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Friday, July 12, 2013

Xc sut - Thng k i hc 1

XXC SUC SUT & THT & THNG KNG KI HI HCC

PHN PHPHN PHI CHNG TRNHI CHNG TRNHSS tititt: 30: 30

------------------------------------------

PHN I. L THUYT XC SUT (Probability theory)

Chng 1. Xc sut ca Bin c Chng 2. Bin ngu nhin Chng 3. Phn phi Xc sut thng dng Vector ngu nhin ri rc Chng 4. nh l gii hn trong Xc sut

PHN II. L THUYT THNG K (Statistical theory)

Chng 5. Mu thng k v c lng tham s Chng 6. Kim nh Gi thuyt Thng k

Ti liu tham kho

1. Nguyn Ph Vinh Gio trnh Xc sut Thng k v ng dng NXB Thng k.

2. inh Ngc Thanh Gio trnh Xc sut Thng k H Tn c Thng Tp.HCM.

3. ng Hng Thng Bi tp Xc sut; Thng k NXB Gio dc.

4. L S ng Xc sut Thng k v ng dng NXB Gio dc.

5. o Hu H Xc sut Thng k NXB Khoa hc & K thut.

6. u Th Cp Xc sut Thng k L thuyt v cc bi tp NXB Gio dc.

7. Phm Xun Kiu Gio trnh Xc sut v Thng k NXB Gio dc.

8. Nguyn Cao Vn Gio trnh L thuyt Xc sut & Thng k NXB Kt Quc dn.

9. Nguyn c Phng Xc sut & Thng k Lu hnh ni b.

10. F.M.Dekking A modern introduction to Probability and Statistics Springer Publication (2005).

BinBin sosonn:: ThSThS. . oonn VngVng NguynNguynDownload Slide Download Slide bbii gigingng XSTKXSTK__HH ttii

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1. Tnh cht ca cc php ton , a) Tnh giao hon:

A B B A= , A B B A= .

b) Tnh kt hp: ( ) ( )A B C A B C= , ( ) ( )A B C A B C= .

c) Tnh phn phi: ( ) ( ) ( )A B C A B A C= , ( ) ( ) ( )A B C A B A C= .

d) Tnh i ngu (DeMorgan): A B A B= , A B A B= .

BB ttcc vv ii ss TT hhpp

2. Quy tc nhn Gi s mt cng vic no c chia thnh k giai on. C n1 cch thc hin giai on th 1,..., c nkcch thc hin giai on th k. Khi ta c:

n = n1nk cch thc hin ton b cng vic. Gi s c k cng vic

1, ...,

kA A khc nhau. C n1 cch

thc hin 1A ,..., c nk cch thc hin kA . Khi ta c:

n = n1nk cch thc hin ton b k cng vic .

3. Quy tc cng Gi s mt cng vic c th thc hin c k cch

(trng hp) loi tr ln nhau: cch th nht cho n1 kt qu,, cch th k cho nk kt qu. Khi vic thc hin cng vic trn cho n = n1 + + nk kt qu.


4. Phn bit cch chn k phn t t tp c n phn t C 4 cch chn ra k phn t t tp c n phn t, n phn

t ny lun c coi l khc nhau mc d bn cht ca chng c th ging nhau. l: Chn 1 ln ra k phn t v khng n th t ca

chng (T hp).

Chn 1 ln ra k phn t v n th t ca chng (Chnh hp).

Chn k ln, mi ln 1 phn t v khng hon li (s cch chn nh Chnh hp).

Chn k ln, mi ln 1 phn t v c hon li (Chnh hp lp).





b) Chnh hp Chnh hp chp k ca n phn t (0 )k n l mt

nhm (b) c th t gm k phn t khc nhau c chn t n phn t cho.

a) T hp T hp chp k ca n phn t (0 )k n l mt nhm

(b) khng phn bit th t gm k phn t khc nhau c chn t n phn t cho.

S t hp chp k ca n phn t c k hiu v tnh theo cng thc:

( )!

! !

k

n

nC

k n k=

. Quy c: 0! = 1.

Tnh cht: k n kn n

C C = ; 11 1

k k k

n n nC C C = + .

BB ttcc vv ii ss TT hhpp S chnh hp chp k ca n phn t c k hiu v

tnh theo cng thc: !

( 1)...( 1)( )!

k

n

nA n n n k

n k= + =

.

c) Chnh hp lp Chnh hp lp k ca n phn t l mt nhm (b) c th

t gm phn k t khng nht thit khc nhau c chn t n phn t cho.

S cc chnh hp lp k ca n phn t l nk. Nhn xt:

T hp Chnh hp Chnh hp lp ( 1)...( 1)k k k

n nC A n n n k n< = + <


PHN I. L THUYT XC SUT (Probability theory)

Chng 1. XC SUT CA BIN C 1. Bin c ngu nhin 2. Xc sut ca bin c 3. Cng thc tnh xc sut

1. BIN C NGU NHIN 1.1. Hin tng ngu nhin Ngi ta chia cc hin tng xy ra trong i sng hng ny thnh hai loi: tt nhin v ngu nhin.

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc Nhng hin tng m khi c thc hin trong cng mt iu kin s cho ra kt qu nh nhau c gi l nhng hin tng tt nhin.

Chng hn, un nc iu kin bnh thng n 1000C th nc s bc hi; mt ngi nhy ra khi my bay ang bay th ngi s ri xung l tt nhin.

Nhng hin tng m cho d khi c thc hin trong cng mt iu kin vn c th s cho ra cc kt qu khc nhau c gi l nhng hin tng ngu nhin.

Chng hn, gieo mt ht la iu kin bnh thng th ht la c th ny mm cng c th khng ny mm.

Hin tng ngu nhin chnh l i tng kho st ca l thuyt xc sut.

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc 1.2. Php th v bin c quan st cc hin tng ngu nhin, ngi ta cho cc hin tng ny xut hin nhiu ln. Vic thc hin mt quan st v mt hin tng ngu nhin no , xem hin tng ny c xy ra hay khng c gi l mt php th (test).

Khi thc hin mt php th, ta khng th d on c kt qu xy ra. Tuy nhin, ta c th lit k tt c cc kt qu c th xy ra.

Tp hp tt c cc kt qu c th xy ra ca mt php th c gi l khng gian mu ca php th . K hiu l .

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc

VD 1. Xt mt sinh vin thi ht mn XSTK, th hnh ng ca sinh vin ny l mt php th.

Mi phn t c gi l mt bin c s cp. Mi tp A c gi l mt bin c (events).

Tp hp tt c cc im s: {0; 0,5; 1; 1,5;...; 9,5; 10} =

m sinh vin ny c th t l khng gian mu. Cc phn t:

10 = ,

20,5 = ,,

2110 =

l cc bin c s cp. Cc tp con ca :





:A sinh vin ny thi t mn XSTK; :B sinh vin ny thi hng mn XSTK.

Trong mt php th, bin c m chc chn s xy rac gi l bin c chc chn. K hiu l .

Bin c khng th xy ra c gi l bin c rng. K hiu l . VD 2. T nhm c 6 nam v 4 n, ta chn ngu nhinra 5 ngi. Khi , bin c chn c t nht 1 nam l chc chn; bin c chn c 5 ngi n l rng.

{4; 4,5;...; 10}A = , {0; 0,5;...; 3,5}B = ,

l cc bin c. Cc bin c A, B c th c pht biu li l:

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc 1.3. Quan h gia cc bin c a) Quan h tng ng Trong 1 php th, bin c A c gi l ko theo bin c B nu khi A xy ra th B xy ra. K hiu l A B .

Hai bin c A v B c gi l tng ng vi nhaunu A B v B A . K hiu l A B= .

VD 3. Cho trc 5 hp trong 2 hp c qu. ng Xm ln lt 3 hp. Gi:

iA : hp c m ln th i c qu ( 1,2, 3i = );

B : ng X m c hp c qu; C : ng X m c 2 hp c qu; D : ng X m c t nht 1 hp c qu. Khi , ta c:

iA B , B C , C B v B D= .

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc b) Tng v tch ca hai bin c

VD 4. Mt ngi th sn bn hai vin n vo mt con th v con th s cht nu n b trng c hai vin n.

Gi :iA vin n th i trng con th (i = 1, 2);

:A con th b trng n; :B con th b cht.

Tng ca hai bin c A v B l mt bin c, bin c ny xy ra khi A xy ra hay B xy ra trong mt php th (t nht mt trong hai bin c xy ra).

K hiu l A B hay A B+ . Tch ca hai bin c A v B l mt bin c, bin c

ny xy ra khi c A v B cng xy ra trong mt php th. K hiu l A B hay AB .

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc Khi , ta c:

1 2A A A= v

1 2B A A= .

VD 5. Xt php th gieo hai ht la. Gi :

iN ht la th i ny mm;

:i

K ht la th i khng ny mm (i = 1, 2); :A c 1 ht la ny mm. Khi , khng gian mu ca php th l:

1 2 1 2 1 2 1 2{ ; ; ; }K K N K K N N N = .

Cc bin c tch sau y l cc bin c s cp: 1 1 2 2 1 2 3 1 2 4 1 2

, , ,K K N K K N N N = = = = .

Bin c A khng phi l s cp v 1 2 1 2

A N K K N= .

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc c) Bin c i lp

VD 6. T 1 l hng cha 12 chnh phm v 6 ph phm, ngi ta chn ngu nhin ra 15 sn phm.

Gi :iA

chn c i chnh phm, 9,10,11,12i = . Ta c khng gian mu l:

9 10 11 12A A A A = ,

v 10 10 9 11 12

\A A A A A= = .

Trong 1 php th, bin c A c gi l bin c i lp(hay bin c b) ca bin c A nu v ch nu khi Axy ra th A khng xy ra v ngc li, khi A khng xy ra th A xy ra. Vy ta c

\A A=

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc 1.4. H y cc bin c a) Hai bin c xung khc Hai bin c A v B c gi l xung khc vi nhautrong mt php th nu A v B khng cng xy ra.

VD 7. Hai sinh vin A v B cng thi mn XSTK. Gi :A sinh vin A thi ; :B ch c sinh vin B thi ; :C ch c 1 sinh vin thi . Khi ,A v B l xung khc; B v C khng xung khc.

Ch Trong VD 7, A v B xung khc nhng khng i lp.




ChngChng 1. 1. XXcc susutt ccaa BiBinn cc b) H y cc bin c

VD 8. Trn ln 4 bao la vo nhau ri bc ra 1 ht. Gi

iA : ht la bc c l ca bao th i, 1, 4i = .

Khi , h 1 2 3 4

{ ; ; ; }A A A A l y . Ch Trong 1 php th, h { ; }A A l y vi A ty .

Trong mt php th, h gm n bin c { }iA , 1,i n=

c gi l h y khi v ch khi c duy nht bin c

0iA ,

0{1; 2;...; }i n ca h xy ra. Ngha l:

1) ,i jA A i j= v 2)

1 2...

nA A A = .

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc2. XC SUT CA BIN C

Quan st cc bin c i vi mt php th, mc d khng th khng nh mt bin c c xy ra hay khng nhng ngi ta c th phng on kh nng xy ra ca cc bin c ny l t hay nhiu. Kh nng xy ra khch quan ca mt bin c c gi l xc sut (probability) ca bin c .

Xc sut ca bin c A, k hiu l ( )P A , c th cnh ngha bng nhiu dng sau:

dng c in; dng thng k; dng tin Kolmogorov; dng hnh hc.

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc 2.1. nh ngha xc sut dng c in Xt mt php th vi khng gian mu

1{ ;...; }

n =

v bin c A c k phn t. Nu n bin c s cpc cng kh nng xy ra (ng kh nng) th xc sutca bin c A c nh ngha

( )k

P An

= =So trng hp A xay ra

So trng hp co the xay ra

VD 1. Mt cng ty cn tuyn hai nhn vin. C 4 ngi n v 2 ngi nam np n ngu nhin (kh nng trng tuyn ca 6 ngi l nh nhau). Tnh xc sut :

1) c hai ngi trng tuyn u l n; 2) c t nht mt ngi n trng tuyn.


VD 2. T 1 hp cha 86 sn phm tt v 14 ph phm ngi ta chn ngu nhin ra 25 sn phm.

