin cn li trong 2 hp em b vo hp III. T hp III ly ngu nhin ra 1 linhkin. Tnh xc sut linh kin ly ra t hp III b hng.

(c) Bit linh kin ly ra t hp III l hng. Tnh xc sut 2 linh kin ly ra t hp I vII lc ban u l hng.

p n. (a) 0.08 (b) 0.2797 (c) 0.0622

Bi 2.65. C 3 ca hng I, II, III cng kinh doanh sn phm Y , trong th phn ca cahng I, III nh nhau v gp i th phn ca ca hng II. T l sn phm loi A trong 3 cahng ln lt l 70%, 75% v 50%. Mt khch hng chn ngu nhin 1 ca hng v t mua mt sn phm.

(a) Tnh xc sut khch hng mua c sn phm loi A.

(b) Gi s khch hng mua c sn phm loi A, hi kh nng ngi y mua c ca hng no l nhiu nht.

p n. (a) 0.63 (b) ca hng I.

Bi 2.66 (*). Cho mt php th ngu nhin vi 3 bin c s cp c th xy ra l A, B vC. Gi s ta tin hnh php th v hn ln v c lp nhau. Tnh theo P (A), P (B) xc sutbin c A xut hin trc bin c B.

p n. P (A)P (A)+P (B)

Chng 3

Bin ngu nhin v hm phn phi

Bi 3.1. Cho bin ngu nhin ri rc X c bng phn phi xc sut sau:

X 2 1 0 1 2P 1/8 2/8 2/8 2/8 1/8

(a) Tm hm phn phi xc sut F (x).

(b) Tnh P (1 X 1) v P (X 1 hoc X = 2).(c) Lp bng phn phi xc sut ca bin ngu nhin Y = X2.

p n. (b) 6/8, 4/8. (c)Y 0 1 4

P 2/8 4/8 2/8

Bi 3.2. Bin ngu nhin ri rc X c hm xc sut cho bi

f(x) =2x+ 1

25, x = 0, 1, 2, 3, 4

(a) Lp bng phn phi xc sut ca X.

(b) Tnh P (2 X < 4) v P (X > 10).p n. (b) 12/25, 1.

Bi 3.3. Gi X l bin ngu nhin ri rc c bng phn phi xc sut sau

X 1 0 3P 0.5 0.2 0.3

(a) Tnh lch chun ca X.

(b) Tnh k vng ca X3.

(c) Tm hm phn phi ca X.

(d) Ta nh ngha Y = X2 +X + 1. Lp bng phn phi xc sut ca Y .

24

25

p n. (a) 1.7436 (b) 7.6 (d)Y 1 13

P 0.7 0.3

Bi 3.4. Hm mt ca bin ngu nhin X l

fX(x) =

x nu 1 x 0x nu 0 < x 10 nu khc

Tnh FX(1/2).

p n. 3/8

Bi 3.5. Bin ngu nhin X c hm mt f(x) nh sau

f(x) =

{kx(2 x) khi 1 < x < 20 ni khc

(a) Xc nh gi tr ca k f(x) l hm mt ca bin ngu nhin X. Vi k va tmc tnh k vng v phng sai ca bin ngu nhin X.

(b) Tm hm phn phi F (x) ca bin ngu nhin X.

(c) Tm hm phn phi G(y) ca bin ngu nhin Y = X3.

p n. (a) k = 3/2, EX = 1.375, DX = 0.0594

(b) F (x) =

0 x 112x3 + 3

2x2 1 1 < x < 2

1 x 2

(c) G(x) =

0 y 112y + 3

2y2/3 1 1 < y < 8

1 y 8

Bi 3.6. Bin ngu nhin lin tc X c hm mt

f(x) =

{ex khi x > 00 khi x 0

(a) Tnh P (3 X).(b) Tm gi tr ca a sao cho P (X a) = 0, 1.(c) Xc nh hm phn phi v mt xc sut ca bin ngu nhin Y =

X.

p n. (a) e3 (b) 0.105 (c) FY (y) =

{0 y 01 ey2 y > 0 , fY (y) =

{0 y 02yey

2

y > 0

Bi 3.7. Tnh P (X 8) nu

fX(x) =

{196x

3ex/2 nu x 00 nu khc

26

p n. 0.4335

Bi 3.8. Cho

fX(x) =

2

pi x2 vi

2

pi x

2

pi

Tnh P (X < 0).

p n. 0.5

Hng dn. Bi ny khng cn phi tnh tch phn.

Bi 3.9. Bin ngu nhin X c hm mt

f(x) =

{a exp

(x2) khi x 00 ni khc

Xc nh:

(a) Hng s a.

(b) Hm phn phi xc sut F (x)

(c) K vng v phng sai ca bin ngu nhin X.

(d) K vng v phng sai ca bin ngu nhin Y = (X/2) 1.

p n. (a) 1/2 (b) F (x) =

{1 ex2 x 00 ni khc

(c) 1/2, 7/4 (d) 3/4, 7/16

Bi 3.10. Cho X l bin ngu nhin c hm mt sau

fX(x) =

{c(1 x2) nu 1 x 1

0 nu |x| > 1

vi c l mt hng s dng. Tm

(a) hng s c

(b) trung bnh ca X

(c) phng sai ca X

(d) hm phn phi FX(x).

p n. (a) 3/4 (b) 0 (c) 1/5 (d) FX(x) =

0 nu x < 1

14x3 + 3

4x+ 1

2nu 1 x 1

1 nu x > 1

Bi 3.11. Cho

fX(x) =

{1/e nu 0 < x < e0 nu khc

Tnh fY (y) vi Y = 2 lnX.

27

Bi 3.12 (*). Bin ngu nhin lin tc X c hm mt

f(x) =

1

2x khi 0 < x < 2

0 ni khc

Tm hm phn phi v hm mt xc sut ca cc bin ngu nhin sau:

(a) Y = X(2X).

(b) Z = 4X3.

(c) T = 3X + 2.

p n. (a) FY (y) =

0 y 011 y 0 < y < 11 y 1

, fY (y) = FY (y) =

{1

21y 0 < y < 1

0 khc

(b) FZ(z) =

0 z 41 1

4( 3

4 z)2 4 < z < 41 z 4

, fZ(z) = FZ(z) =

{1

6 34z 4 < z < 4

0 khc

(c) FT (t) =

0 t 214

(t23

)22 < t < 8

1 t 8, fT (t) = F

T (t) =

{118

(t 2) 2 < t < 80 khc

Bi 3.13. Tnh phng sai caX nu

pX(x) =

1/4 nu x = 01/2 nu x = 11/4 nu x = 4

p n. 1/2.

Bi 3.14. Tnh V ar(eX) nu X l bin ngu nhin c hm xc sut c cho nh sau

pX(x) =

1/4 nu x = 01/4 nu x = 11/2 nu x = 40 nu khc

p n. 695.7198

Hng dn. t Y = eX . Tnh EY , E(Y 2).

Bi 3.15. Cho

fX(x) =

{4x2e2x nu x 0

0 nu x < 0

Tnh phng sai ca X.

p n. 3/4

28

Bi 3.16. Tnh phn v mc 25% (tc l gi tr x0.25 sao cho P (X < x0.25) = 0.25) ca binngu nhin lin tc X c hm mt sau:

fX(x) =

{xex2/2 nu x 0

0 nu x < 0

p n. 0.4999

Hng dn. Xt 2 trng hp x0.25 < 0 v x0.25 0.

Bi 3.17. Ta nh ngha Y = |X|, vi X l mt bin ngu nhin lin tc c hm mt

fX(x) =

3/4 nu 1 x 01/4 nu 1 x 20 nu khc

Tm phn v mc 95% ca Y .

p n. 1.8

Bi 3.18 (*). Cho

FX(x) =

0 nu x < 0x/2 nu 0 x 1

x/6 + 1/3 nu 1 < x < 41 nu x 4

l hm phn phi ca bin ngu nhin lin tc X.

(a) Tnh hm mt ca X.

(b) Tm phn v mc 75% ca X (tc l tm x0.75 sao cho P (X < x0.75) = 0.75).

(c) Tnh k vng ca X.

(d) Tnh E(1/X).

(e) Ta nh ngha

Y =

{1 nu X 11 nu X > 1

(i) Tm FY (0).

(ii) Tnh phng sai ca Y .

p n. (a) fX(x) =

1/2 nu 0 < x < 1

1/6 nu 1 < x < 4

0 nu x < 0 hoc x > 4

(b) 2.5 (c) 1.5 (d) (e)-(i) 0.5 -(ii) 1

Bi 3.19 (*). Cho hm mt ca bin ngu nhin X nh sau:

fX(x) =

{6x(1 x) nu 0 x 1

0 nu khc

29

(a) Tnh k vng ca 1/X.

(b) Tm hm phn phi ca X.

(c) Ta nh ngha

Y =

{2 nu X 1/40 nu X < 1/4

Tnh E(Y k) vi k l mt s t nhin.

(d) t Z = X2. Tm hm mt ca Z.

p n. (a) 3 (c) (27/32)2k

Bi 3.20. Bin ngu nhin lin tc X c hm mt xc sut

f(x) =

3

4x(2 x) khi 0 x 2

0 ni khc

(a) Xc nh hm phn phi xc sut F (x) ca bin ngu nhin X.

(b) Tnh E(X), Var (X) v trung v ca bin ngu nhin X.