Tnh xc sut chn c: 1) c 25 sn phm u tt; 2) ng 20 sn phm tt.

VD 3. Trong mt vng dn c, t l ngi mc bnh tim l 9%; mc bnh huyt p l 12%; mc c bnh tim v huyt p l 7%. Chn ngu nhin 1 ngi trong vng . Tnh xc sut ngi ny khng mc bnh tim v khng mc bnh huyt p?

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc 2.2. nh ngha xc sut dng thng k Nu khi thc hin mt php th no n ln, thy c

k ln bin c A xut hin th t s kn

c gi l tn

sut ca bin c A.

Khi n thay i, tn sut cng thay i theo nhng lun

dao ng quanh mt s c nh limn

kp

n= .

S p c nh ny c gi l xc sut ca bin c Atheo ngha thng k.

Trong thc t, khi n ln th ( ) kP An

.

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc VD 4. Pearson gieo mt ng tin cn i, ng cht

12.000 ln thy c 6.019 ln xut hin mt sp (tn sut l 0,5016); gieo 24.000 ln thy c 12.012 ln xut hin mt sp (tn sut l 0,5005).

Laplace nghin cu t l sinh trai gi London, Petecbua v Berlin trong 10 nm v a ra tn sutsinh b gi l 21/43.

Cramer nghin cu t l sinh trai gi Thy in trong nm 1935 v kt qu c 42.591 b gi c sinh ra trong tng s 88.273 tr s sinh, tn sut l 0,4825.





2.3. nh ngha xc sut dng hnh hc (tham kho) Cho min . Gi o ca l di, din tch, th tch (ng vi l ng cong, min phng, khi). Xt im M ri ngu nhin vo min .

Gi A: im M ri vo min S , ta c:

( ) .P A =

o o S

o o

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc VD 5. Tm xc sut ca im M ri vo hnh trn ni tip tam gic u c cnh 2 cm. Gii. Gi A: im M ri vo hnh trn ni tip. Din tch ca tam gic l:

222 . 3( ) 3

4dt cm = = .

Bn knh ca hnh trn l:

1 2 3 3.

3 2 3r cm= =

2

3( ) ( ) 0, 6046

3 3 3 3dt S P A

= = = = .

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc VD 6. Hai ngi bn hn gp nhau ti 1 a im xc nh trong khong t 7h n 8h. Mi ngi n (v chc chn n) im hn mt cch c lp, nu khng gp ngi kia th i 30 pht hoc n 8 gi th khng i na. Tm xc sut hai ngi gp nhau.

Gii. Chn mc thi gian 7h l 0. Gi ,x y (gi) l thi gian tng ng ca mi ngi i n im hn, ta c: 0 1, 0 1x y .

Suy ra l hnh vung c cnh l 1 n v.


0,5 0,5 00,5

0,5 0,5 0.

x y x yx y

x y x y

+

Suy ra, min gp nhau gp nhau ca hai ngi l S : {0 1,0 1, 0,5 0, 0,5 0}x y x y x y + .

Vy ( ) 3 75%( ) 4

dt Sp

dt= = =

.

T iu kin, ta c:

2.5. Tnh cht ca xc sut 1) 0 ( ) 1P A , mi bin c A . 2) ( ) 0P = , 3) ( ) 1P = . 4) Nu A B th ( ) ( )P A P B .

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc3. CNG THC TNH XC SUT

3.1. Cng thc cng xc sut Xt mt php th, ta c cc cng thc cng xc sut sau Nu A v B l hai bin c ty th

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

Nu A v B l hai bin c xung khc th

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

Nu h { }iA ( 1,..., )i n= xung khc tng i th

( )1 2 1 2... = ( )+ ( )+...+ ( )n nP A A A P A P A P A


VD 1. Mt nhm c 30 nh u t cc loi, trong c: 13 nh u t vng; 17 nh u t chng khon v 10 nh u t c vng ln chng khon. Mt i tc gpngu nhin mt nh u t trong nhm. Tm xc sut ngi gp c nh u t vng hoc chng khon?

VD 2. Mt hp phn c 10 vin trong c 3 vin mu . Ly ngu nhin t hp ra 3 vin phn. Tnh xc sut ly c t nht 1 vin phn mu .

c bit

( ) 1 ( ); ( ) ( . ) ( . )P A P A P A P AB P AB= = +





3.2. XC SUT C IU KIN Xt php th: 3 ngi A, B v C thi tuyn vo mt

cng ty. Gi A: ngi A thi , B : ngi B thi , C : ngi C thi , H : c 2 ngi thi . Khi , khng gian mu l: { , , , , , , , }ABC ABC ABC ABC ABC ABC ABC ABC .

Ta c: 4

{ , , , } ( )8

A ABC ABC ABC ABC P A= = ;

3{ , , } ( )

8H ABC ABC ABC P H= = .


Lc ny, bin c: 2 ngi thi trong c A l:

{ , }AH ABC ABC= v 2( )8

P AH = .

By gi, ta xt php th l: A, B , C thi tuyn vo mt cng ty v bit thm thng tin c 2 ngi thi .

Khng gian mu tr thnh H v A tr thnh AH .

Gi A H : A thi bit rng c 2 ngi thi th ta

c: ( ) 2 ( )3 ( )

P AHP A H

P H= = .

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc 3.2.1. nh ngha xc sut c iu kin Trong mt php th, xt hai bin c bt k A v B vi

( ) 0P B > . Xc sut ca bin c A sau khi bin c B xy ra c gi l xc sut ca A vi iu kin B , k hiu v cng thc tnh l

( ) ( )( )

P A BP A B

P B=

VD 3. T 1 hp cha 3 bi v 7 bi xanh ngi ta bc ngu nhin ra 2 bi.

Gi A: bc c bi ; B : bc c bi xanh. Hy tnh ( | ), ( | )P A B P B A ?


Nhn xt Khi tnh ( )P A B vi iu kin B xy ra, ngha l ta hn ch khng gian mu xung cn B v hn ch A xung cn A B .

Tnh cht 1) ( )0 1P A B , A ;

2) nu A C th ( ) ( )P A B P C B ;

3) ( ) ( )1P A B P A B= .

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc 3.2.2. Cng thc nhn xc sut a) S c lp ca hai bin c Trong mt php th, hai bin c A v B c gi l c lp nu B c xy ra hay khng cng khng nh hng n kh nng xy ra A v ngc li.

Ch Nu A v B c lp vi nhau th cc cp bin c:

A v B , A v B , A v B cng c lp vi nhau.

b) Cng thc nhn Nu A v B l hai bin c khng c lp th

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

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc Nu A v B l hai bin c c lp th

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

Nu n bin c , 1, ...,iA i n= khng c lp th

1 2 1 2 1 1 1( ... ) ( ) ( | )... ( | ... )

n n nP AA A P A P A A P A A A =

VD 4. Mt ngi c 5 bng n trong c 2 bng b hng. Ngi th ngu nhin ln lt tng bng n (khng hon li) cho n khi chn c 1 bng tt.

Tnh xc sut ngi th n ln th 2. VD 5. Mt sinh vin hc h nin ch c thi li 1 ln nu ln thi th nht b rt (2 ln thi c lp). Bit rngxc sut sinh vin ny thi ln 1 v ln 2 tng ng l 60% v 80%. Tnh xc sut sinh vin ny thi ?




ChngChng 1. 1. XXcc susutt ccaa BiBinn cc VD 6. C hai ngi A v B cng t lnh (c lp) mua c phiu ca mt cng ty vi xc sut mua c tng ng l 0,8 v 0,7. Bit rng c ngi mua c, xc sut ngi A mua c c phiu ny l:

A. 1947

; B. 1219

; C. 4047

; D. 1019

.

VD 7. ng A bn ln lt 2 vin n vo 1 mc tiu v mc tiu s b ph hy nu b trng c 2 vin n. Xc sut vin n th nht trng mc tiu l 0,8. Nu vin th nht trng mc tiu th xc sut vin th hai trng l 0,7. Nu vin th nht khng trng th xc sut vin th hai trng mc tiu l 0,3. Bit rng ng A bn trng, tnh xc sut mc tiu b ph hy ?

ChngChng 1. 1. XXcc susutt ccaa BiBinn cc VD 8. Trong dp tt, ng A em bn 1 cy mai ln v 1 cy mai nh. Xc sut bn c cy mai ln l 0,9. Nu bn c cy mai ln th xc sut bn c cy mai nh l 0,7. Nu cy mai ln khng bn c th xc sut bn c cy mai nh l 0,2. Bit rng ng A bn c t nht 1 cy mai, xc sut ng A bn c c hai cy mai l: A. 0,6342; B. 0,6848; C. 0,4796; D. 0,8791.

VD 9. Hai ngi A v B cng chi tr chi nh sau: C hai lun phin ly mi ln 1 vin bi t mt hp ng 2 bi trng v 4 bi en (bi c ly ra khng tr li hp). Ngi no ly c bi trng trc th thng cuc.

Gi s A ly trc, tnh xc sut A thng cuc ?


a) Cng thc xc sut y Xt h n bin c { }

iA ( 1,2,...,i n= ) y v B l

mt bin c bt k trong php th, ta c ( ) ( )

( )1 1

1

( ) ( ) ... ( )

( )

n n

n

i ii

P B P A P B A P A P B A

P A P B A=

= + +

=

VD 10. Mt ca hng bn hai loi bng n cng kch c gm: 70 bng mu trng vi t l bng hng l 1% v 30 bng mu vng vi t l hng 2%. Mt khch hng chn mua ngu nhin 1 bng n t ca hng ny.

Tnh xc sut ngi ny mua c bng n tt ?

3.2.3. Cng thc xc sut y v Bayes ChngChng 1. 1. XXcc susutt ccaa BiBinn cc

Gii. Gi B : khch chn c bng n tt,

1A : khch chn c bng n mu trng,

2A : khch chn c bng n mu vng.

Suy ra h 1 2

{ , }A A l y . Ta c:

1 1 2 2( ) ( ) ( | ) ( ) ( | )P B P A P B A P A P B A= +

70 30.0,99 .0,98 0,987

70 30 70 30= + =

+ +.

Ch . Trong trc nghim ta gii nhanh nh sau: Nhnh 1: P(n tt mu trng) = 0,7.0,99. Nhnh 2: P(n tt mu vng) = 0,3.0,98. Suy ra P(n tt) = tng xc sut ca 2 nhnh = 0,987.


VD 11. Chung th I c 3 con th trng v 4 con th en, chung II c 5 th trng v 3 th en. Quan st thy c 1 con th chy t chung I sang chung II, sau c 1 con th chy ra t chung II. Tnh xc sut con th chy ra t chung II l th trng ?

VD 12. C mt kho bia km cht lng cha cc thng ging nhau (24 lon/thng) gm 2 loi: loi I ln mi thng 5 lon qu hn s dng v loi II ln mi thng 3 lon qu hn. Bit rng s thng bia loi I bng 1,5 ln s thng bia loi II. Chn ngu nhin 1 thng trong kho v t thng ly ra 10 lon. Tnh xc sut chn phi 2 lon bia qu hn s dng ?


VD 13. Xt tip VD 10. Gi s khch hng chn mua c bng n tt. Tnh xc sut ngi ny mua c bng n mu vng ?

b) Cng thc Bayes Xt h n bin c { }

iA ( 1,2,...,i n= ) y v B l

mt bin c bt k trong php th. Khi , xc sut bin c

iA xy ra sau khi B xy ra l

( ) ( )

( )

( )

1

( ) ( )

( )( )

i i i i

i n

i ii

P A P B A P A P B AP A B

P BP A P B A

=

= =




ChngChng 1. 1. XXcc susutt ccaa BiBinn cc Gii. t tn bin c nh VD 10, ta c:

( ) ( )2 22( ) 0, 3.0,98 14

( ) 0,987 47

P A P B AP A B

P B= = = .

Ch . Nu ta dng s nh VD 10. Khi : P(n vng | tt) = (nhnh 2) chia (tng 2 nhnh).

Phn bit cc bi ton p dng cng thcNhn y Bayes

Trong 1 bi ton, ta xt 3 bin c1 2, , .A A B

1) Nu bi ton yu cu tm xc sut ca1

,A B2A B

th y l bi ton cng thc nhn.