(c) t Y =X, xc nh hm phn phi v hm mt xc sut ca bin ngu nhin Y .

p n. (a) F (x) =

0 x < 034x2 1

4x3 0 x 2

1 x > 2

(b) 1; 2; 1 (c) FY (y) =

0 y < 034y4 1

4y6 0 y 2

1 y >

2

,

fY (y) = FY (y) =

{3y3 3

2y5 0 y 2

0 khc

Bi 3.21. Tui th ca mt loi cn trng no l mt bin ngu nhin lin tc X (nv thng) c hm mt

f(x) =

{kx2(4 x) khi 0 x 40 ni khc

(a) Tm hng s k.

(b) Tm F (x).

(c) Tm E (X), Var (X) v Mod(X).

(d) Tnh xc sut cn trng cht trc mt thng tui.

p n. (a) 3/64 (b) F (x) =

0 x < 0116x3 3

256x4 0 x 4

1 x > 4

(c) 2.4; 0.64; 8/3 (d) 0.0508

Bi 3.22. Bin ngu nhin lin tc X c hm mt

f(x) =

{kx2e2x khi x 00 ni khc

30

(a) Tm hng s k.

(b) Tm hm phn phi xc sut F (x).

(c) Tm E (X), Var (X) v Mod(X).

p n. (a) 4 (c) 3/2; 3/4; 1.

Bi 3.23 (*). Mt bin ngu nhin X c hm mt sau:

fX(x) =

{xkex2/2k nu x > 00 nu khc

vi k l mt hng s.

(a) Tnh trung bnh v phng sai ca X.

(b) nh hng ca hng s k ln hnh dng ca hm fX?

p n. (pik/2)1/2; 2k(1 pi/4)

Bi 3.24. Trong mt hp c 20 vin , 10 vin loi basalt v 10 vin loi granite. Nmvin c rt ra ngu nhin v khng hon li thc hin cc phn tch ha hc. Gi X ls vin loi basalt trong mu.

(a) Tm phn phi xc sut ca X.

(b) Tnh xc sut mu ch cha cc vin cng loi.

p n. (b) 0.0326

Bi 3.25. C hai thng thuc A v B, trong :

- thng A c 20 l gm 2 l hng v 18 l tt

- thng B c 20 l gm 3 l hng v 17 l tt.

(a) Ly mi thng 1 l. Gi X l s l hng trong hai l ly ra. Tm hm mt ca X.

(b) Ly thng B ra 3 l. Gi Y l s l hng trong 3 l ly ra. Tm hm mt ca Y .

p n. (a) f(x) =

0.765 khi x = 0

0.22 khi x = 1

0.015 khi x = 2

0 khi x 6= 0, 1, 2(b) f(y) =

0.596 khi y = 0

0.358 khi y = 1

0.045 khi y = 2

0.001 khi y = 3

0 khi y 6= 0, 1, 2, 3

Bi 3.26. Mt thng ng 10 l thuc trong c 1 l hng. Ta kim tra tng l (khnghon li) cho ti khi pht hin c l hng th dng. Gi X l s ln kim tra. Tm hmmt ca X. Tnh k vng v phng sai.

p n. EX = 5.5, V ar(X) = 8.25

31

Bi 3.27. Mt bin ngu nhin lin tc c hm mt xc sut sau:

fX(x) =

{cxex/2 nu x 00 nu x < 0

(a) Tm hng s c.

(b) Tm hm phn phi xc sut FX(x).

(c) Tm trung bnh ca X

(d) Tm lch chun ca X.

(e) Tm Med(X).

p n. (a) 1/4 (c) 4 (d)

2 (e) 1.5361

Bi 3.28. Gi X l tui th ca con ngi. Mt cng trnh nghin cu cho bit hm mt ca X l

f(x) =

{cx2(100 x)2 khi 0 x 1000 khi x < 0 hay x > 100

(a) Xc nh hng s c.

(b) Tnh k vng v phng sai ca X.

(c) Tnh xc sut ca mt ngi c tui th 60

(d) Tnh xc sut ca mt ngi c tui th 60, bit rng ngi hin nay hn 50tui.

p n. (a) 3/109 (b) 50; 357.143 (c) 0.317 (d) 0.643

Bi 3.29. Mt thit b gm 3 b phn hot ng c lp vi nhau, xc sut trong khongthi gian t cc b phn hng tng ng bng 0.2; 0.3; 0.25. Gi X l s b phn b hngtrong khong thi gian t.

(a) Lp bng phn phi xc sut ca X.

(b) Vit biu thc hm phn phi ca X.

(c) Tnh P (0 < X 4) theo hai cch.

p n. (a)X 0 1 2 3

P 0.42 0.425 0.14 0.015(c) 0.58

Bi 3.30. Mt mu 4 sn phm c rt ra khng hon li t 10 sn phm. Bit rng trong10 sn phm ny c 1 th phm. Tnh xc sut th phm c trong mu.

p n. 0.4

Bi 3.31. Mt ci hp cha 100 transistor loi A v 50 transistor loi B.

32

(a) Cc transistor c rt ra ln lt, ngu nhin v c hon li, cho n khi ly ctransistor loi B u tin. Tnh xc sut 9 hoc 10 transistor c rt ra.

(b) S lng cc transistor t nht phi rt ra, ngu nhin v c hon li, l bao nhiunu ta mun xc sut ly c ch loi A nh hn 1/3?

p n. (a) 0.0217 (b) 3

Bi 3.32. Gi X l s ln mt nht xut hin sau ba ln tung mt con xc xc.

(a) Lp bng phn phi xc sut ca X.

(b) Tnh xc sut c t nht mt ln c mt nht.

(c) Tnh xc sut c ti a hai ln mt nht.

(d) Tnh EX,V ar(X)

p n. (a)X 0 1 2 3

P 0.579 0.347 0.069 0.005(b) 0.421 (c) 0.995 (d) 0.5; 0.417

Bi 3.33. Xt tr chi, tung mt con xc xc ba ln: nu c ba ln c 6 nt th lnh 6ngn , nu hai ln 6 nt th lnh 4 ngn , mt ln 6 nt th lnh 2 ngn , v nu khngc 6 nt th khng lnh g ht. Mi ln chi phi ng A ngn . Hi :

(a) A l bao nhiu th ngi chi v lu v di hu vn (gi l tr chi cng bng).

(b) A l bao nhiu th trung bnh mi ln ngi chi mt 1 ngn .

p n. (a) 1000 (b) 2000

Bi 3.34. Mt h thng an ninh gm c 10 thnh phn hot ng c lp ln nhau. Hthng hot ng nu t nht 5 thnh phn hot ng. kim tra h thng c hot nghay khng, ngi ta kim tra nh k 4 thnh phn c chn ngu nhin (khng hon li).H thng c bo co l hot ng nu t nht 3 trong 4 thnh phn c kim tra hotng. Nu tht s ch c 4 trong 10 thnh phn hot ng, th xc xut h thng c boco l hot ng l bao nhiu?

p n. 0.1191

Bi 3.35. Trong mt tr chi nm phi tiu, ngi chi hng v mt tm bia ln c v mtvng trn c bn knh 25 cm. Gi X l khong cch (theo cm) gia u phi tiu cm vo biav tm vng trn. Gi s rng

P (X x) ={cpix2 nu 0 x < 25

1 nu x 25

vi c l mt hng s no .

(a) Tnh

(i) hng s c

33

(ii) hm mt , fX(x), ca X

(iii) trung bnh ca X

(iv) xc sut P (X 10|X 5).(b) Ngi chi s mt 1 (n v: ngn ng) cho mi ln phng v thng

10 nu X r1 nu r < X 2r0 nu 2r < X < 25

Vi gi tr no ca r th s tin trung bnh ngi chi t c bng 0.25?

p n. (a)-(i) 1pi252

-(iii) 50/3 -(iv) 1/8 (b) 7.7522

Bi 3.36. Cho X l mt i lng ngu nhin c phn phi xc sut nh sau

X 0 1 2 3 4 5 6 7

P 0 a 2a 2a 3a a2 2a2 7a2 + a

(a) Xc nh a

(b) Tnh P (X 5), P (X < 3).

(c) Tnh k nh nht sao cho P (X k) 12

p n. (a) 1/10 (b) 0.2; 0.3 (c) 3

Bi 3.37. Cho hm mt ca bin ngu nhin X c dng

(a)

f(x) =

{Ax khi x [0, 1]0 khi x / [0, 1]

(b)

f(x) =

{A sinx khi x [0, pi]0 khi x / [0, pi]

(c)

f(x) =

{A cospix khi x [0, 12 ]0 khi x / [0, 12 ]

(d)

f(x) =

{Ax4

khi x 10 khi x < 1

Hy xc nh A. Tm hm phn phi xc sut ca X. Tnh EX,V ar(X) nu c.

34

p n. (a) 2; 2/3; 0.055 (b) 0.5; pi/2; pi2/4 2 (c) pi; 1/2 1/pi; (pi 3)/pi2 (d) 3; 3/2; 3/4

Bi 3.38. Cho bin ngu nhin lin tc X c hm phn phi

F (x) =

0 khi x < pi2a+ b sinx khi pi2 x pi21 khi x > pi2

vi a, b l hng s.

(a) Tm a v b.

(b) Vi a v b tm c cu a), tnh hm mt f(x) ca Xv Mod(X),Med(X), P (X > pi4 )

p n. (a) a = 1/2; b = 1/2.

Bi 3.39. Cho X v Y l hai bin ngu nhin c lp v c phn phi xc sut tng ngl

X 1 0 1 2P 0.2 0.3 0.3 0.2

Y 1 0 1P 0.3 0.4 0.3

Tm phn phi xc sut ca X2, X + Y . Tnh k vng, phng sai ca X, X + Y .