Xc sut l xc sut tch ca tng nhnh.

2) Nu bi ton yu cu tm xc sut ca vB1 2

{ , }A A

y th y l bi ton p dng cng thc y


Xc sut bng tng 2 nhnh.

3) Nu bi ton yu cu tm xc sut ca v cho1 2,A A

bit xy ra, ng thi h y th y

l bi ton p dng cng thc Bayes. Xc sut l t sgia nhnh cn tm vi tng ca hai nhnh.

B 1 2{ , }A A

VD 14. C 20 thng hng ging nhau gm 3 loi: 8 thng loi I, 7 thng loi II v 5 thng loi III. Mi thng hng c 10 sn phm v s sn phm tt tng ng cho mi loi ln lt l 8, 7 v 5. Chn ngu nhin 1 thng hng v t thng ly ra 3 sn phm.


VD 15. Nh my X c 3 phn xng A, B , C tng ng sn xut ra 20%, 30% v 50% tng sn phm ca nh my. Gi s t l sn phm hng do cc phn xng A, B , C tng ng sn xut ra l 1%, 2% v 3%.

Chn ngu nhin 1 sn phm do nh my X sn xut ra.

1) Tnh xc sut (t l) sn phm ny l hng.

1) Tnh xc sut c 2 sn phm ly ra l tt. 2) Tnh xc sut c 2 sn phm ly ra l tt v ca

thng hng loi II. 3) Gi s c 2 sn phm ly ra l tt, tnh xc sut 2

sn phm ny l ca thng hng loi II.


VD 16. T l t ti, t con v xe my i qua ng Xc trm bm du l 5 : 2 : 13. Xc sut t ti, t con v xe my i qua ng ny vo bm du ln lt l 0,1; 0,2 v 0,15. Bit rng c 1 xe i qua ng Xvo bm du, tnh xc sut l t con ?

A. 1157

; B. 1057

; C. 857

; D. 757

.

2) Tnh xc sut sn phm ny hng v do phn xng A sn xut ra.

3) Bit rng sn phm c chn l hng, tnh xc sut sn phm ny l do phn xng A sn xut ra.

ChngChng 2. 2. BiBinn ngnguu nhinnhin 1. Bin ngu nhin v hm mt 2. Hm phn phi xc sut 3. Tham s c trng ca bin ngu nhin

1. BIN NGU NHIN V HM MT 1.1. Khi nim bin ngu nhin Xt mt php th vi khng gian mu . Gi s, ng

vi mi bin c s cp , ta lin kt vi 1 s thc ( )X , th X c gi l mt bin ngu nhin.

Tng qut, bin ngu nhin (BNN) X ca mt php th vi khng gian mu l mt nh x

:X ( )X x = .

Gi tr x c gi l mt gi tr ca bin ngu nhin X .

ChngChng 2. 2. BiBinn ngnguu nhinnhin VD 1. Ngi A mua mt loi bo him tai nn trong 1 nm vi ph l 70 ngn ng. Nu b tai nn th cng ty s chi tr 3 triu ng. Gi X l s tin ngi A c c sau 1 nm mua bo him ny. Khi , ta c

Nu ( )X l 1 tp hu hn 1 2

{ , ,..., }n

x x x hay v hn m c th X c gi l bin ngu nhin ri rc.

cho gn, ta vit l 1 2

{ , ,..., ,...}n

X x x x= .

Php th l: mua bo him tai nn. Bin c l T : ngi A b tai nn. Khng gian mu l { , }T T = .

Vy ( ) 2,93X T = (triu), ( ) 0, 07X T = (triu).




ChngChng 2. 2. BiBinn ngnguu nhinnhin

Ch Trong thc nghim, cc bin ngu nhin thng l ri

rc. Khi bin ngu nhin ri rc X c cc gi tr nhiu trn 1 khong ca , th ta xem X l bin ngu nhin lin tc. Thc cht l, cc bin ngu nhin lin tc c dng lm xp x cho cc bin ngu nhin ri rc khi tp gi tr ca bin ngu nhin ri rc ln.

Cho bin ngu nhin X v hm s ( )y x= . Khi , bin ngu nhin ( )Y X= c gi l hm

ca bin ngu nhin X .

Nu ( )X l 1 khong ca (hay c ) th X c gi l bin ngu nhin lin tc.


a) Bin ngu nhin ri rc Cho BNN ri rc :X ,

1 2{ , ,..., , ...}

nX x x x= .

Gi s 1 2

... ...n

x x x< < < < vi xc sut tng ng l ({ : ( ) }) ( ) , 1,2,...

i i iP X x P X x p i = = = =

Ta nh ngha

1.2. Hm mt

Bng phn phi xc sut ca X l X 1x 2x nx

P 1p 2p np

Hm mt ca X l ,

( )0 , .i i

i

p khi x xf x

khi x x i

==

ChngChng 2. 2. BiBinn ngnguu nhinnhin Ch 0

ip ; 1, 1, 2, ...

ip i= =

Nu 1 2

{ , ,..., ,...}n

x x x x th ( ) 0P X x= = .

( )

i

ia x b

P a X b p<

< = .

VD 2. Cho BNN ri rc X c bng phn phi xc sut X 1 0 1 3 5 P 3a a 0,1 2a 0,3

1) Tm a v tnh ( 1 3)P X < . 2) Lp bng phn phi xc sut ca hm 2Y X= .


VD 3. Mt x th c 4 vin n, bn ln lt tng vin vo mt mc tiu mt cch c lp. Xc sut trng mctiu mi ln bn l 0,8. Bit rng, nu c 1 vin trng mc tiu hoc ht n th dng. Gi X l s vin n x th bn, hy lp bng phn phi xc sut ca X ?

VD 4. Mt hp c 3 vin phn trng v 2 vin phn . Mt ngi ly ngu nhin mi ln 1 vin (khng tr li) t hp ra cho n khi ly c 2 vin phn . Gi X l s ln ngi ly phn. Hy lp bng phn phi xc sut ca X ?

ChngChng 2. 2. BiBinn ngnguu nhinnhin b) Bin ngu nhin lin tc

Ch . ( )f x l hm mt ca bin ngu nhin lin tc X khi v ch khi ( ) 0,f x x v ( ) 1f x dx

+

= . Nhn xt

Khi ( )f x lin tc trn ln cn ca im a , ta c:

( ) ( )

a

a

P a X a f x dx

+

+ =

Hm s ( )f x khng m, xc nh trn c gi l hm mt ca bin ngu nhin lin tc X nu

( ) ( ) ,

A

P X A f x dx A =


0( ) lim ( ) 0

a

a

P X a f x dx

+

= = = .

Vy ( ) ( )P a X b P a X b < = <

( ) ( ) .

b

a

P a X b f x dx= < < =

ngha hnh hc, xc sut ca bin ngu nhin X nhn gi tr trong [ ; ]a b bng din tch hnh thang cong gii hn bi , , ( )x a x b y f x= = = v Ox .

( )f xS

( ) ( )b

a

P a X b f x dx =





VD 5. Chng t 34 , [0; 1]

( ) 0, [0; 1]

x xf x

x

= l hm mt

ca bin ngu nhin X v tnh (0,5 3)P X < ?

VD 6. Cho bin ngu nhin X c hm mt

2

0, 2

( ), 2.

x

f x kx

x





Gi s BNN lin tc X c hm mt 0,

( )( ), .

x af x

x x a


th ca ( )F x :

xO

( )F x

2 1 3 4

0,10, 30, 5

1

VD 1. Cho BNN X c bng phn phi xc sut l X 2 1 3 4 P 0,1 0,2 0,2 0,5

Lp hm phn phi xs ca X v v th ca ( )F x ?


th ca ( )F x :

VD 2. Cho BNN X c hm mt l

2

0, [0; 1]( )

3 , [0; 1].

xf x

x x

/=

Tm hm phn phi xs ca X v v th ca ( )F x ?


2.2. Tnh cht ca hm phn phi xc sut 1) Hm ( )F x xc nh vi mi x . 2) 0 ( ) 1,F x x ; ( ) 0; ( ) 1F F = + = .

4) ( ) ( ) ( )P a X b F b F a < = .

3) ( )F x khng gim v lin tc tri ti mi x . c bit, vi X lin tc th ( )F x lin tc x .

VD 3. Cho BNN X c hm mt l

2

0, 100

( ) 100, 100.

x

f xx

x





VD 5. Cho BNN X c hm mt 23 , [ 1; 3]

( ) 280, [ 1; 3].

x xf x

x

= /

Hm phn phi xc sut ca X l:

A. 3

0, 1

( ) , 1 3281, 3 .

x

xF x x

x




ChngChng 2. 2. BiBinn ngnguu nhinnhin 3.2. K VNG

3.2.1. nh ngha

K vng (Expectation) ca bin ngu nhin X , k hiu EX hay ( )M X , l mt s thc c xc nh nh sau:

Nu X l ri rc vi xc sut ( )i i

P X x p= = th

i ii

EX x p=

Nu X l lin tc c hm mt ( )f x th

. ( )EX x f x dx

+

=


VD 5. Mt l hng gm 10 sn phm tt v 2 ph phm. Ly ngu nhin 4 sn phm t l hng , gi X l s sn phm tt trong 4 sn phm ly ra.

Tm phn phi xc sut v tnh k vng ca X ?

c bit Nu bin ngu nhin ri rc

1 2{ ; ;...; }

nX x x x= c

xc sut tng ng l 1 2, ,...,

np p p th

1 1 2 2...

n nEX x p x p x p= + + +

VD 4. Cho BNN X c bng phn phi xc sut X 1 0 2 3 P 0,1 0,2 0,4 0,3

Tnh k vng ca X ?


VD 7. Tm k vng ca BNN X c hm mt , 0

( )0, 0.

kxe xf x

x

= >

Ch Nu X l BNN lin tc trn [ ; ]a b th [ ; ]EX a b . Nu

1{ ,..., }

nX x x= th:

1 1[min{ ,..., }; max{ ,..., }]

n nEX x x x x .

VD 6. Tm k vng ca BNN X c hm mt 23 ( 2 ), [0; 1]

( ) 40, [0; 1].

x x xf x

x

+ =


VD 8. Cho BNN X c bng phn phi xc sut X

1 2 4 5 7 P

a 0,2 b 0,2 0,1 Tm gi tr ca tham s a v b 3,5EX = ?

VD 9. Cho bin ngu nhin X c hm mt 2, [0; 1]

( )0, [0; 1].

ax bx xf x

x

+ =

Cho bit 0,6EX = hy tnh ( 0,5)P X < ?

ChngChng 2. 2. BiBinn ngnguu nhinnhin 3.2.2. Tnh cht ca K vng

1) ,EC C C= . 2) ( ) . ,E CX C EX C= . 3) ( )E X Y EX EY = .

4) ( . ) .E XY EX EY= nu ,X Y c lp.

VD 10. Cho hai BNN ,X Y c lp c bng ppxs: X 1 1 3 Y 1 2 P 0,3 0,1 0,6 P 0,6 0, 4

Tnh 2( . 3 5 7)E X Y XY Y + + .


K vng ca bin ngu nhin X l gi tr trung bnh(tnh theo xc sut) m X nhn c, n phn nh gi tr trung tm phn phi xc sut ca X .

Trong thc t sn xut hay kinh doanh, khi cn chnphng n cho nng sut hay li nhun cao, ngi ta thng chn phng n sao cho k vng nng sut hay k vng li nhun cao.

3.2.3. ngha ca K vng

VD 11. Thng k cho bit t l tai nn xe my thnh ph H l 0,001. Cng ty bo him A ngh bn loi bo him tai nn xe my cho ng B thnh ph H trong 1 nm vi s tin chi tr l 10 (triu ng), ph bo him l 0,1 (triu ng). Hi trung bnh cng ty A li bao nhiu khi bn bo him cho ng B ?





VD 12. Mt ca hng in my li 2,3 triu ng khi bn c 1 my git, nhng nu my git b hng trc thi hn bo hnh th b l 4,5 triu. Bit rng ca hng li trung bnh 1,96 triu ng khi bn c 1 my git. Tnh t l my git phi bo hnh ?