Bi 3.40 (*). Mt mu gm 4 bin ngu nhin X1, X2, X3, X4 c lp vi nhau tng imt. Mi bin ngu nhin Xi, i = 1, . . . , 4 c hm mt nh sau:

f(x) =

{2x khi 0 < x < 10 ni khc

t Y = max{X1, X2, X3, X4} v Z = min{X1, X2, X3, X4}. Tm hm mt ca Y v Z.Hng dn. Ch rng

max{x, y} < z x < z v y < z min{x, y} < z x < z hoc y < z.

Bi 3.41 (*). Cho FX l hm phn phi xc sut ca bin ngu nhin X. Tm hm phnphi xc sut ca bin ngu nhin

Y =

{X|X| nu X 6= 01 nu X = 0

Hng dn. Ch rng x/|x| ={1 nu x < 01 nu x > 0

.

Bi 3.42 (*). Tm hm phn phi ca1

2(X + |X|) nu hm phn phi ca X l FX .

35

Bi 3.43 (*). Gi s X c hm phn phi lin tc F (x). Xc nh hm phn phi caY = F (X).

p n. Y U(0, 1)

Bi 3.44 (*). Gi s F (x) l hm phn phi ca bin ngu nhin dng lin tc X, c tnhcht

P (X < t+ x|X > t) = P (X < x) vi x, t > 0

Chng minh rng F (x) = 1 ex vi x > 0.

Chng 4

Mt s phn phi xc sut thng

dng

4.1 Phn phi Bernoulli, nh thc

Bi 4.1. C 8000 sn phm trong c 2000 sn phm khng t tiu chun k thut. Lyngu nhin (khng hon li) 10 sn phm. Tnh xc sut trong 10 sn phm ly ra c 2sn phm khng t tiu chun.

p n. 0.282

Hng dn. Gi X l s sn phm khng t tiu chun trong 10 sn phm ly ra.

Ta c, X B(10, 20008000

) = B(10, 0.25)

Bi 4.2. Khi tim truyn mt loi huyt thanh, trung bnh c mt trng hp phn ngtrn 1000 trng hp. Dng loi huyt thanh ny tim cho 2000 ngi. Tnh xc sut

(a) c 3 trng hp phn ng,

(b) c nhiu nht 3 trng hp phn ng,

(c) c nhiu hn 3 trng hp phn ng.

p n. (a) 0.18 (b) 0.86 (c) 0.14

Bi 4.3. Gi s t l sinh con trai v con gi l bng nhau v bng1

2. Mt gia nh c 4

ngi con. Tnh xc sut 4 a con gm

(a) 2 trai v 2 gi.

(b) 1 trai v 3 gi.

(c) 4 trai.

p n. (a) 0.375 (b) 0.25 (c) 0.0625

36

4.1. PHN PHI BERNOULLI, NH THC 37

Bi 4.4. Mt nh my sn xut vi t l ph phm l 7%.

(a) Quan st ngu nhin 10 sn phm. Tnh xc sut

i) c ng mt ph phm.

ii) c t nht mt ph phm.

iii) c nhiu nht mt ph phm.

(b) Hi phi quan st t nht bao nhiu sn phm xc sut nhn c t nht mt phphm 0.9

p n. (a)-(i) 0.364 -(ii) 0.516 -(iii) 0.848 (b) 32

Bi 4.5. T l mt loi bnh bm sinh trong dn s l p = 0.01. Bnh ny cn s chm scc bit lc mi sinh. Mt nh bo sinh thng c 20 ca sinh trong mt tun. Tnh xc sut

(a) khng c trng hp no cn chm sc c bit,

(b) c ng mt trng hp cn chm sc c bit,

(c) c nhiu hn mt trng hp cn chm sc c bit.

Tnh bng quy lut nh thc ri dng quy lut Poisson so snh kt qu khi ta xp x phnphi nh thc B(n; p) bng phn phi Poisson P (np).

p n. (a) 0.818 (b) 0.165 (c) 0.017

Bi 4.6. T l c tri ng h ng c vin A trong mt cuc bu c l 60%. Ngi ta hi kin 20 c tri c chn mt cch ngu nhin. Gi X l s ngi b phiu cho A trong 20ngi .

(a) Tm gi tr trung bnh, lch chun v Mod ca X.

(b) Tm P (X 10)

(c) Tm P (X > 12)

(d) Tm P (X = 11)

p n. (a) 12; 2.191; 12 (b) 0.245 (c) 0.416 (d) 0.16

Bi 4.7. Gi s t l dn c mc bnh A trong vng l 10%. Chn ngu nhin 1 nhm 400ngi.

(a) Vit cng thc tnh xc sut trong nhm c nhiu nht 50 ngi mc bnh A.

(b) Tnh xp x xc sut bng phn phi chun.

p n. (b) 0.953

4.1. PHN PHI BERNOULLI, NH THC 38

Bi 4.8. Mt chic my bay mun bay c th phi c t nht mt na s ng c hotng. Nu mi ng c hot ng, c lp nhau, vi xc sut 0.6, th mt my bay c 4 ngc c ng tin cy hn mt my bay c 2 ng c hay khng? Gii thch?

p n. khng

Hng dn. Gi X, Y ln lt l s ng c hot ng trong 4 ng c v trong 2 ng c. So snh P (X 2)v P (Y 1).

Bi 4.9. S lng X cc phn t pht ra t mt ngun phng x no trong 1 gi l mtbin ngu nhin c phn phi Poisson vi tham s = ln 5. Hn na, ta gi s rng s phtx ny c lp nhau qua mi gi.

(a) Tnh xc sut c t nht 30 gi, trong 168 gi ca mt tun no , khng c phn tno c pht ra.

(b) S dng phn phi Poisson tnh xp x xc sut trong cu (a).

p n. (a) 0.7549 (b) 0.7558

Bi 4.10. Mt my sn xut ra sn phm loi A vi xc sut 0.485. Tnh xc sut sao ctrong 200 sn phm do my sn xut ra c t nht 95 sn phm loi A.

p n. 0.61

Bi 4.11. Da vo s liu trong qu kh, ta c lng rng 85% cc sn phm ca mtmy sn xut no l th phm. Nu my ny sn xut 20 sn phm mi gi, th xc sut8 hoc 9 th phm c sn xut trong mi khong thi gian 30 pht l bao nhiu?

p n. 0.6233

Bi 4.12. Mi mu c kch thc 10 c rt ra ngu nhin v khng hon li t ccthng cha 100 sn phm, trong mi thng c 2 ph phm. Mt thng sn phm c chpnhn nu c nhiu nht mt th phm c pht hin trong mu tng ng. Hi xc sutc t hn chn thng c chp nhn l bao nhiu?

p n. 0.0036

Bi 4.13. Xc sut mt sn phm c sn xut bi mt my no ph hp vi ccyu cu k thut l 0.95, c lp vi cc sn phm khc. Ta tin hnh ly ra cc sn phmc sn xut bi my ny cho n khi c sn phm t cc yu cu k thut. Th nghimngu nhin ny c lp li trong 15 ngy lin tip (c lp). Gi X l s ngy, trong 15ngy th nghim, m ta phi ly t nht 2 sn phm nhn c mt sn phm ph hpvi cc yu cu k thut.

(a) Tm gi tr trung bnh ca X.

(b) S dng phn phi Poisson tnh xp x xc sut c iu kin P (X = 2|X 1).p n. (a) 0.75 (b) 0.2519

4.1. PHN PHI BERNOULLI, NH THC 39

Bi 4.14. Xc sut trng s l 1%. Mi tun mua mt v s. Hi phi mua v s lin tiptrong ti thiu bao nhiu tun c khng t hn 95% hy vng trng s t nht 1 ln.

p n. 299

Bi 4.15. Trong tr chi "bu cua c ba con xc sc, mi con c su mt hnh l: bu,cua, hu, nai, tm v g. Gi s c hai ngi, mt ngi chi v mt ngi lm ci. Nu mivn ngi chi ch t mt (mt trong cc hnh: bu, cua, hu, nai, tm v g) sau khichi nhiu vn th ngi no s thng trong tr chi ny. Gi s thm mi vn ngi chit 1000 nu thng s c 5000 , nu thua s mt 1000 . Hi trung bnh mi vn ngithng s thng bao nhiu?

p n. 972.2222

Bi 4.16. C ba l ging nhau: hai l loi I, mi l c 3 bi trng v 7 bi en; mt l loi IIc 4 bi trng v 6 bi en. Mt tr chi c t ra nh sau: Mi vn, ngi chi chn ngunhin mt l v ly ra hai bi t l . Nu ly c ng hai bi trng th ngi chi thng,ngc li ngi chi thua.

(a) Ngi A chi tr chi ny, tnh xc sut ngi A thng mi vn.

(b) Gi s ngi A chi 10 vn, tnh s vn trung bnh ngi chi thng c v s vnngi A thng tin chc nht.

(c) Ngi A phi chi t nht bao nhiu vn xc sut thng t nht mt vn khng di0,99.

p n. (a) 0.0889 (b) 0.889; 0 (c) 50

Bi 4.17 (*). Cho X v Y l hai i lng ngu nhin c lp.

(a) Gi s X B(1, 15), Y B(2, 15). Lp bng phn phi xc sut ca X + Y v kim trarng X + Y B(3, 15)

(b) Gi s X B(1, 12), Y B(2, 15). Tm phn b xc sut ca X + Y . Chng minh rngX + Y khng c phn b nh thc.

(c) Gi s X B(n1, p1), Y B(n2, p2). Chng minh rng X + Y c phn phi nh thckhi v ch khi p1 = p2.

Bi 4.18. 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 X l sln trng r ca c hai cu th. Lp bng phn phi xc sut ca X, bit rng kt qu cacc ln nm r l c lp vi nhau.

p n.X 0 1 2 3

P 0.048 0.256 0.444 0.252

Bi 4.19. Bu in dng mt my t ng c a ch trn b th phn loi tng khuvc gi i, my c kh nng c c 5000 b th trong 1 pht. Kh nng c sai 1 a chtrn b th l 0,04% (xem nh vic c 5000 b th ny l 5000 php th c lp).

4.1. PHN PHI BERNOULLI, NH THC 40

(a) Tnh s b th trung bnh mi pht my c sai.

(b) Tnh s b th tin chc nht trong mi pht my c sai.

(c) Tnh xc sut trong mt pht my c sai t nht 3 b th.p n. (a) 2 (b) 2 (c) 0.323

Bi 4.20. Mt bi thi trc nghim gm c 10 cu hi, mi cu c 4 phng n tr li, trong ch c mt phng n ng. Gi s mi cu tr li ng c 4 im v cu tr li sai btr 2 im. Mt sinh vin km lm bi bng cch chn ngu nhin mt phng n cho micu hi.

(a) Tnh xc sut hc sinh ny c 4 im.

(b) Tnh xc sut hc sinh ny b im m.

(c) Gi X l s cu tr li ng, tnh E(X) v V ar(X).

(d) Tnh s cu sinh vin ny c kh nng tr li ng ln nht.p n. (a) 0.146 (b) 0.2503 (c) 2.5; 1.875 (d) 2

Bi 4.21. 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 snphm. Qu trnh sn xut c bo co l t yu cu nu c khng qu mt sn phm lth phm.

(a) Nu tt c cc hp c kim tra u cha chnh xc hai th phm, th xc sut qutrnh sn xut c bo co t yu cu t nht 7 ln trong mt ngy lm vic 8 gi lbao nhiu?

(b) S dng phn phi Poisson xp x xc sut c tnh trong cu (a).

(c) Bit rng ln kim tra cht lng cui cng trong cu (a), qu trnh sn xut c boco t yu cu. Hi xc sut mu 10 sn phm tng ng khng cha th phm l baonhiu?

p n. (a) 0.6572 (b) 0.6626 (c) 0.4118

Bi 4.22. Mt cng ty bo him c 20 nhn vin kinh doanh. Mi ngi, ti mt thi imno , c th vn phng hoc ang trn ng giao dch. Bit rng nhn vin kinh doanhlm vic vn phng vo lc 14h30, vo mt ngy lm vic trong tun, vi xc sut l 0.2,c lp vi cc ngy lm vic khc v nhng nhn vin khc.

(a) Cng ty mun b tr mt s lng t nht cc bn lm vic sao cho mt nhn vin kinhdoanh bt k c th tm thy mt bn trng lm vic trong t nht 90% trng hp.Tm s lng bn t nht ny.

(b) Tnh s lng bn t nht trong phn (a) bng cch s dng xp x Poisson.

(c) Mt ngi ph n gi in n cng ty vo lc 14h30 vo 2 ngy lm vic cui cngtrong tun ni chuyn vi mt nhn vin kinh doanh no . Cho rng c ta khngsp xp cuc hn t trc. Tm xc sut c ta phi gi t nht hai ln na vi gi srng c ta lun gi vo 14h30.

p n. (a) 6 (b) 7 (c) 0.8

4.2. PHN PHI POISSON 41

4.2 Phn phi Poisson

Bi 4.23. Mt trung tm bu in nhn c trung bnh 3 cuc in thoi trong mi pht.Tnh xc sut trung tm ny nhn c 1 cuc, 2 cuc, 3 cuc gi trong 1 pht, bit rngs cuc gi trong mt pht c phn phi Poisson.

p n. 0.149; 0.224; 0.224

Bi 4.24. Tnh P (X 1|X 1) nu X P (5)p n. 5/6

Bi 4.25 (*). Cho X, Y l cc bin ngu nhin c lp, X P (1), Y P (2)

(a) Tnh xc sut P (X + Y = n)

(b) Tnh xc sut P (X = k|X + Y = n)

Bi 4.26. Mt ca hng cho thu xe t nhn thy rng s ngi n thu xe t vo ngyth by cui tun l mt i lng ngu nhin X c phn phi Poisson vi tham s = 2.Gi s ca hng c 4 chic t.

(a) Tm xc sut khng phi tt c 4 chic t u c thu.

(b) Tm xc sut tt c 4 chic t u c thu.

(c) Tm xc sut ca hng khng p ng c yu cu.

(d) Trung bnh c bao nhiu t c thu.

(e) Ca hng cn c t nht bao nhiu t xc sut khng p ng c nhu cu thu bhn 2%

p n. (a) 0.857 (b) 0.143 (c) 0.053 (d) 2 (e) 5

Bi 4.27. Mt tng i bu in c cc cuc in thoi gi n xut hin ngu nhin, clp vi nhau v c tc trung bnh 2 cuc gi trong 1 pht. Tm xc sut

(a) c ng 5 cuc in thoi trong 2 pht,

(b) khng c cuc in thoi no trong khong thi gian 30 giy,

(c) c t nht 1 cuc in thoi trong khong thi gian 10 giy.

p n. (a) 0.156 (b) 0.368 (c) 0.283

Bi 4.28. Cc cuc gi in n tng i tun theo phn phi Poisson vi mc trn mipht. T kinh nghim c c trong qu kh, ta bit rng xc sut nhn c chnh xcmt cuc gi trong mt pht bng ba ln xc sut khng nhn c cuc gi no trong cngthi gian.

4.2. PHN PHI POISSON 42

(a) Gi X l s cuc gi nhn c trong mi pht. Tnh xc sut P (2 X 4).

(b) Ta xt 100 khong thi gian mt pht lin tip v gi U l s khong thi gian mt phtkhng nhn c cuc gi in no. Tnh P (U 1).

p n. (a) 0.6161 (b) 0.0377

Hng dn. U B(100, 0.0498)

Bi 4.29. Ti mt im bn v my bay, trung bnh trong 10 pht c 4 ngi n mua v.Tnh xc sut :

(a) Trong 10 pht c 7 ngi n mua v.

(b) Trong 10 pht c khng qu 3 ngi n mua v.

p n. (a) 0.06 (b) 0.433

Bi 4.30. Cc khch hng n quy thu ngn, theo phn phi Poisson, vi s lng trungbnh 5 ngi mi pht. Tnh xc sut xut hin t nht 10 khch hng trong khong thigian 3 pht.

p n. 0.9301

Bi 4.31. S khch hng n quy thu ngn tun theo phn phi Poisson vi tham s = 1trong mi khong 2 pht. Tnh xc sut thi gian i n khi khch hng tip theo xuthin (t khch hng trc ) nh hn 10 pht.

p n. 0.9933

Bi 4.32. S lng nho kh trong mt ci bnh quy bt k c phn phi Poisson vi thams . Hi gi tr l bao nhiu nu ta mun xc sut c nhiu nht hai bnh quy, trong mthp c 20 bnh, khng cha nho kh l 0.925?

p n. 2.9977

Bi 4.33. Mt trm cho thu xe Taxi c 3 chic xe. Hng ngy trm phi np thu 8 USDcho 1 chic xe (bt k xe c c thu hay khng). Mi chic c cho thu vi gi 20USD.Gi s s xe c yu cu cho thu ca trm trong 1 ngy l i lng ngu nhin c phnphi Poisson vi = 2.8.

(a) Tnh s tin trung bnh trm thu c trong mt ngy.

(b) Gii bi ton trn trong trng hp trm c 4 chic xe.

(c) Theo bn, trm nn c 3 hay 4 chic xe?

p n. (a) 32 USD (b) 24 USD (c) 3

Bi 4.34 (*). Ta c 10 my sn xut (c lp nhau), mi my sn xut ra 2% th phm(khng t chun).

4.3. PHN PHI CHUN 43

(a) Trung bnh c bao nhiu sn phm c sn xut bi my u tin trc khi n to rath phm u tin?

(b) Ta ly ngu nhin mt sn phm t mi my sn xut. Hi xc sut nhiu nht hai thphm trong 10 sn phm ny l bao nhiu?

(c) Lm li cu (b) bng cch s dng xp x Poisson.

(d) Phi ly ra t nht bao nhiu sn phm c sn xut bi my u tin xc sut tc t nht mt th phm khng nh hn 1/2 (gi s rng cc sn phm l c lp vinhau)?

p n. (a) 49 (b) 0.9991 (c) 0.9989 (d) 35

Bi 4.35. S li nh my trong mt quyn sch 500 trang c phn phi Poisson vi thams = 2 mi trang, c lp trn tng trang.

(a) Hi xc sut phi ly t nht 10 trang, ngu nhin v c hon li, t c 3 trangtrong mi trang cha t nht 2 li l bao nhiu?

(b) Gi s rng tht s c 20 trang, trong 500 trang, mi trang cha chnh xc 5 li.

(i) Nu 100 trang c ly, ngu nhin v khng hon li, th xc sut nhiu nht 5trang cha chnh xc 5 li mi trang l bao nhiu?