VD 13. ng A tham gia mt tr chi , en nh sau: Trong mt hp c 4 bi v 6 bi en. Mi ln ng A ly ra 1 bi: nu l th c thng 100 (ngn ng), nu l en th b mt 70 (ngn ng). Hi trung bnh mi ln ly bi ng A b mt bao nhiu tin?


VD 14. Ngi th chp tranh mi tun chp hai bc tranh c lp A v B vi xc sut hng tng ng l 0,03 v 0,05. Nu thnh cng th ngi th s kim li t bc tranh A l 1,3 triu ng v B l 0,9 triu ng, nhng nu hng th b l do bc tranh A l 0,8 triung v do B l 0,6 triu ng. Hi trung bnh ngi th nhn c bao nhiu tin chp tranh mi tun?

A. 2,185 triu ng; B. 2,148 triu ng.

C. 2,116 triu ng; D. 2,062 triu ng.


VD 15. Nhu cu hng ngy ca 1 khu ph v 1 loi thc phm ti sng c bng phn phi xc sut

Nhu cu (kg) 31 32 33 34 P 0,15 0,25 0,45 0,15

Mt ca hng trong khu ph nhp v mi ngy 34 kg loi thc phm ny vi gi 25.000 ng/kg v bn ra vi gi 40.000 ng/kg. Nu b , cui ngy ca hng phi h gi cn 15.000 ng/kg mi bn ht. Gi s ca hng lun bn ht hng, tnh tin li trung bnh ca ca hng ny v loi thc phm trn trong 1 ngy ?


Gi s ( )Y X= l hm ca bin ngu nhin X .

Ch Khi bin ngu nhin X l ri rc th ta nn lp bng phn phi xc sut ca Y , ri tnh EY .

Nu X l bin ngu nhin ri rc th: . ( ).i i i i

i i

EY y p x p= =

Nu X l bin ngu nhin lin tc th:

. ( ) ( ). ( )EY y f x dx x f x dx

+ +

= =

3.2.4. K vng ca hm ca bin ngu nhin


VD 16. Cho BNN X c bng phn phi xc sut X 1 0 1 2 P 0,1 0,3 0,35 0,25

Tnh EY vi 2 3Y X= ?

VD 17. Cho BNN X c hm mt xc sut

2

2, [1; 2]

( )0, [1; 2].

xf x x

x

=

Tnh EY vi 5 2Y XX

= ?


3.3. PHNG SAI 3.3.1. nh ngha

Phng sai (Variance hay Dispersion) ca bin ngu nhin X , k hiu VarX hay ( )D X , l mt s thc khng m c xc nh bi

2 2 2( ) ( ) ( )VarX E X EX E X EX= =

Nu BNN X l ri rc v ( )i i

P X x p= = th 2

2. .i i i i

i i

VarX x p x p =




ChngChng 2. 2. BiBinn ngnguu nhinnhin Nu BNN X l lin tc v c hm mt ( )f x th

2

2. ( ) . ( )VarX x f x dx x f x dx

+ +

=

VD 18. Cho BNN X c bng phn phi xc sut

X 1 2 3 P 0,2 0,7 0,1

Ta c:

2 2 2(1 .0,2 2 .0,7 3 .0,1)VarX = + +

2(1.0, 2 2.0,7 3.0,1) 0,29 + + = .


VD 19. Tnh phng sai ca X , bit hm mt 23 ( 2 ), [0; 1]

( ) 40, [0; 1].

x x xf x

x

+ =

VD 20. Cho BNN X c hm mt xc sut 23 (1 ), 1

( ) 40, 1.

x xf x

x

= >

Tnh phng sai ca Y , cho bit 22Y X= .

ChngChng 2. 2. BiBinn ngnguu nhinnhin 3.3.2. Tnh cht ca Phng sai 1) 0,VarC C= ; 2) 2( ) .Var CX C VarX= ; 3) ( )Var X Y VarX VarY = + nu X v Y c lp. 3.3.3. ngha ca Phng sai

2( )X EX l bnh phng sai bit gia gi tr ca Xso vi trung bnh ca n. V phng sai l trung bnh ca sai bit ny, nn phng sai cho ta hnh nh v s phn tn ca cc s liu: phng sai cng nh th s liu cng tp trung xung quanh trung bnh ca chng.

Trong k thut, phng sai c trng cho sai s ca thit b. Trong kinh doanh, phng sai c trng cho ri ro u t.

ChngChng 2. 2. BiBinn ngnguu nhinnhin Do n v o ca VarX bng bnh phng n v o

ca X nn so snh c vi cc c trng khc,ngi ta a vo khi nim lch tiu chun(standard deviation) l

.VarX =

VD 21. Nng sut (sn phm/pht) ca hai my tng ng l cc BNN X v Y , c bng phn phi xc sut:

X

1 2 3 4 Y

2 3 4 5 P

0,3 0,1 0,5 0,1

P

0,1 0,4 0,4 0,1

T bng phn phi xc sut, ta tnh c: 2, 4EX = ; 1, 04VarX = ; 3,5EY = ; 0,65VarY = .

V , EX EY VarX VarY< > nn nu phi chn mua mt trong hai loi my ny th ta chn mua my Y .

ChngChng 2. 2. BiBinn ngnguu nhinnhin Ch Trong trng hp

EX EY

VarX VarY

th ta khng th so snh c. gii quyt vn ny,

trong thc t ngi ta dng t s tng i .100%

(l trung bnh) so snh s n nh ca cc BNN X v Y . T s tng i cng nh th n nh cng cao.

Ta c: .100% 17,89%xEX

= ; .100% 15,06%y

EY

= .

Vy lp B hc u (n nh) hn lp A.

VD 22. im thi ht mn XSTK ca lp A v B tng ng l cc BNN X v Y . Ngi ta tnh c:

6, 25EX = ; 1, 25VarX = ; 5, 75EY = ; 0, 75VarY = .

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng

1.1. nh ngha Xt tp c N phn t gm

AN phn t c tnh cht A

v A

N N phn t c tnh cht A. T tp , ta chn ra n phn t.

Gi X l s phn t c tnh cht A ln trong n phn t chn th X c phn phi Siu bi (Hypergeometric distribution) vi 3 tham s N ,

AN , n .

K hiu l: ( , , )A

X H N N n hay ( , , ).

AX H N N n

1. Phn phi Siu bi 2. Phn phi Nh thc

3. Phn phi Poisson 4. Phn phi Chun 5. Vector ngu nhin ri rc

1. PHN PHI SIU BI




ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng Xc sut trong n phn t chn ra c k phn t A l

( ) A Ak n k

N N N

k n

N

C Cp P X k

C

= = =

Trong : 0 k n v ( )A A

n N N k N .

VD 1. Mt hp phn gm 10 vin, trong c 7 vin mu trng. Ly ngu nhin 5 vin phn t hp ny. Gi X l s vin phn trng ly c.

Lp bng phn phi xc sut v tnh k vng ca X ? Gii. Ta c: {2; 3; 4; 5}X = v

10, 7, 5 (10, 7, 5)A

N N n X H= = = .

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng Bng phn phi xc sut ca X l

X 2 3 4 5

P

2 37 3

510

C C

C

3 27 3

510

C C

C

4 17 3

510

C C

C

5 07 3

510

C C

C

2 3 3 2 4 1 5 07 3 7 3 7 3 7 3

5 5 5 510 10 10 10

72 3 4 5

2

C C C C C C C CEX

C C C C= + + + = .

1.2. Cc s c trng ca X ~ H(N, NA, n) ;

1

N nEX np VarX npq

N

= =

trong : , 1 .AN

p q pN

= =


VD 2. Mt ca hng bn 100 bng n, trong c 12 bng hng. Mt ngi chn mua ngu nhin 15 bng n t ca hng ny. Hi trung bnh ngi mua c bao nhiu bng n tt ?

VD 3. Ti mt cng trnh c 100 ngi ang lm vic, trong c 70 k s. Chn ngu nhin 40 ngi t cng trnh ny. Gi X l s k s chn c.

1) Tnh xc sut chn c t 27 n 29 k s ? 2) Tnh EX v VarX ?


2. PHN PHI NH THC 2.1. Phn phi Bernoulli a) nh ngha Php th Bernoulli l mt php th m ta ch quan tm n 2 bin c A v A , vi ( )P A p= .

Bng phn phi xc sut ca X l: X

0 1 P q p

Xt bin ngu nhin: 1

( ) 10

AX P A p q

A

= = =

khi xay ra,

khi xay ra,.

Khi , ta ni X c phn phi Bernoulli vi tham s p . K hiu l ( )X B p hay ( )X B p .


VD 1. Mt cu hi trc nghim c 4 phng n tr li, trong ch c 1 phng n ng. Mt sinh vin chn ngu nhin 1 phng n tr li cu hi .

Gi A: sinh vin ny tr li ng. Khi , vic tr li cu hi ca sinh vin ny l mt php th Bernoulli v ( ) 0,25p P A= = , 0,75q = .

Gi BNN 1

0X

=

khi sinh vien nay tra li ung,

khi sinh vien nay tra li sai,

th (0,25)X B v 0,25, 0,1875EX VarX= = .

b) Cc s c trng ca X ~ B(p)

; EX p VarX pq= =


2.2. Phn phi Nh thc a) nh ngha Xt dy n php th Bernoulli c lp. Vi php th

th i , ta xt bin ngu nhin ( )i

X B p ( 1,..., )i n= .

Ngha l, 1

0ii A

Xi A

=

khi lan th xay ra,

khi lan th khong xay ra.

Gi X l s ln bin c A xut hin trong n php th. Khi ,

1...

nX X X= + + v ta ni X c phn phi

Nh thc (Binomial distribution) vi tham s n , p . K hiu l ( , )X B n p hay ( , )X B n p .





VD 2. Mt thi XSTK gm 20 cu hi trc nghimnh trong VD 1. Sinh vin B lm bi mt cch ngu nhin. Bit rng, nu tr li ng 1 cu th sinh vin Bc 0,5 im v nu tr li sai 1 cu th b tr 0,125 im. Tnh xc sut sinh vin B t im 5 ?

Xc sut trong n ln th c k ln A xy ra l

( ) ( 0,1,..., )k k n kk np P X k C p q k n= = = =

b) Cc s c trng ca X ~ B(n, p)

0 0

;

Mod : 1

EX np VarX npq

X x np q x np q

= =

= +


VD 4. Mt nh vn trng 126 cy lan qu, xc sut n hoa ca mi cy trong 1 nm l 0,67.

1) Gi bn 1 cy lan qu n hoa l 2 triu ng. Gi s nh vn bn ht nhng cy lan n hoa th mi nm nh vn thu c chc chn nht l bao nhiu tin?

2) Nu mun trung bnh mi nm c nhiu hn 100 cy lan qu n hoa th nh vn phi trng ti thiu my cy lan qu ?

VD 3. ng B trng 100 cy bch n vi xc sut cy cht l 0,02. Gi X l s cy bch n cht.

1) Tnh xc sut c t 3 n 5 cy bch n cht ? 2) Tnh trung bnh s cy bch n cht v VarX ? 3) Hi ng B cn phi trng ti thiu my cy bch

n xc sut c t nht 1 cy cht ln hn 50% ?


VD 5. C 10 hp phn mu ging nhau, mi hp cha 20 vin phn gm hai loi: 3 hp loi I, mi hp c 12 vin phn ; 7 hp loi II, mi hp c 8 vin phn . Chn ngu nhin 1 hp v t hp ly ln lt ra 5 vin phn (ly vin no xong th tr li vo hp vin ). Tnh xc sut chn c 3 vin phn ?

VD 6. Mt l hng cha 20 sn phm trong c 4 ph phm. Chn lin tip 3 ln t l hng (mi ln chn c hon li), mi ln chn ra 4 sn phm. Tnh xc sut trong 3 ln chn c t nht 1 ln chn phi 2 ph phm ?


3. PHN PHI POISSON 3.1. Bi ton dn n phn phi Poisson Gi s cc v tai nn giao thng vng A xy ra mt

cch ngu nhin, c lp vi nhau v trung bnh 1 ngy c v tai nn. Gi X l s v tai nn giao thng xy ra trong 1 ngy vng A.