(ii) Ta xt 50 bn sao ca quyn sch ny. Nu th nghim ngu nhin trong phn (i)c lp li cho mi bn sao, th xc sut c chnh xc 30 trong 50 bn sao m muly ra c nhiu nht 5 trang vi 5 li mi trang l bao nhiu?

p n. (a) 0.0273 (b)-(i) 0.8083 -(ii) 0.000357

4.3 Phn phi chun

Bi 4.36. Cc kt qu ca bi kim tra ch s thng minh (IQ) cho cc hc sinh ca mttrng tiu hc cho thy im IQ ca cc hc sinh ny tun theo phn phi chun vi cctham s l = 100 v 2 = 225. T l hc sinh c im IQ nh hn 91 hoc ln hn 130 lbao nhiu?

p n. 0.2971

Bi 4.37. Gi s chiu di X (n v tnh m) ca mt ni xe bt k tun theo phn phichun N(, 0.012).

(a) Mt ngi n ng s hu mt chic xe hi cao cp c chiu di ln hn 15% chiu ditrung bnh ca mt ch u xe. Hi t l ch u xe c th s dng l bao nhiu?

(b) Gi s rng = 4. Hi chiu di ca xe l bao nhiu nu ta mun ch ca n c th sdng 90% ch u xe?

p n. (a) 0.0668 (b) 3.49

4.3. PHN PHI CHUN 44

Bi 4.38. ng knh ca mt chi tit my do mt my tin t ng sn xut c phn phichun vi trung bnh = 50 mm v lch chun = 0.05 mm. Chi tit my c xem lt yu cu nu ng knh khng sai qu 0.1 mm.

(a) Tnh t l sn phm t yu cu.

(b) Ly ngu nhin 3 sn phm. Tnh xc sut c t nht mt sn phm t yu cu.

p n. (a) 95.4% (b) 0.999

Bi 4.39. Trng lng X (tnh bng gam) mt loi tri cy c phn phi chun N(, 2),vi = 500 (gam) v 2 = 16 (gam2). Tri cy thu hoch c phn loi theo trng lngnh sau:

(a) loi 1 : trn 505 gam,

(b) loi 2 : t 495 n 505 gam,

(c) loi 3 : di 495 gam.

Tnh t l mi loi.

p n. (a) 0.106 (b) 0.788 (c) 0.106

Bi 4.40. Mt cng ty kinh doanh mt hng A d nh s p dng mt trong 2 phng nkinh doanh. K hiu X1 l li nhun thu c khi p dng phng n th 1, X2 l li nhunthu c khi p dng phng n th 2. X1, X2 u c tnh theo n v triu ng/ thng)v X1 N(140, 2500), X2 N(200, 3600). Nu bit rng, cng ty tn ti v pht trinth li nhun thu c t mt hng kinh doanh A phi t t nht 80 triu ng/thng. Hycho bit cng ty nn p dng phng n no kinh doanh mt hng A? V sao?

p n. phng n 2

Bi 4.41. Nghin cu chiu cao ca nhng ngi trng thnh, ngi ta nhn thy rngchiu cao tun theo quy lut phn b chun vi trung bnh l 175 cm v lch tiuchun 4 cm. Hy xc nh:

(a) t l ngi trng thnh c tm vc trn 180 cm.

(b) t l ngi trng thnh c chiu cao t 166 cm n 177 cm.

(c) tm h0, nu bit rng 33% ngi trng thnh c tm vc di mc h0.

(d) gii hn bin ng chiu cao ca 90% ngi trng thnh xung quanh gi tr trung bnhca n.

p n. (a) 0.106 (b) 0.68 (c) 173.24 (d) 6.6

Bi 4.42. Ta quan tm n tui th X (theo nm) ca mt thit b. T kinh nghim trongqu kh, ta c lng xc sut thit b loi ny cn hot ng tt sau 9 nm l 0.1.

4.3. PHN PHI CHUN 45

(a) Ta a ra m hnh sau cho hm mt ca X

fX(x) =a

(x+ 1)bvi x 0

trong a > 0 v b > 1. Tm hai hng s a, b.

(b) Nu ta a ra mt phn phi chun vi trung bnh = 7 cho X, th gi tr tham s l bao nhiu?

(c) Ta xt 10 thit b loi ny mt cch c lp. Tnh xc sut 8 hoc 9 thit b loi ny ctui i hot ng t hn 9 nm.

p n. (a) 1; 2 (b) 1.5601 (c) 0.5811

Bi 4.43. EntropyH ca mt bin ngu nhin lin tcX c nh ngha lH = E[ ln fX(X)]vi fX l hm mt xc sut ca bin ngu nhin X v ln l logarit t nhin. Tnh entropyca bin ngu nhin Gauss vi trung bnh 0 v phng sai 2 = 2.

p n. 1.766

Bi 4.44 (*). Mt nh sn xut bn sn phm vi mt mc gi c nh s. Nh sn xuts hon li tin cho khch hng nu khch hng pht hin trng lng sn phm nh hntrng lng cho trc w0 v thu li sn phm, c gi tr ti ch l r(< s). Trng lng Wtun theo phn phi chun vi trung bnh v phng sai 2. Mt ci t thch hp chophp nh sn xut c nh gi tr bng mt gi tr mong mun, nhng khng th c nhgi tr . Chi ph sn xut C l mt hm theo trng lng ca sn phm: C = + W , vi v l cc hng s dng.

(a) Hy xc nh biu thc cho li nhun Z theo W .

(b) Chng minh rng li nhun trung bnh, z(), c xc nh bi

z() = s (s r)P [W < w0]

Tm gi tr 0 ca lm cc i z().

Chng 5

L thuyt mu

Bi 5.1. S liu v chiu cao ca cc sinh vin n (n v: inch) trong mt lp hc nh sau:

62 64 66 67 65 68 61 65 67 65 64 63 67

68 64 66 68 69 65 67 62 66 68 67 66 65

69 65 70 65 67 68 65 63 64 67 67

(a) Tnh chiu cao trung bnh v lch tiu chun.

(b) Trung v ca chiu cao sinh vin lp ny l bao nhiu?

p n. (a) 65.811; 2.106 (b) 66

Bi 5.2. Cho b d liu sau:

4.2 4.7 4.7 5.0 3.8 3.6 3.0 5.1 3.1 3.8

4.8 4.0 5.2 4.3 2.8 2.0 2.8 3.3 4.8 5.0

Tnh trung bnh mu, phng sai mu v lch tiu chun.

p n. 4; 0.866; 0.931

Bi 5.3. Cho b d liu sau:

43 47 51 48 52 50 46 49

45 52 46 51 44 49 46 51

49 45 44 50 48 50 49 50

Tnh trung bnh mu, phng sai mu v lch tiu chun.

p n. 48.125; 7.245; 2.692

Bi 5.4. Xt biu thc y =n

i=1(xi a)2. Vi a no th y t gi tr nh nht?p n. x

Bi 5.5. Xt yi = a + bxi, i = 1, . . . , n v a, b l cc hng s khc 0. Hy tm mi lin hgia x v y, sx v sy.

46

47

p n. y = a+ bx; sy = |b|sx

Bi 5.6 (*). Gi s ta c mu c n gm cc gi tr quan trc x1, x2, . . . , xn v tnh ctrung bnh mu xn v phng sai mu s2n. Quan trc thm gi tr th (n + 1) l xn+1, gixn+1 v s2n+1 ln lt l trung bnh mu v phng sai mu ng vi mu c (n + 1) quantrc.

(a) Tnh xn+1 theo xn v xn+1.

(b) Chng t rng

ns2n+1 = (n 1)s2n +n(xn+1 xn)2

n+ 1

Bi 5.7. T bng cc s ngu nhin ngi ta ly ra 150 s. Cc s c phn thnh 10khong nh sau:

xi 1 11 21 31 41 51 61 71 81 9110 20 30 40 50 60 70 80 90 100

ni 16 15 19 13 14 19 14 11 13 16

Xc nh trung bnh mu v phng sai mu.

p n. 48.97; 834.9

Bi 5.8. Kho st thu nhp ca cng nhn mt cng ty, cho bi bng sau (n v ngnng).

Thu nhp [500, 600] [600, 700] [700, 800] [800, 900] [900, 1000] [1000, 1100][1100, 1200]S ngi 2 10 15 30 25 14 4

Xc nh thu nhp trung bnh, lch chun.

p n. 874; 136.4

Bi 5.9. o lng huyt tng ca 8 ngi mnh kho, ta c

2, 863, 372, 752, 623, 503, 253, 123, 15

Hy xc nh cc c trng mu.

p n. n = 8, x = 3.0775, s2 = 0.096

Bi 5.10. Quan st thi gian cn thit sn xut mt chi tit my, ta thu c s liucho bng sau:

Khong thi gian (pht) S ln quan st20-25 225-30 1430-35 2635-40 3240-45 1445-50 850-55 4

48

Tnh trung bnh mu x, phng sai mu s2.

p n. 36.6; 45.14

Bi 5.11. o di ca mt loi trc xe, ta c kt qu

Nhm 18.4-18.6 18.6-18.8 18.8-19 19-19.2 19.2-19.4 19.4-19.6 19.6-19.8ni 1 4 20 41 19 8 4

Hy tnh di trung bnh v phng sai mu.

p n. 19.133; 0.054

Chng 6

c lng tham s thng k

6.1 c lng trung bnh tng th

Bi 6.1. Trn tp mu gm 100 s liu, ngi ta tnh c x = 0.1 s = 0.014. Xc nhkhong tin cy 95% cho gi tr trung bnh tht.

p n. (0.0973, 0.1027)

Bi 6.2. Chn ngu nhin 36 cng nhn ca x nghip th thy lng trung bnh l 380 ngn/thng. Gi s lng cng nhn tun theo phn phi chun vi = 14 ngn ng. Vi tin cy 95%, hy c lng mc lng trung bnh ca cng nhn trong ton x nghip.

p n. (375.423, 384.573) ngn /thng

Bi 6.3. o sc bn chu lc ca mt loi ng th nghim, ngi ta thu c b s liu sau

4500, 6500, 5200, 4800, 4900, 5125, 6200, 5375

T kinh nghim ngh nghip, ngi ta cng bit rng sc bn c phn phi chun vi lch chun = 300. Hy xy dng khong tin cy 90% cho sc bn trung bnh ca loi ngtrn.

p n. (5149.991, 5500.009)

Bi 6.4. Sn lng mi ngy ca mt phn xng l bin ngu nhin tun theo lut chun.Kt qu thng k ca 9 ngy cho ta:

27, 26, 21, 28, 25, 30, 26, 23, 26

Hy xc nh cc khong tin cy 95% cho sn lng trung bnh.

p n. (23.755, 27.805)

Bi 6.5. Quan st chiu cao X (cm) ca mt s ngi, ta ghi nhn

x (cm) 140-145 145-150 150-155 155-160 160-165 165-170S ngi 1 3 7 9 5 2

49

6.1. C LNG TRUNG BNH TNG TH 50

(a) Tnh x v s2

(b) c lng tin cy 0.95

p n. (a) 156.2; 37.68 (b) (153.77, 158.63)

Bi 6.6. im trung bnh mn ton ca 100 th sinh d thi vo trng A l 5 vi lchchun l 2.5.

(a) c lng im trung bnh mn ton ca ton th th sinh vi tin cy l 95%.

(b) Vi sai s c lng im trung bnh cu a) l 0.25 im, hy xc nh tin cy.

p n. (a) (4.51, 5.49) (b) 68.26%

Bi 6.7. Tui th ca mt loi bng n c bit theo quy lut chun vi lch chun100 gi.

(a) Chn ngu nhin 100 bng n th nghim, thy mi bng tui th trung bnh l1000 gi. Hy c lng tui th trung bnh ca bng n x nghip A sn xut vi tin cy l 95%.

(b) Vi dung sai ca c lng tui th trung bnh l 15 gi, hy xc nh tin cy.

(c) dung sai ca c lng tui th trung bnh khng qu 25 gi vi tin cy l 95%th cn phi th nghim t nht bao nhiu bng.

p n. (a) (980.4, 1019.6) (b) 86.64% (c) 62

Bi 6.8. Khi lng cc bao bt m ti mt ca hng lng thc tun theo phn phi chun.Kim tra 20 bao, thy khi lng trung bnh ca mi bao bt m l 48kg, v phng sai mus2 = (0.5 kg)2.

(a) Vi tin cy 95% hy c lng khi lng trung bnh ca mt bao bt m thuc cahng.

(b) Vi dung sai ca c lng cu a) l 0.284 kg, hy xc nh tin cy.

(c) dung sai ca c lng cu a) khng qu 160 g vi tin cy l 95%, cn phikim tra t nht bao nhiu bao?

p n. (a) (47.766, 48.234) (b) 0.98 (c) 38

Bi 6.9. o ng knh ca mt chi tit my do mt my tin t ng sn xut, ta ghinhn c s liu nh sau:

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 3

vi n ch s trng hp tnh theo tng gi tr ca X (mm).

6.1. C LNG TRUNG BNH TNG TH 51

(a) Tnh trung bnh mu x v lch chun s ca mu.

(b) c lng ng knh trung bnh tin cy 0.95.

(c) Nu mun sai s c lng khng qu = 0.02 mm tin cy 0.95 th phi quan stt nht my trng hp.

p n. (a) 12.21; 0.103 (b) (12.18, 12.24) (c) 102

Bi 6.10. Ngi ta o ion Na+ trn mt s ngi v ghi nhn li c kt qu nh sau

129, 132, 140, 141, 138, 143, 133, 137, 140, 143, 138, 140

(a) Tnh trung bnh mu x v phng sai mu s2.

(b) c lng trung bnh ca tng th tin cy 0.95.

(c) Nu mun sai s c lng trung bnh khng qu = 1 vi tin cy 0.95 th phi quanst mu gm t nht my ngi?

p n. (a) 137.83; 19.42 (b) (135.01, 140.63) (c) 75

Bi 6.11. Quan st tui th x (gi) ca mt s bng n do x nghip A sn xut, ta ghinhn

x 1000 1100 1200 1300 1400 1500 1600 1700 1800n 10 14 16 17 18 16 16 12 9

vi n ch s trng hp theo tng gi tr ca x.

(a) Tnh trung bnh mu x v lch chun mu s.

(b) c lng tui th trung bnh ca bng n tin cy 0.95.

(c) Nu mun sai s c lng khng qu = 30 gi vi tin cy 0.99 th phi quan stmu gm t nht my bng n?

p n. (a) 1391.41; 234.45 (b) (1350.79, 1432.03) (c) 235

Bi 6.12. Chiu di ca mt loi sn phm c xut khu hng lot l bin ngu nhinphn phi chun vi = 100 mm v 2 = 42 mm2. Kim tra ngu nhin 25 sn phm. Khnng chiu di trung bnh ca s sn phm kim tra nm trong khong t 98mm n 101mml bao nhiu?

p n. 88.82%

6.2. C LNG T L TNG TH 52

6.2 c lng t l tng th

Bi 6.13. 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 baonhiu phn trm phiu bu?

p n. 66.97%

Bi 6.14. Mt loi bnh c t l t vong l 0.01. Mun chng t mt loi thuc c hiunghim (ngha l h thp c t l t vong nh hn 0.005) tin cy 0.95 th phi ththuc trn t nht bao nhiu ngi?

p n. 1522

Bi 6.15. c lng xc sut mc bnh gan vi tin cy 90% v sai s khng vt qu2% th cn phi khm t nht bao nhiu ngi, bit rng t l mc bnh gan thc nghim cho bng 0,9.

p n. 613

Bi 6.16. Gi s quan st 100 ngi thy c 20 ngi b bnh st xut huyt. Hy c lngt l bnh st xut huyt tin cy 97%. Nu mun sai s c lng khng qu 3% tin cy 95% th phi quan st t nht bao nhiu ngi?

p n. (0.1132, 0.2868); 683

Bi 6.17. Mt loi thuc mi em iu tr cho 50 ngi b bnh B, kt qu c 40 ngi khibnh.

(a) c lng t l khi bnh p nu dng thuc iu tr vi tin cy 0.95 v 0.99.

(b) Nu mun sai s c lng khng qu 0.02 tin cy 0.95 th phi quan st t nhtmy trng hp?

p n. (a) (0.69, 0.91); (0.65, 0.946) (b) 1537

Bi 6.18. Ta mun c lng t l vin thuc b sc m p trong mt l thuc ln.

(a) Nu mun sai s c lng khng qu 0.01 vi tin cy 0.95 th phi quan st t nhtmy vin?

(b) Quan st ngu nhin 200 vin, thy c 18 vin b st m. Hy c lng p tin cy0.95.

(c) Khi , nu mun sai s c lng khng qu 0.01 vi tin cy 0.95 th phi quan stt nht my vin?

p n. (a) 9604 (b) (0.051, 0.13) (c) 3147

6.3. TNG HP 53

Bi 6.19. Mun bit trong ao c bao nhiu c, ngi ta bt ln 2000 con, nh du xongli th xung h. Sau mt thi gian, ngi ta bt ln 500 con v thy c 20 con c c nhdu ca ln bt trc. Da vo kt qu hy c lng s c c trong h vi tin cy95%.

p n. (34965.03, 877719.3)

Bi 6.20. c th d on c s lng chim thng ngh ti vn nh mnh, ngi chbt 89 con, em eo khoen cho chng ri th i. Sau mt thi gian, ng bt ngu nhin c120 con v thy c 7 con c eo khoen. Hy d on s chim gip ng ch vn tin cy99%.

p n. (785.1688, 27396.59)

6.3 Tng hp

Bi 6.21. Cn th 100 qu cam, ta c b s liu sau:

Khi lng (g) 32 33 34 35 36 37 38 39 40S qu 2 3 15 26 28 6 8 8 4

(a) Hy c lng khi lng trung bnh cc qu cam tin cy 95%.

(b) Cam c khi lng di 34 g c coi l cam loi 2. Tm khong c lng cho t l loi2 vi tin cy 90%.

p n. (a) (35.539, 36.241) (b) (0.014, 0.086)

Bi 6.22. em cn mt s tri cy va thu hoch, ta c kt qu sau:

X (gam) 200-210 210-220 220-230 230-240 240-250S tri 12 17 20 18 15

(a) Tm khong c lng ca trng lng trung bnh ca tri cy vi tin cy 0.95 v0.99.

(b) Nu mun sai s c lng khng qu = 2 gam tin cy 99% th phi quan st tnht bao nhiu tri?

(c) Tri cy c khi lng X 230 gam c xp vo loi A. Hy tm khong c lng chot l p ca tri cy loi A tin cy 0.95 v 0.99. Nu mun sai s c lng khngqu 0.04 tin cy 0.99 th phi quan st t nht my trng hp?

p n. (a) (222.98, 228.72); (222.08, 229.63) (b) 293 (c) (0.2963, 0.5085); (0.2627,0.5421); 1001

Chng 7

Kim nh gi thuyt thng k

7.1 So snh k vng vi mt s cho trc

Bi 7.1. Gim c mt x nghip cho bit lng trung bnh ca 1 cng nhn thuc xnghip l 380 ngn /thng. Chn ngu nhin 36 cng nhn thy lng trung bnh l 350ngn /thng, vi lch chun s = 40. Li bo co ca gim c c tin cy c khng,vi mc c ngha l = 5%.