Chia 24 gi trong ngy thnh n khong thi gian sao cho ta c th coi rng trong mi khong thi gian c nhiu nht 1 v tai nn xy ra, v kh nng xy ra

tai nn giao thng trong mi khong thi gian bng n

.

Khi , ,X B nn

.


Ta c: ( ) 1k n k

k

nP X k C

n n

= =

( )! 1

. . . 1( ) .! !

nk

k k k

n

nn n nk n k

=

( 1)...( 1). . 1 .! ( )

nk

k

n n n k

k nn

+ =

Suy ra:

( ) . .!

knP X k e

k

=

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng 3.2. nh ngha phn phi Poisson Bin ngu nhin X c gi l c phn phi Poisson tham s 0 > , k hiu l ( )X P hay ( )X P , nu X nhn cc gi tr 0, 1, 2,, n , vi xc sut

.( ) ( 0,1,..., ,...)

!

k

k

ep P X k k n

k

= = = =

l trung bnh s ln xut hin bin c ta quan tm trong mt khong xc nh (khong thi gian hoc khong n v tnh).

VD. Quan st ti mt sn bay thy trung bnh 16 pht c 2 my bay h cnh. Suy ra trong 1gi trung bnh c:





3.3. Cc s c trng ca X ~ P()

0 0Mod : 1

EX VarX

X x x

= =

=

60.27,5

16 = = my bay h cnh.

VD. Trung bnh c 100 sinh vin thi ht mn XSTK c 71 sinh vin thi t. Suy ra 120 sinh vin thi ht mn XSTK th trung bnh c 85,2 sinh vin thi t.

VD 1. Quan st ti siu th A thy trung bnh 5 pht c 18 khch n mua hng.

1) Tnh xc sut trong 7 pht c 25 khch n siu th A ?

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng 2) Tnh xc sut trong 2 pht c t 3 n 5 khch

n siu th A ? 3) Tnh s khch chc chn nht s n siu th A

trong 1 gi ? VD 2. Quan st thy trung bnh 2 pht c 6 t i qua trm thu ph. Bit xc sut c t nht 1 t i qua trm thu ph trong t pht bng 0,9. Gi tr ca t (pht) l:

A. 0,9082 B. 0,8591 C. 0,8514 D. 0,7675. VD 3. C mi ln i cu c th ng A chn ngu nhin 1 trong 2 ni cu. Nu i cu a im I th trung bnh c 10 ln mc mi, ng A cu c 2 con c; cu a im II th trung bnh c 12 ln mc mi, ng A cu c 3 con c.


Hm nay ng A i cu, ng mc mi 20 ln v cu c 5 con c. Tnh xc sut ng A cu c 5 con c a im II ?

VD 4. Ti mt xng dt, trung bnh dt 10 m vi loi B

th b li 13 ch. Chn ln lt 5 xp vi loi B ca xng, mi xp di 6 m. Tnh xc sut 3 trong 5 xp vi y, mi xp vi c ng 7 ch b li ?

VD 5. Quan st thy trung bnh 1 ngy (24 gi) c 12 chuyn tu vo cng A. Chn ngu nhin 6 gi trong 1 ngy. Tnh xc sut 2 trong 6 gi y, mi gi c ng 1 tu vo cng A ?


4. PHN PHI CHUN 4.1. Phn phi chun

a) nh ngha

Bin ngu nhin lin tc X c gi l c phn phi chun (Normal distribution) vi hai tham s v 2 ( 0) > , k hiu l 2( ; )X N hay 2( ; )X N , nu hm mt xc sut ca X c dng

2

2

( )

21

( ) , 2

x

f x e x

=

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng b) Cc s c trng ca X ~ N(, 2)

2Mod ; aX EX V rX= = =

c) Xc sut ca X ~ N(, 2)

2

2

( )

21

( ) ( )2

xb b

a a

P a X b f x dx e dx

= =

Nhn xt. i bin xz =

, ta c:

2 2

2

( )

2 21 1

2 2

bxb z

a a

e d x e d z

=

.

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng 4.2. Phn phi chun tc

a) nh ngha

Bin ngu nhin Z c phn phi chun vi hai tham s 0 = v 2 1 = c gi l c phn phi chun tc

(hay phn phi Gauss), k hiu l (0; 1)Z N hay (0; 1)Z N . Hm mt xc sut ca Z l

2

21( ) ,

2

z

f z e z

=

(Gi tr hm ( )f z c cho trong bng ph lc A).





b) Xc sut ca Z ~ N(0; 1)

Hm Laplace

Hm s 0

( ) ( )

x

x f z dz = c gi l hm Laplace.

(Gi tr hm ( )x c cho trong bng ph lc B ).

Tnh cht ca hm Laplace

Hm ( )x ng bin trn ;

( ) ( )x x = (hm ( )x l);

( ) 0,5 = ; ( ) 0,5 + = .

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng Cng thc tnh xc sut

( ) ( ) ( ) ( )P Z f z dz

= =

Ch ( ) 0,5 ( )P Z < = + ; ( ) 0,5 ( )P Z > = . Nu 4x th ( ) 0,5x .

Nu 2( ; )X N th (0; 1)XZ N=

Vy, ta c cng thc tnh xc sut ca pp chun l

( )b a

P a X b =

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng VD 1. Thi gian X (thng) t chun chiu cao ca loi cy ging A ti mt vn m l bin ngu nhin c phn phi (8; 3)N . T l (xc sut) t chun chiu cao ca loi cy ging A ti vn m ny trong khong t 6 thng n 8,2 thng l:

A. 27,65% B. 31,15% C. 42,27% D. 45,78%. VD 2. Mt k thi u vo trng chuyn A quy nh im l tng s im cc mn thi khng c thp hn 15 im. Gi s tng im cc mn thi ca hc sinh l bin ngu nhin c phn phi chun vi trung bnh 12 im. Bit rng t l hc sinh thi l 25,14%. lch chun l:

A. 4 im; B. 4,5 im; C. 5 im; D. 5,5 im.

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng VD 3. Tc chuyn d liu t my ch ca k tc x n my tnh ca sinh vin vo bui sng ch nht c phn phi chun vi trung bnh 60Kbits/s v lch chun 4Kbits/s. Xc sut tc chuyn d liu ln hn 63Kbits/s l:

A. 0,2266; B. 0,2144; C. 0,1313; D. 0,1060.

VD 4. Cho BNN X c phn phi chun vi 10EX = v (10 20) 0,3P X< < = . Tnh (0 15)P X< ?

VD 5. Thi gian khch phi ch c phc v ti mt ca hng l BNN X (pht), (4,5; 1,21)X N .

1) Tnh xc sut khch phi ch t 3,5 pht n 5 pht ?


2) Tnh thi gian ti thiu t nu xc sut khch phi ch vt qu t l khng qu 5% ?

VD 6. Tui th ca 1 loi my lnh A l BNN X (nm) c phn phi (10; 6,25)N . Khi bn 1 my lnh A th li c 1,4 triu ng nhng nu my lnh phi bo hnh th l 3,8 triu ng. Vy c tin li trung bnh khi bn mi my lnh loi ny l 1 triu ng th cn phi quy nh thi gian bo hnh l bao nhiu ?

5. PHN PHI XC SUT CA VECTOR NGU NHIN RI RC


5.1 Bng phn phi xc sut ng thi ca (X, Y)

Y

X 1y

2y

jy ny Tng dng

1x

11p

12p 1 jp 1np 1p

2x

21p

22p 2 jp 2np 2p

ix

1ip

2ip ijp inp ip

mx

1mp

2mp mjp mnp mp

Tng ct 1p 2p jp np 1




Trong ( );i j ijP X x Y y p= = = v 1 1

1m n

iji j

p= =

= .

T bng phn phi xc sut ng thi ca ( , )X Y ta c: Bng phn phi xc sut ca X

X 1x 2x mx

P 1p 2p mp

Trong 1 2i i i in

p p p p= + + + (tng dng i ca bng phn phi xc sut ng thi).

K vng ca X l: 1 1 2 2

.m m

EX x p x p x p= + + +


5.2. Phn phi xc sut thnh phn (phn phi l)

Bng phn phi xc sut ca Y Y 1y 2y ny

P 1p 2p np

Trong 1 2j j j mjp p p p= + + +

(tng ct j ca bng phn phi xc sut ng thi). K vng ca Y l:

1 1 2 2 .

n nEY y p y p y p= + + +

Y

X 1 2 3

6 0,10 0,05 0,15 7 0,05 0,15 0,10 8 0,10 0,20 0,10

VD 1. Phn phi xc sut ng thi ca vector ngu nhin ( , )X Y cho bi bng:


Gii 1) ( )6 0,1 0, 05 0,15 0, 3P X = = + + = .

1) Tnh ( )6P X = v ( )7, 2P X Y . 2) Lp bng phn phi xs thnh phn v tnh EX , EY .

( )7, 2 {(7,2)}+ {(7,3)}+ {(8,2)}{(8,3)} 0,15 0,1 0,2 0,1 0,55.

P X Y P P P

P

=+ = + + + =

2) Bng phn phi ca X l: X 6 7 8 P 0,3 0,3 0,4

6.0,3 7.0, 3 8.0,4 7,1EX = + + = .

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng Bng phn phi ca Y l:

Y 1 2 3 P 0,25 0,40 0,35

1.0,25 2.0, 4 3.0,35 2,1EY = + + = .


5.3. Phn phi xc sut c iu kin

T cng thc xc sut c iu kin, ta c:

( )

( = , = )= = ,

( )

i j ij

i jj j

P X x Y y pP X x Y y

P Y y p= =

= 1,i m= .

( )

( = , = )= = ,

( )

i j ij

j ii i

P X x Y y pP Y y X x

P X x p= =

= 1,j n= .

Bng phn phi xc sut ca X vi iu kin j

Y y= : X 1x 2x mx

( )= =i jP x YX y 1

j

j

p

p

2

j

j

p

p

mj

j

p

p

K vng ca X vi iu kin

jY y= l:

1 1 2 2

1( ... ).

j j m mjj

EX x p x p x pp

= + + +

Bng phn phi xc sut ca Y vi iu kin i

X x= : Y 1y 2y ny

( )= =j iP Y y X x 1 /i ip p 2 /i ip p /in ip p


VD 2. Cho bng phn phi xs ng thi ca ( , )X Y : Y

X

1 2 3

6 0,10 0,05 0,15 7 0,05 0,15 0,10 8 0,20 0,10 0,10

1) Lp bng phn phi xc sut ca X vi iu kin 2Y =

v tnh k vng ca X . 2) Lp bng phn phi xc sut ca Y vi iu kin

8X = v tnh k vng ca Y .

K vng ca Y vi iu kin i

X x= l:

1 1 2 2

1( ... ).

i i n ini

EY y p y p y pp

= + + +





Gii. 1) Ta c: ( ) 0,05 16

0,05 0,15 0,1 6| 2X YP = = =

+ += .

( ) 0,15 170,05 0,15 0,1 2

| 2X YP = = =+ +

= .

( ) 0,1 180,05 0,15

|0 1 3

2,

YP X = = =+ +

= .

Bng phn phi xc sut ca X vi iu kin 2Y = l: X 6 7 8

( )| 2= =iP YX x 16

1

2

1

3

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng1 1 1 43

6. 7. 8.6 2 3 6

EX = + + = .

2) Bng phn phi xc sut ca Y vi iu kin 8X = : Y 1 2 3

( )| 8= =jP XY y 0,50 0,25 0,25 1.0,5 2.0,25 3.0,25 1,75EY = + + = .

VD 3. Cho vector ngu nhin ri rc ( , )X Y c bng phn phi xc sut ng thi nh sau:

( , )X Y (0; 0) (0; 1) (1; 0) (1; 1) (2; 0) (2; 1)

ijp 1

18

3

18

4

18

3

18

6

18

1

18


Gii. 1) Ta c: 4 1 5

( 1) {(1,0)}+ {(2,1)} +18 18 18

P X Y P P = = = = .

1) Tnh xc sut ( )1P X Y = . 2) Tnh xc sut ( 0 | 1)P X Y> = . 3) Tnh trung bnh ca X v Y . 4) Tnh trung bnh ca Y khi 1X = .

2) ( 0 | =1) ( =1 | =1) ( =2 | =1)P X Y P X Y P X Y> = +

{(1,1)} {(2,1)} 4

( 1) ( 1) 7

P P

P Y P Y= + =

= =.

ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng 3) Bng phn phi thnh phn ca X v Y l:

X 0 1 2 Y 0 1

P

4

18

7

18

7

18

P

11

18

7

18

Vy 4 7 7 210. 1. 2.18 18 18 18

EX = + + = v 718

EY = .

4) Bng phn phi xc sut ca Y khi 1X = l: Y 0 1

|( = )=1jXP Y y 4

7

3

7

Vy 37

EY = .


ChngChng 3. 3. PhnPhn phphii xxcc susutt thngthng ddngng VD 4. Chi ph qung co X (triu ng) v doanh thu

Y (triu ng) ca mt cng ty c bng phn phi xc sut ng thi nh sau:

Y

X

500 (400 600)

700 (600 800)

900 (800 1000)

30 0,10 0, 05 0

50 0,15 0, 20 0, 05

80 0, 05 0, 05 0, 35 Nu doanh thu l 700 triu ng th chi ph qung co trung bnh l:

A. 60,5 triu ng; B. 48,3333 triu ng; C. 51,6667 triu ng; D. 76,25 triu ng.

.

ChngChng 4.4. nhnh ll gigiii hhnn trongtrong xxcc susutt

1. Mt s loi hi t trong xc sut v cc nh l 2. Cc loi xp x phn phi xc sut

1. MT S LOI HI T TRONG XC SUT V CC NH L

(tham kho) 1.1. Hi t theo xc sut Lut s ln a) nh ngha Dy cc bin ngu nhin { }

iX ( 1,..., ,...i n= ) c gi

l hi t theo xc sut n bin ngu nhin X nu: ( ), 0 : lim ( ) ( ) 0.n

nP X X

> =

K hiu: ( ).Pn

X X n




Dy cc bin ngu nhin { }i

X ( 1,..., ,...i n= ) c gi l tun theo lut s ln (dng Tchbyshev) nu:

1 1

1 10 : lim 1

n n

i in

i i

P X EXn n = =

> < = .

b) nh l (Bt ng thc Tchbyshev) Nu bin ngu nhin X c EX = v 2VarX = th:

( )2

20 :P X

>

( )2

21P X

<

.


Chng minh Nu X l bin ngu nhin ri rc, ta c:

2 2( ) ( )x

x f x =

2 2( ) ( ) ( ) ( )x x

x f x x f x

= .


H qu Nu dy cc BNN { }

iX ( 1,..., ,...i n= ) c lp tng i

c i

EX = v 2i

VarX = th 1

1 n Pi

i

Xn =

.

ngha ca nh l Th hin tnh n nh ca trung bnh cc BNN c lp

cng phn phi v c phng sai hu hn. o mt i lng vt l no , ta o n ln v ly

trung bnh cc kt qu lm gi tr thc ca i lng cn o.

p dng trong thng k l: da vo mt mu kh nh kt lun tng th.


1.2. Hi t yu nh l gii hn trung tm a) nh ngha Dy cc bin ngu nhin { }

iX ( 1,..., ,...i n= ) c gi

l hi t yu hay hi t theo phn phi n bin ngu nhin X nu lim ( ) ( ), ( ).

nn

F x F x x C F

=

Trong , ( )C F l tp cc im lin tc ca ( )F x . K hiu: d

nX X hay .d

nF F

Ch Nu P

nX X th d

nX X .





ngha ca nh l S dng nh l gii hn trung tm Liapounop

tnh xp x (gn ng) xc sut. Xc nh cc phn phi xp x gii quyt cc vn ca l thuyt c lng, kim nh,


b) nh l gii hn trung tm (nh l Liapounop)

Cho 1,...,

nX X l cc BNN c lp c cng phn phi

xc sut, vi k vng v phng sai 2 hu hn. Nu

1...

n nS X X= + + th 2,

n nES n VarS n= =

v khi n th 2( ; )dnS X N n n .

2. CC LOI XP X PHN PHI XC SUT ChngChng 4.4. nhnh ll gigiii hhnn trongtrong xxcc susutt

2.1. Xp x phn phi Siu bi bi Nh thc

Xt BNN X c phn phi Siu bi ( ; ; )A

H N N n .

Nu N kh ln v n rt nh so vi N th

( ; ), AN

X B n p pN

=

Ch

Khi c mu n kh nh so vi kch thc N (khong 5%N ) ca tng th th vic ly mu c hon li hay khng hon li l nh nhau.


VD 1. Trong kho, ngi ta ln 500 sn phm loi B vi 1500 sn phm loi A. Chn ngu nhin 40 sn phm t kho ny. Tnh xc sut chn c 30 sn phm loi A ?

VD 2. Mt vn lan c 10.000 cy sp n hoa, trong c 1.000 cy hoa mu .

1) Tnh xc sut khi chn ngu nhin 50 cy lan th c 10 cy c hoa mu . 2) C th tnh xc sut khi chn ngu nhin 300 cy lan th c 45 cy hoa mu c khng ?


2.2. Xp x phn phi Nh thc bi Poisson

Xt bin ngu nhin X c phn phi Nh thc ( ; )B n p . Nu n ln v p gn bng 0 (hoc gn bng 1) th

( ), X P np =

VD 3. Mt l hng tht ng lnh ng gi nhp khu c cha 3% b nhim khun. Tm xc sut khi chn ngu nhin 2.000 gi tht t l hng ny c t 40 n 42 gi b nhim khun ?

VD 4. Gii cu 2) trong VD 2.

Tm tt cc loi xp x ri rc

( , , )A

X H N N n

AN

pN

=

( , )X B n p

( )X P

np =. AN

nN

=

Sai s rt ln

ChngChng 4.4. nhnh ll gigiii hhnn trongtrong xxcc susutt ChngChng 4.4. nhnh ll gigiii hhnn trongtrong xxcc susutt

2.3. Xp x phn phi Nh thc bi phn phi Chun

Xt bin ngu nhin ( ; )X B n p . Nu n ln, p khng qu gn 0 v 1 th

2( ; )X N vi 2, np npq = = .

Khi : 1

( ) .k

P X k f =

(gi tr c cho trong bng A vi ( ) ( )f x f x = )





2 11 2

( )k k

P k X k

(gi tr c cho trong bng B vi ( ) ( )x x = ). Ch

Khi k = , ta s dng cng thc hiu chnh

( ) ( 0,5 0,5)P X k P k X k= +

VD 5. Trong mt t thi tuyn cng chc mt thnh ph c 1.000 ngi d thi vi t l thi t l 80%.

Tnh xc sut : 1) c 172 ngi khng t; 2) c khong 170 n 180 ngi khng t.


VD 6. Mt kho cha 10.000 sn phm trong c 2.000 sn phm khng c kim tra cht lng. Chn ngu nhin t kho ra 400 sn phm. Tnh xc sut trong 400 sn phm :

1) c 80 sn phm khng c kim tra; 2) c t 70 n 100 sn phm khng c kim tra.

VD 7. Ngi ta pht ra 480 giy mi d hi ngh khch hng. Bit rng sc cha ca khn phng l 400 khch v thng ch c 80% khch hng n d. Tnh xc sut tt c khch hng n d u c ch ngi ?

ChngChng 4.4. nhnh ll gigiii hhnn trongtrong xxcc susutt VD 8. Mt khch sn nhn t ch ca 325 khch hng cho 300 phng vo ngy 1/1 v theo kinh nghim ca nhng nm trc cho thy c 10% khch t ch nhng khng n. Bit mi khch t 1 phng, tnh xc sut:

1) c 300 khch n vo ngy 1/1 v nhn phng; 2) tt c khch n vo ngy 1/1 u nhn c phng. VD 9. Mt ca hng bn c ging c 20.000 con c loi da trn trong ln 4.000 con c tra. Mt khch hng chn ngu nhin 1.000 con t 20.000 con c da trn . Tnh xc sut khch hng chn c t 182 n 230 con c tra ?

A. 0,8143; B. 0,9133; C. 0,9424; D. 0,9765.

Tm tt xp x Chun cho Nh thc

( , )X B n pnp =

2( , )X N EX np=

VarX npq=2 npq =

EX =2VarX =

1( ) ,

kP X k f

= =

( ) .b a

P a X b

< < =


PHN II. L THUYT THNG K (Statistical theory)

1. L THUYT MU

Tp hp tt c phn t l cc i tng m ta nghincu c gi l tng th. S phn t ca tng th cgi l kch thc ca tng th (thng rt ln).

Chng V. MU THNG K V C LNG THAM S 1. L thuyt mu

2. c lng khong

1.1. Tng th v Mu

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ss

T tng th ta chn ra n phn t th n phn t c gi l mt mu c kch thc n (c mu). Mu c chn ngu nhin mt cch khch quan c gi l mu ngu nhin.

Khi mu c kch thc ln th ta khng phn bit muc hon li hay khng hon li.

C hai cch ly mu: Mu c hon li: phn t va quan st xong c

tr li cho tng th trc khi quan st ln sau. Mu khng hon li: phn t va quan st xong

khng c tr li cho tng th.




Mu nh tnh l mu m ta ch quan tm n cc phn t ca n c tnh cht A no hay khng.

Mu nh lng l mu m ta quan tm n cc yu t v lng (nh chiu di, cn nng,) ca cc phn t c trong mu.

Gi 1 2, ,...,

nX X X l nhng kt qu quan st. Ta xem

nh quan st n ln, mi ln ta c mt bin ngu nhin ( 1,..., )

iX i n= .

Do ta thng ly mu trong tng th c rt nhiu phn t nn

1 2, ,...,

nX X X c xem l c lp v c cng

phn phi xc sut.

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ss 1.2. Sp xp mu da vo s liu thc nghim a) Sp xp theo dng bng VD 1. Kim tra ngu nhin 50 sinh vin. Ta sp xp im s X thu c theo th t tng dn v s sinh vin n c im tng ng vo bng nh sau:

X (im) 2 4 5 6 7 8 9 10 n (s SV) 4 6 20 10 5 2 2 1

b) Sp xp theo dng khong VD 2. o chiu cao X (cm) ca 100n = thanh nin. V chiu cao khc nhau nn tin vic sp xp, ngita chia chiu cao thnh nhiu khong.


ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ss Cc thanh nin c chiu cao trong cng 1 khong c xem l cao nh nhau. Khi , ta c bng s liu dng khong nh sau: X 148-152 152-156 156-160 160-164 164-168 n 5 20 35 25 15

Khi cn tnh ton, ngi ta chn s trung bnh ca mi khong a s liu trn v dng bng:

X

150 154 158 162 166 n 5 20 35 25 15

Ch i vi trng hp s liu c cho di dng lit k th ta sp xp li dng bng.

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ss 1.3. Cc c trng mu Xt mt mu ngu nhin

1 2( , ,..., )

nX X X , ta c cc

c trng mu nh sau a) Trung bnh mu

1

1 n

n ii

X Xn =

=

n gin, ta dng k hiu n

X X= . b) Phng sai mu Phng sai mu

( )22 21

1 n

n ii

S S X Xn =

= =

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ss Phng sai mu hiu chnh

( )22 21

1

1

n

n ii

S S X Xn =

= =

Trong tnh ton c th, ta s dng cng thc

2 2 2 2( ) ( )1 1

n nS X X S

n n

= =

trong 2 21

1( )

n

ii

X Xn =

= . c) T l mu Xt mu nh tnh vi cc bin ngu nhin

iX

( 1,..., )i n= c phn phi Bernoulli (1; )B p :

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ss0,

i

AX

A

=

1, neu phan t khong co tnh chatneu phan t co tnh chat .

Nu mu c m phn t c tnh cht A th t l mu l

1 2...

nn

X X X mF F

n n

+ + += = =

d) Lin h gia c trng ca mu v tng th Cc c trng mu X , 2S , F l cc thng k dng nghin cu cc c trng 2, , p tng ng ca tng th. T lut s ln ta c:

2 2, , F p X S (theo xc sut).