Hng dn. Ta cn kim nh cc gi thuyt{H0 : = 380

H1 : 6= 380

y l trng hp n = 36 30 v 2 cha bit, nn ta dng

z =

n(x )s

=

36(350 380)

40= 4.5

Ta thy |z| > z12

= z0.975 = 1.96. Do ta bc b gi thuyt H0. Ngha l li bo co ca gim c khng

ng tin cy.

Bi 7.2. Trong thp nin 80, trng lng trung bnh ca thanh nin l 48 kg. Nay xcnh li trng lng y, ngi ta chn ngu nhin 100 thanh nin o trng lng trung bnhl 50 kg v phng sai mu s2 = (10 kg)2. Th xem trng lng thanh nin hin nay phichng c thay i, vi mc c ngha l 1%?

p n. z = 2. Trng lng thanh nin hin nay khng thay i so vi trc kia.

Bi 7.3. Mt ca hng thc phm nhn thy thi gian va qua trung bnh mt khchhng mua 25 ngn ng thc phm trong ngy. Nay ca hng chn ngu nhin 15 khchhng thy trung bnh mt khch hng mua 24 ngn ng trong ngy v phng sai mu ls2 = (2 ngn ng)2.Vi mc ngha l 5%, kim nh xem c phi sc mua ca khch hng hin nay thc sgim st hay khng. Bit rng sc mua ca khch hng c phn phi chun.

p n. t = 1.9365. Sc mua ca khch hng hin nay thc s gim st.

54

7.1. SO SNH K VNG VI MT S CHO TRC 55

Bi 7.4. i vi ngi Vit Nam, lng huyt sc t trung bnh l 138.3 g/l. Khm cho80 cng 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 nhnnh my ho cht ny thp hn mc chung hay khng? Kt lun vi = 0.05.

p n. z = 10.912. Lng huyt t trung bnh ca cng nhn nh my thp hn mc chung.

Bi 7.5. Trong iu kin chn nui bnh thng, lng sa trung bnh ca 1 con b l14 kg/ngy. Nghi ng iu kin chn nui km i lm cho lng sa gim xung, ngi taiu tra ngu nhin 25 con v tnh c lng sa trung bnh ca 1 con trong 1 ngy l 12.5v lch chun s = 2.5. Vi mc ngha = 0.05. hy kt lun iu nghi ng ni trn.Gi thit lng sa b l 1 bin ngu nhin chun.

p n. t = 3. iu kin chn nui km i lm cho lng sa gim xung.

Bi 7.6. 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 tnhc x = 404.8 ngn /thng v s = 20 ngn /thng. Vi = 1%

(a) Nu lp gi thit 2 pha v gi thit 1 pha th kt qu kim nh nh th no?

(b) Ging cu a, vi x = 406 ngn /thng v s = 20 ngn /thng.

Bi 7.7. Mt my ng gi cc sn phm c khi lng 1 kg. Nghi ng my hot ngkhng bnh thng, ngi ta chn ra mt mu ngu nhin gm 100 sn phm th thy nhsau:

Khi lng 0.95 0.97 0.99 1.01 1.03 1.05S gi 9 31 40 15 3 2

Vi mc ngha 0.05, hy kt lun v nghi ng trn.

p n. z = 6.9204. My hot ng khng bnh thng.

Bi 7.8. Trng lng trung bnh khi xut chung mt tri chn nui trc l 3.3 kg/con.Nm nay ngi ta s dng mt loi thc n mi, cn th 15 con khi xut chung ta ccc s liu nh sau:

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, 3.50

Gi thit trng lng g l i lng ngu nhin phn phi theo quy lut chun.

(a) Vi mc ngha = 0.05. Hy cho kt lun v tc dng ca loi thc n ny?

(b) Nu tri chn nui bo co trng lng trung bnh khi xut chung l 3.5 kg/con th cchp nhn c khng? ( = 0.05).

p n. (a) t = 3.0534. Thc n mi ny lm thay i trng lng g.

(b) t = 1.1409. Tri chn nui bo co trng lng trung bnh khi xut chung l chp nhn c.

7.1. SO SNH K VNG VI MT S CHO TRC 56

Bi 7.9. o cholesterol (n v mg%) cho mt nhm ngi, ta ghi nhn li c

Chol. 150 160 160 - 170 170 - 180 180 - 190 190 - 200 200 - 210S ngi 3 9 11 3 2 1

Cho rng cholesterol tun theo phn phi chun.

(a) Tnh trung bnh mu x v phng sai mu s2.

(b) Tm khong c lng cho trung bnh cholesterol trong dn s tin cy 0.95.

(c) C ti liu cho bit lng cholesterol trung bnh l 0 = 175 mg%. Gi tr ny c phhp vi mu quan st khng? (kt lun vi = 0.05).

p n. (a) 173.2759; 143.3498 (b) (168.7226, 177.8292) (c) t = 0.7755. Gi tr mu ph hp vi ti liu.

Bi 7.10. Quan st s hoa hng bn ra trong mt ngy ca mt ca hng bn hoa sau mtthi gian, ngi ta ghi c s liu sau:

S hoa hng (o) 12 13 15 16 17 18 19S ngy 3 2 7 7 3 2 1

Gi thit rng s hoa bn ra trong ngy c phn phi chun.

(a) Tm trung bnh mu x, phng sai mu s2.

(b) Sau khi tnh ton, ng ch ca hng ni rng nu trung bnh mt ngy khng bn c15 o hoa th chng th ng ca cn hn. Da vo s liu trn, anh (ch) hy kt lungip ng ch ca hng xem c nn tip tc bn hay khng mc ngha = 0.05.

(c) Gi s nhng ngy bn c t 13 n 17 o hng l nhng ngy bnh thng. Hyc lng t l ca nhng ngy bnh thng ca ca hng tin cy 90%.

p n. (a) 15.4; 3.5 (b) t = 1.069. ng ch ca hng nn tip tc bn. (c) (0.6191, 0.9009)

Bi 7.11. Mt x nghip c mt s rt ln cc sn phm bng thp vi s khuyt tt trungbnh mi sn phm l 3. Ngi ta ci tin cch sn xut v kim tra 36 sn phm. Kt qunh sau:

S khuyt tt trn sn phm 0 1 2 3 4 5 6S sn phm tng ng 7 4 5 7 6 6 1

Gi s s khuyt tt ca cc sn phm c phn phi chun.

(a) Hy c lng s khuyt tt trung bnh mi sn phm sau khi ci tin, vi tin cy90%.

(b) Hy cho kt lun v hiu qu ca vic ci tin sn xut mc ngha 0.05.

p n. (a) (2.1333, 3.1445) (b) z = 1.1785. Ci tin khng hiu qu.

7.2. SO SNH HAI K VNG 57

Bi 7.12. nh gi tc dng ca mt ch n bi dng m du hiu quan st l s hngcu. Ngi ta m s hng cu ca 20 ngi trc v sau khi n bi dng:

xi 32 40 38 42 41 35 36 47 50 30yi 40 45 42 50 52 43 48 45 55 34xi 38 45 43 36 50 38 42 41 45 44yi 32 54 58 30 60 35 50 48 40 50

Vi mc ngha = 0.05, c th kt lun g v tc dng ca ch n bi dng ny?

p n. t = 3.0386. Ch n bi dng lm thay i hng cu.

Hng dn. t Z = Y X ch s lng hng cu thay i sau khi n bi dng.

Bi 7.13. Gi s ta mun xc nh xem hiu qu ca ch n king i vi vic gimtrng lng nh th no. 20 ngi qu bo thc hin ch n king. Trng lng catng ngi trc khi n king (X kg) v sau khi n king (Y kg) c cho nh sau:

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 81

Kim tra xem ch n king c tc dng lm thay i trng lng hay khng ( = 0.05).

p n. t = 3.3002. Ch n king c tc dng lm thay i trng lng.

7.2 So snh hai k vng

Bi 7.14. Mt nh pht trin sn phm quan tm n vic gim thi gian kh ca sn. Vvy hai cng thc sn c em th nghim. Cng thc 1 l cng thc c cc thnh phnchun v cng thc 2 c thm mt thnh phn lm kh mi c cho rng s lm gim thigian kh ca sn. T cc th nghim ngi ta thy rng 1 = 2 = 8 pht. 10 vt csn vi cng thc 1 v 10 vt khc c sn vi cng thc 2. Thi gian kh trung bnhca tng mu l x1 = 121 pht v x2 = 112 pht. Nh pht trin sn phm c th rt rakt lun g v nh hng ca thnh phn lm kh mi? Vi mc ngha 5%.

Hng dn. Ta cn kim nh cc gi thuyt{H0 : 1 = 2H1 : 1 > 2

Ta tnh c

z =x1 x221n1

+22n2

=121 112

82

10+ 8

2

10

= 2.5156

Ta thy z > z1 = z0.95 = 1.65. Do , ta bc b gi thuyt H0 ngha l thnh phn lm kh mi lm gimthi gian kh.

7.2. SO SNH HAI K VNG 58

Bi 7.15. Tc chy ca hai loi cht n lng c dng lm nhin liu trong tu v trc nghin cu. Ngi ta bit rng lch chun ca tc chy ca hai loi nhin liubng nhau v bng 3 cm/s. Hai mu ngu nhin kch thc n1 = 20 v n2 = 20 c thnghim; trung bnh mu tc chy l x1 = 18 cm/s v x2 = 24 cm/s. Vi mc ngha = 0.05 hy kim nh gi thuyt hai loi cht n lng ny c cng tc t chy.

p n. z = 6.3246. Hai loi cht n lng ny c tc t chy khc nhau.