DDngng mmyy ttnhnh bb ttii ttnhnh cc trngtrng mmuuS DNG MY TNH B TI TNH

CC C TRNG CA MU 1. S liu n (khng c tn s) VD 1. Cho mu c c mu l 5n = :

12; 13; 11; 14; 11.

a) My fx 500 570 MS Xa b nh: SHIFT MODE 3 = = Vo ch thng k nhp d liu: MODE 2 (chn SD i vi fx500MS); MODE MODE 1 (chn SD i vi fx570MS). Nhp cc s:

12 M+ 13 M+ 11 M+ 14 M+ 11 M+

DDngng mmyy ttnhnh bb ttii ttnhnh cc trngtrng mmuu

Xut kt qu: SHIFT 2 1 =

(kt qu x l trung bnh mu). SHIFT 2 2 =

(kt qu x n l lch chun ca mu s ). SHIFT 2 3 =

( 1x n l lch chun ca mu c hiu chnh s ).

b) My fx 500 570 ES Xa b nh: SHIFT 9 3 = = Vo ch thng k nhp d liu: SHIFT MODE dch chuyn mi tn tm chn mc Stat 2 (ch khng tn s).

DDngng mmyy ttnhnh bb ttii ttnhnh cc trngtrng mmuu MODE 3 (stat) 1 (1-var) (nhp cc s):

12= 13= 11= 14= 11= AC Xut kt qu: SHIFT 1 5 (var) 1 = (n : c mu) SHIFT 1 5 (var) 2 = (x ) SHIFT 1 5 (var) 3 = ( x n s = ). SHIFT 1 5 (var) 4 = ( 1x n s = ).

2. S liu c tn s VD 2. Cho mu c c mu l 9n = nh sau:

X 12 11 15 n 3 2 4


a) My fx 500 570 MS

Xa b nh: SHIFT MODE 3 = = Vo ch thng k nhp d liu: MODE 2 (chn SD i vi fx500MS); MODE MODE 1 (chn SD i vi fx570MS). Nhp cc s:

12 SHIFT , 3 M+ 11 SHIFT , 2 M+ 15 SHIFT , 4 M+

Xut kt qu, ta lm nh 1a).

DDngng mmyy ttnhnh bb ttii ttnhnh cc trngtrng mmuu b) My fx 500 570 ES

Xa b nh: SHIFT 9 3 = = Vo ch thng k nhp d liu: SHIFT MODE (SETUP) dch chuyn mi tn 4 1 MODE 3 (stat) 1 (1-var) Nhp cc gi tr v tn s vo 2 ct trn mn hnh:

X FREQ 12 3 11 2 15 4 AC

Xut kt qu, lm nh 1b).


VD 3. iu tra nng sut ca 100 ha la trong vng A, ta c bng s liu sau:

Nng sut (tn/ha)

3 - 3,5

3,5 - 4

4 - 4,5

4,5 - 5

5 - 5,5

5,5 - 6

6 - 6,5

6,5 - 7

Din tch(ha) 7 12 18 27 20 8 5 3 Nhng tha rung c nng sut t hn 4,4 tn/ha l c

nng sut thp. Dng my tnh b ti tnh: 1) t l din tch la c nng sut thp; 2) nng sut la trung bnh, phng sai mu cha hiu

chnh v lch chun ca mu c hiu chnh.




DDngng mmyy ttnhnh bb ttii ttnhnh cc trngtrng mmuuGii

Bng s liu c vit li: Nng sut

(tn/ha)

3,25

3,75

4,25

4,75

5,25

5,75

6,25

6,75

Din tch(ha)

7

12

18

27

20

8

5

3

1) 7 12 18 37%100

mf

n

+ += = = .

2) 24,75; 0,685; 0, 8318x s s= = = .

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ssKHI NIM CHUNG V C LNG

c lng l phng on mt gi tr cha bit ca tng th da vo quan st trn mu ly ra t tng th . Thng thng, ta cn c lng v trung bnh, t l, phng sai, h s tng quan ca tng th.

C hai hnh thc c lng: c lng im: kt qu cn c lng c cho

bi mt tr s. c lng khong: kt qu cn c lng c cho

bi mt khong. c lng im c u im l cho ta mt gi tr c

th, c th dng tnh cc kt qu khc, nhng nhc im l khng cho bit sai s ca c lng.

c lng khong th ngc li.


Bi ton i tm khong c lng cho c gi l bi ton c lng khong.

Xc sut 1 c gi l tin cy ca c lng,

2 12 = c gi l di ca khong c lngv c gi l chnh xc ca c lng.

2. C LNG KHONG 2.1. nh ngha Xt thng k T c lng tham s v mt c tnh X no ca tng th, khong

1 2( ; ) c gi l

khong c lng nu vi xc sut 1 cho trc th

1 2( ) 1P < < = .


Trng hp 1. C mu 30n > v phng sai tng th 2 bit.

T mu ta tnh trung bnh mu x .

T /2 /2

11 ( )

2

Bz z

= tra bang .

2.2. c lng khong cho trung bnh tng th

Xt c tnh X ca tng th c trung bnh cha bit. Vi tin cy 1 cho trc, ta i tm khong c lng cho l

1 2( ; ) tha

1 2( ) 1P < < = .

Trong thc hnh, ta c 4 trng hp sau

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ss Khong c lng l ( ; )x x + , trong

/2z

n

=

Trng hp 2. C mu 30n > v phng sai tng th 2 cha bit. Tnh x v s ( lch chun ca mu hiu chnh). T

/2 /2

11 ( )

2

Bz z

= tra bang .

Khong c lng l ( ; )x x + , trong

/2

sz

n =

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ss Ch

Mi lin h gia lch chun mu hiu chnh s v cha hiu chnh s l:

2 2 2 1 1

n ns s s s

n n= =

Trng hp 3. C mu 30n , 2 bit v X c phn phi chun, ta lm nh trng hp 1.

Trng hp 4. C mu 30n , 2 cha bit v X c phn phi chun.

T mu ta tnh ,x s .





T 1/2

1 C nt tra bang

(nh gim bc thnh 1n ri mi tra bng!) Khong c lng l ( ; )x x + , trong

1/2n st

n

=

Sai s chun

Nu chng ta chn mu ngu nhin N ln (N l s rt ln), mi ln vi n i tng th chng ta s c N s trung bnh. lch chun ca N s trung bnh ny c gi l sai s chun. Sai s chun phn nh dao ng hay bin thin ca cc s trung bnh mu.

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ss ngha ca c lng khong

Xt c tnh X ca tng th c trung bnh cha bit. Ta chn mu ngu nhin gm n phn t ca tng th v tnh c x , s . Khi

1 s phn t ca tng th c c tnh X dao ng trong khong t

/2x z s n /2x z s+ ;

s trung bnh v c tnh X ca tt c cc phn t ca tng th dao ng trong khong t

/2

sx z

n n /2

sx z

n+ vi xc sut l 1 .

lch chun phn nh bin thin ca mt s phn t trong mt tng th. Cn sai s chun phn nh dao ng ca cc s trung bnh chn t tng th.

CC BI TON V C LNG KHONGBi 1. c lng khong

Ty theo bi ton thuc trng hp no, ta s dngtrc tip cng thc ca trng hp .

Bi 2. Tm tin cy (ta khng xt TH4)

/2 /2.

s nz z

sn= =

/2 /2

1( ) 1 2 ( ).

2z z

= =


/2 /2;

nz z

n= =

Gii phng trnh:

hay

Tra bng B, ta suy ra:

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ssBi 3. Tm c mu (ta ch xt TH1 v TH2)

Ta c nh s (hay ) tm c mu N.

a) Nu > th ta gii bt ng thc:2

/2 /2 max.

s sz N z N

N

> <

b) Nu < th ta gii bt ng thc:2

/2 /2 min.

s sz N z N

N

< >

VD 1. Lng Vitamin c trong tri cy A l bin ngu nhin X (mg) c lch chun 3,98 mg. Phn tch 250 tri cy A th thu c lng Vitamin trung bnh l 20mg. Vi tin cy 95%, hy c lng lng Vitamin trung bnh c trong mi tri cy A ?


VD 2. Bit chiu cao ca con ngi l bin ngu nhin X (cm) c phn phi chun ( ; 100)N . Vi tin cy 95%, nu mun c lng chiu cao trung bnh ca dn s c chnh xc khng qu 1 cm th phi cn o t nht my ngi ?

VD 3. Kim tra tui th (tnh bng gi) ca 50 bng n do nh my A sn xut ra, ngi ta c bng s liu

Tui th 3.300 3.500 3.600 4.000 S bng n 10 20 12 8

1) Hy c lng tui th trung bnh ca loi bng n do nh my A sn xut vi tin cy 97% ?


2) Da vo mu trn c lng tui th trung bnhca loi bng n do nh my A sn xut c chnh xc 59,02 gi th m bo tin cy l bao nhiu ?

3) Da vo mu trn, nu mun c lng tui thtrung bnh ca loi bng n do nh my A sn xut c chnh xc nh hn 40 gi vi tin cy 98% th cn phi kim tra ti thiu bao nhiu bng n na ?

VD 4. Chiu cao ca loi cy A l bin ngu nhin c phn phi chun. Ngi ta o ngu nhin 20 cy A th thy chiu cao trung bnh 23,12 m v lch chun ca mu cha hiu chnh l 1,25 m.

Tm khong c lng chiu cao trung bnh ca loi cy A vi tin cy 95%?




ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ssVD 5. nghin cu nhu cu v loi hng X phng A ngi ta tin hnh kho st 400 trong ton b 2500 gia nh. Kt qu kho st l:

Nhu cu (kg/thng) 0,5 1,5 2,5 3,5 S gia nh 10 35 86 132

Nhu cu (kg/thng) 4,5 5,5 6,5 7,5 S gia nh 78 31 18 10

1) Hy c lng nhu cu trung bnh v loi hng X ca ton b gia nh phng A trong 1 nm vi tin cy 95%?

2) Vi mu kho st trn, nu c lng nhu cu trung bnh v loi hng X ca phng A vi chnh xc ln hn 3 tn/nm v tin cy 99% th cn kho st ti a bao nhiu gia nh trong phng A ?


VD 6. o ng knh ca 100 trc my do 1 nh my sn xut th c bng s liu: ng knh (cm) 9,75 9,80 9,85 9,90

S trc my 5 37 42 16 1) Hy c lng trung bnh ng knh ca trc my

vi tin cy 97% ? 2) Da vo mu trn c lng trung bnh ng

knh ca trc my c chnh xc 0,006cm th m bo tin cy l bao nhiu ?

3) Da vo mu trn, nu mun c lng trung bnh ng knh ca trc my c chnh xc ln hn 0,003cm vi tin cy 99% th cn phi o ti a bao nhiu trc my na ?


VD 7. Tin hnh kho st 420 trong tng s 3.000 gia nh mt phng th thy c 400 gia nh dng loi sn phm X do cng ty A sn xut vi bng s liu: S lng (kg/thng) 0,75 1,25 1,75 2,25 2,75 3,25

S gia nh 40 70 110 90 60 30 Hy c lng trung bnh tng khi lng sn phm X do cng ty A sn xut c tiu th phng ny trong mt thng vi tin cy 95%? A. (5612,7kg; 6012,3kg); B. (5893,3kg; 6312,9kg); C. (5307,3kg; 5763,9kg); D. (5210,4kg; 5643,5kg).


T c mu n v s phn t c tnh cht A trong mu

l m , ta tnh c t l mu mfn

= .

Khong c lng cho p l ( ; )f f + , trong

/2

(1 )f fz

n

=

Gi s t l p cc phn t c tnh cht A ca tng th cha bit. Vi tin cy 1 cho trc, khong c lng p l

1 2( ; )p p tha mn

1 2( ) 1P p p p< < = .

2.3. c lng khong cho t l tng th


VD 9. c lng s c c trong mt h ngi ta bt ln 10.000 con, nh du ri th li xung h. Sau mt thi gian, li bt ln 8.000 con c thy 564 con c nh du. Vi tin cy 97%, hy c lng t l c c nh du v s c c trong h ?