Bi 7.16. Theo di gi c phiu ca 2 cng ty A v B trong vng 31 ngy ngi ta tnhc cc gi tr sau

x sCng ty A 37.58 1.50Cng ty B 38.24 2.20

Gi thit rng gi c phiu ca hai cng ty A v B l hai bin ngu nhin phn phi theoquy lut chun. Hy cho bit ngha k vng ca cc bin ngu nhin ni trn? Hy cho bitc s khc bit thc s v gi c phiu trung bnh ca hai cng ty A v B khng? Vi mc ngha = 5%

p n. t = 1.3801. Gi c phiu trung bnh ca hai cng ty A v B bng nhau.

Bi 7.17. Hm lng ng trong mu ca cng nhn sau 5 gi lm vic vi my siu caotn o c hai thi im trc v sau 5 gi lm vic. Ta c kt qu sau:

Trc: n1 = 50 x = 60 mg% sx = 7Sau: n2 = 40 y = 52 mg% sy = 9.2

Vi mc ngha = 0.05, c th khng nh hm lng ng trong mu sau 5 gi lm vic gim i hay khng?

p n. t = 4.6851. Hm lng ng trong mu sau 5 gi lm vic gim i.

Bi 7.18. Trng cng mt ging la trn hai tha rung nh nhau v bn hai loi phnkhc nhau. n ngy thu hoch ta c kt qu nh sau:

Tha th nht ly mu 1000 bng la thy s ht trung bnh ca mi bng l x = 70ht v sx = 10.

Tha th hai ly mu 500 bng thy s ht trung bnh mi bng l y = 72 ht vsy = 20.

Hi s khc nhau gia X v Y l ngu nhin hay bn cht, vi = 0.05?

p n. t = 2.5824. S khc nhau gia X v Y l do bn cht.

Bi 7.19. so snh trng lng trung bnh ca tr s sinh thnh th v nng thn, ngita th cn trng lng ca 10000 chu v thu c kt qu sau y:

7.2. SO SNH HAI K VNG 59

Vng S chu c cn Trng lng trung bnh lch chun muNng thn 8000 3.0 kg 0.3 kgThnh th 2000 3.2 kg 0.2 kg

Vi mc ngha = 0.05 c th coi trng lng trung bnh ca tr s sinh thnh th caohn nng thn hay khng? (Gi thit trng lng tr s sinh l bin ngu nhin chun).

p n. t = 28.2885. Trng lng trung bnh ca tr s sinh thnh th cao hn nng thn.

Bi 7.20. so snh nng lc hc ton v vt l ca hc sinh, ngi ta kim tra ngu nhin8 em bng hai bi ton v vt l. Kt qu cho bi bng di y (X l im ton, Y l iml):

X 15 20 16 22 24 18 20 14Y 15 22 14 25 19 20 24 16

Gi s X v Y u c phn phi chun. Hy so snh im trung bnh gia X v Y , mc ngha 5%.

p n. t = 0.3913. im trung bnh ca X v Y l nh nhau.

Bi 7.21. Hai my c s dng rt nc vo cc bnh. Ngi ta ly mu ngu nhin10 bnh do my th nht v 10 bnh do my th hai th c kt qu sau:

My 1 16.03 16.01 16.04 15.96 16.05 15.98 16.05 16.02 16.02 15.99My 2 16.02 16.03 15.97 16.04 15.96 16.02 16.01 16.01 15.99 16.00

Vi mc ngha = 0.05 c th ni rng hai my rt nc vo bnh nh nhau khng?

p n. t = 0.7986. Hai my rt nc vo bnh nh nhau.

Bi 7.22. nghin cu nh hng ca mt loi thuc, ngi ta cho 10 bnh nhn ungthuc. Ln khc h cng cho bnh nhn ung thuc nhng l thuc gi. Kt qu th nghimthu c nh sau:

Bnh nhn 1 2 3 4 5 6 7 8 9 10S gi ng c thuc 6.1 7.0 8.2 7.6 6.5 8.4 6.9 6.7 7.4 5.8

S gi ng vi thuc gi 5.2 7.9 3.9 4.7 5.3 5.4 4.2 6.1 3.8 6.3

Gi s s gi ng ca bnh nhn tun theo phn phi chun. Vi mc ngha 5%, hy ktlun v nh hng ca loi thuc trn.

p n. t = 3.7134. Loi thuc trn nh hng n s gi ng ca bnh nhn.

Bi 7.23. Quan st sc nng ca b trai (X) v b gi (Y) lc s sinh (n v gam), ta ckt qu

Trng lng 3000-3200 3200-3400 3400-3600 3600-3800 3800-4000S b trai 1 3 8 10 3S b gi 2 10 10 5 1

7.3. SO SNH T L VI MT S CHO TRC 60

(a) Tnh x, y, s2x, s2y.

(b) So snh cc k vng X , Y (kt lun vi = 5%).

(c) Nhp hai mu li. Tnh trung bnh v lch chun ca mu nhp. Dng mu nhp c lng sc nng trung bnh ca tr s sinh tin cy 95%.

p n. (a) 3588; 3450; 40266.67; 37407.41

(b) t = 2.5476. Trng lng b trai v b gi lc s sinh khc nhau.

(c) 3515.094; 206.9896; (3459.367, 3570.821)

7.3 So snh t l vi mt s cho trc

Bi 7.24. Mt my sn xut t ng vi t l chnh phm l 98%. Sau mt thi gian hotng, ngi ta nghi ng t l trn b gim. Kim tra ngu nhin 500 sn phm thy c28 ph phm, vi = 0.05 hy kim tra xem cht lng lm vic ca my c cn c nhtrc hay khng?

Hng dn. Gi p l t l chnh phm ca my sn xut t ng sau mt thi gian hot ng.

Ta cn kim nh cc gi thuyt {H0 : p = 0.98

H1 : p < 0.98

Ta c n = 500, f =500 28

500= 0.944, nf = 472 5 v n(1 f) = 28 5.

Do , ta dng

z =

n(f p)

pq

=

500(0.944 0.98)

0.98 0.02= 5.7499

Ta thy z < z = z0.05 = z0.95 = 1.65. Do ta bc b gi thuyt H0. Ngha l cht lng lm vic camy khng cn tt nh trc.

Bi 7.25. Trong mt vng dn c c 18 b trai v 28 b gi mc bnh B. Hi rng t lnhim bnh ca b trai v b gi c nh nhau khng? (kt lun vi = 0.05 v gi s rngs lng b trai v b gi trong vng tng ng nhau, v rt nhiu).

p n. z = 1.4745. T l mc bnh ca b trai v b gi l nh nhau.

Bi 7.26. 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 thkt lun cng nhn nng trng c t l huyt sc t di 110 g/l cao hn mc chung haykhng?

p n. z = 4.6291. Cng nhn nng trng c t l huyt sc t di 110 g/l cao hn mc chung.

Bi 7.27. Theo mt ngun tin th t l h dn thch xem dn ca trn Tivi l 80%. Thm d36 h dn thy c 25 h thch xem dn ca. Vi mc c ngha l 5%. Kim nh xem nguntin ny c ng tin cy khng?

7.4. SO SNH HAI T L 61

p n. z = 1.584. Ngun tin ny ng tin cy.

Bi 7.28. Mt my sn sut t ng, lc u t l sn phm loi A l 20%. Sau khi pdng mt phng php ci tin sn xut mi, ngi ta ly 40 mu, mi mu gm 10 snphm kim tra. Kt qu kim tra cho bng sau:

S sn phm loi A trong mu 1 2 3 4 5 6 7 8 9 10S mu 2 0 4 6 8 10 4 5 1 0

Vi mc ngha 5%. Hy cho kt lun v phng php sn sut ny.

p n. z = 16.875. Phng php ci tin sn xut mi thay i t l sn phm loi A.

Bi 7.29. T l ph phm ca mt nh my trc y l 5%. Nm nay nh my p dngmt bin php k thut mi. nghin cu tc dng ca bin php k thut mi, ngi taly mt mu gm 800 sn phm kim tra v thy c 24 ph phm.

(a) Vi = 0.01. Hy cho kt lun v bin php k thut mi ny?

(b) Nu nh my bo co t l ph phm sau khi p dng bin php k thut mi l 2% thc chp nhn c khng? ( = 0.01).

p n. (a) z = 2.5955. Bin php k thut mi lm thay i t l ph phm.(b) z = 2.0203. Nh my bo co t l ph phm l chp nhn c.

7.4 So snh hai t l

Bi 7.30. Trong 90 ngi dng DDT nga bnh ngoi da th c 10 ngi nhim bnh;trong 100 ngi khng dng DDT th c 26 ngi mc bnh. Hi rng DDT c tc dngnga bnh ngoi da khng? (kt lun vi = 0.05)

Hng dn. Gi

p1 : t l ngi mc bnh dng DDT

p2 : t l ngi mc bnh khng dng DDT

Ta cn kim nh cc gi thuyt {H0 : p1 = p2H1 : p1 < p2

Ta c

n1 = 90 30n2 = 100 30f1 =

10

90= 0.1111

f2 =26

100= 0.26

p =n1f1 + n2f2n1 + n2

=10 + 26

90 + 100= 0.1895

7.4. SO SNH HAI T L 62

Ta tnh c

z =f1 f2

pq(

1n1

+ 1n2

)=

0.1111 0.260.1895(1 0.1895) ( 1

90+ 1

100

)= 2.61