VD 8. Mt tri g ty ang nui 250.000 con g trng 22 tun tui. Cn th 160 con g trng ny th thy c 138 con t chun (nng hn 12 kg). Vi tin cy 95%, hy c lng s g trng ca tri t chun ?

ChngChng 5.5. MMuu ththngng kk & & cc llngng thamtham ss VD 10. Ngi ta chn ngu nhin 500 chic tivi trong

mt kho cha TV th thy c 27 TV Sony. 1) Da vo mu trn, c lng t l TV Sony trong

kho c chnh xc l 1,77% = th m bo tin cy ca c lng l bao nhiu?

2) Da vo mu trn, nu mun c chnh xc ca c lng t l TV Sony nh hn 0,01 vi tin cy 95% th cn chn thm t nht bao nhiu TV na?

VD 11. Ly ngu nhin 200 sn phm trong kho hng A th thy c 21 ph phm.

1) Da vo mu trn, c lng t l ph phm trong kho A c chnh xc l 3,5% = th m bo tin cy ca c lng l bao nhiu?





2) Nu mun chnh xc c lng t l ph phm nh hn 1% vi tin cy 93% th cn kim tra thm t nht bao nhiu sn phm na?

VD 12. Tha rung c nng sut la trn 5,5 (tn/ha) l rung c nng sut cao. Kho st nng sut X (tn/ha) ca 100 ha la huyn A, ta c bng s liu

X

3,25 3,75 4,25 4,75 5,25 5,75 6,25 6,75 S (ha) 7 12 18 27 20 8 5 3

c lng t l din tch la c nng sut cao huyn A c chnh xc l 8,54% = th m bo tin cy l bao nhiu? A. 92%; B. 94%; C. 96%; D. 98%.

ChngChng 6.6. KiKimm nhnh GiGi thuythuytt ThThngng kk 1. Khi nim v kim nh gi thuyt thng k 2. Kim nh so snh c trng vi mt s 3. Kim nh so snh hai c trng

1. KHI NIM V KIM NH GI THUYT THNG K

1.1. Khi nim chung M hnh tng qut ca bi ton kim nh l: ta nu ln

hai mnh tri ngc nhau, mt mnh c gi l gi thuyt H v mnh cn li c gi l nghch thuyt (hay i thuyt) H .

Gii quyt mt bi ton kim nh l: bng cch da vo quan st mu, ta nu ln mt quy tc hnh ng, ta chp nhn gi thuyt H hay bc b gi thuyt H .

ChngChng 6.6. KiKimm nhnh GiGi thuythuytt ThThngng kk

Khi ta chp nhn gi thuyt H , ngha l ta tin rng Hng; khi bc b H , ngha l ta tin rng H sai. Do chda trn mt mu quan st ngu nhin, nn ta khng th khng nh chc chn iu g cho tng th.

Trong chng ny, ta ch xt loi kim nh tham s (so snh c trng vi 1 s, so snh hai c trng ca hai tng th).

1.2. Cc loi sai lm trong kim nh Khi thc hin kim nh gi thuyt, ta da vo quan

st ngu nhin mt s trng hp ri suy rng ra cho tng th. S suy rng ny c khi ng, c khi sai.

Thng k hc phn bit 2 loi sai lm sau:


a) Sai lm loi I Sai lm loi 1 l loi sai lm m ta phm phi trong

vic bc b gi thuyt H khi H ng. Xc sut ca vic bc b H khi H ng l xc sut

ca sai lm loi 1 v c k hiu l .

b) Sai lm loi II Sai lm loi 2 l loi sai lm m ta phm phi trong

vic chp nhn gi thuyt H khi H sai. Xc sut ca vic chp nhn gi thuyt H khi H sai l

xc sut ca sai lm loi 2 v c k hiu l .


c) Mi lin h gia hai loi sai lm Khi thc hin kim nh, ta lun mun xc sut phm

phi sai lm cng t cng tt. Tuy nhin, nu h thp th s tng ln v ngc li.

Trong thc t, gia hai loi sai lm ny, loi no tc hi hn th ta nn trnh.

Trong thng k, ngi ta quy c rng sai lm loi 1 tc hi hn loi 2 nn cn trnh hn. Do , ta ch xt cc php kim nh c khng vt qu mt gi tr n nh trc, thng thng l 1%; 3%; 5%;

Gi tr cn c gi l mc ngha ca kim nh.


1.3. C s l thuyt ca kim nh gii quyt bi ton kim nh, ta quan st mu ngu

nhin 1,...,

nX X v a ra gi thuyt H .

T mu trn, ta chn thng k 1 0

( ,..., ; )n

T f X X=

sao cho nu khi H ng th phn phi xc sut ca Thon ton xc nh.

Vi mc ngha , ta tm c khong tin cy (hay khong c lng) [ ; ]a b cho T tin cy 1 .

Khi : nu [ ; ]t a b th ta chp nhn gi thuyt H ; nu [ ; ]t a b th ta bc b gi thuyt H .





( ) ( )2

P T t P T t

= = .

Nu th hm mt T i xng qua trc xc sut th ta chn khong i xng [ ; ]t t , vi:

Vy, khi xt na bn phi ca trc xc sut th ta c: nu t t th ta chp nhn gi thuyt H ; nu t t> th ta bc b gi thuyt H .

Nu th hm mt T khng i xng th ta chn khong tin cy [0; ]C , vi ( )P T C = . Nu t C th ta chp nhn gi thuyt H , v nu t C> th ta bc b gi thuyt H .

ChngChng 6.6. KiKimm nhnh GiGi thuythuytt ThThngng kk2. KIM NH SO SNH C TRNG

CA TNG TH VI MT S 2.1. Kim nh so snh trung bnh vi mt s Xt c tnh X ca tng th. Gi s ta cn so snh trung bnh ca tng th v c tnh X vi s

0 , ta

t gi thuyt 0

:H = . Ta c 4 trng hp sau

Trng hp 1. C mu 230, n > bit.

T mc ngha /2 /2

1( )

2

Bz z

= .

Tnh gi tr thng k 0| |x

z

n

=


Nu /2

z z th ta chp nhn H , ngha l 0 = ; nu

/2z z> th ta bc b H , ngha l 0 .

Trng hp 2. C mu 230, n > cha bit. Ta lm nh trng hp 1 nhng thay bi s .

Trng hp 3. C mu 230, n bit v X c phn phi chun, ta lm nh trng hp 1.

Trng hp 4. C mu 230, n cha bit v X c phn phi chun.

T c mu n v mc ngha 1/2

C nt tra bang

.


Tnh gi tr thng k 0

| |xt

s

n

=

Nu 1/2nt t th ta chp nhn gi thuyt H ;

1/2nt t > th ta bc b gi thuyt H .

Ch Trong tt c cc trng hp bc b, ta so snh x v

0 :

nu 0

x > th ta kt lun 0

> ; nu

0x < th ta kt lun

0 < .

ChngChng 6.6. KiKimm nhnh GiGi thuythuytt ThThngng kk VD 1. S in lc A bo co rng: trung bnh mi h hng thng phi tr 250 ngn ng tin in, vi lch chun l 20 ngn. Ngi ta kho st ngu nhin 500 h th tnh c trung bnh hng thng mi h tr 252 ngn ng tin in. Trong kim nh gi thuyt H : trung bnh mi h phi tr hng thng l 250 ngn ng tin in vi mc ngha 1%= , hy cho bit gi tr thng k v kt lun ?

VD 2. Nh Gio dc hc B mun nghin cu xem s gi t hc trung bnh hng ngy ca sinh vin c thay i khng so vi mc 1 gi/ngy cch y 10 nm. ng B kho st ngu nhin 120 sinh vin v tnh c trung bnh l 0,82 gi/ngy vi 0,75s = gi/ngy.

ChngChng 6.6. KiKimm nhnh GiGi thuythuytt ThThngng kk Vi mc ngha 3%, hy cho bit kt lun ca ng B ? VD 3. Trong mt nh my go, trng lng ng bao theo quy nh ca 1 bao go l 50 kg v lch chun l 0,3 kg. Cn th 296 bao go ca nh my ny th thy trng lng trung bnh l 49,97 kg. Kim nh gi thuyt H : trng lng trung bnh mi bao go ca nh my ny l 50 kg c gi tr TK z v kt lun l:

A. 1,7205z = ; chp nhn H vi mc ngha 6%. B. 1,7205z = ; bc b H , trng lng thc t ca

bao go nh hn 50 kg vi mc ngha 6%. C. 1,9732z = ; chp nhn H vi mc ngha 4%. D. 1,9732z = ; bc b H , trng lng thc t ca

bao go nh hn 50 kg vi mc ngha 4%.




ChngChng 6.6. KiKimm nhnh GiGi thuythuytt ThThngng kk VD 4. Mt cng ty cho bit mc lng trung bnh ca 1 k s cng ty l 5,7 triu ng/thng vi lch chun 0,5 triu ng/thng. K s A thm d 18 k s cng ty ny th thy lng trung bnh l 5,45 triu ng/thng. K s A quyt nh rng: nu mc lng trung bnh thc s bng hay cao hn mc lng cng ty a ra th np n xin lm. Vi mc ngha 5%, cho bit kt lun ca k s A ?

VD 5. Ngi ta kim tra ngu nhin 38 ca hng ca cng ty A v c bng doanh thu trong 1 thng l

X (triu ng/thng) 200 220 240 260 S ca hng 8 16 12 2

ChngChng 6.6. KiKimm nhnh GiGi thuythuytt ThThngng kk Kim nh gi thuyt H : doanh thu trung bnh hng thng ca mi ca hng cng ty l 230 triu ng, mc ngha ti a gi thuyt H c chp nhn l: A. 3,4%; B. 4,2%; C. 5,6%; D. 7,8%.

VD 6. im trung bnh mn Ton ca sinh vin nm trc l 5,72. Nm nay, theo di 100 SV c s liu

im 3 4 5 6 7 8 9

S sinh vin 3 5 27 43 12 6 4 Kim nh gi thuyt H : im trung bnh mn Ton ca sinh vin nm nay bng nm trc, mc ngha ti a H c chp nhn l:

A. 13,94%; B. 13,62%; C. 11,74%; D. 11,86%.


VD 7. Thi gian X (pht) gia hai chuyn xe bus trong thnh ph l bin ngu nhin c phn phi chun. Cng ty xe bus ni rng: trung bnh c 5 pht li c 1 chuyn xe bus. Ngi ta chn ngu nhin 8 thi im v ghi li thi gian (pht) gia hai chuyn xe bus l:

5,3; 4,5; 4,8; 5,1; 4,3; 4,8; 4,9; 4,7.

Vi mc ngha 7%, hy kim nh li ni trn ?

VD 8. Chiu cao cy ging X trong mt vm m l bin ngu nhin c phn phi chun. Ngi ta o ngu nhin 25 cy ging ny v c bng s liu:

ChngChng 6.6. KiKimm nhnh GiGi thuythuytt ThThngng kk Theo quy nh ca vn m, khi no chiu cao trung bnh ca cy hn 1 m th em ra trng. Vi mc ngha 5%, kim nh gi thuyt H : cy ging ca vn m cao trung bnh 1 m c gi tr thng k v kt lun l:

A. 2,7984t = , khng nn em cy ra trng. B. 2,7984t = , nn em cy ra trng. C. 1, 9984t = , khng nn em cy ra trng. D. 1, 9984t = , nn em cy ra trng.

2.2. Kim nh so snh t l vi mt s

Gi s ta cn so snh t l p v tnh cht A no ca tng th X vi s

0p , ta t gi thuyt

0:H p p= .


T mc ngha /2 /2

1( )

2

Bz z

= .

Tnh t l mu mfn

= v gi tr thng k

0

0 0(1 )

f pz

p p

n

=

Nu /2

z z th ta chp nhn H , ngha l 0p p= . Nu

/2z z> th ta bc b H , ngha l 0p p .

Khi : 0 0

f p p p> > ; 0 0

f p p p< < .

ChngChng 6.6. KiKimm nhnh GiGi thuythuytt ThThngng kk VD 9. Mt bo co cho bit c 58% ngi tiu dng Vit Nam quan tm n hng Vit. Kho st ngu nhin1.000 ngi dn Vit Nam thy c 612 ngi c hi l c quan tm n hng Vit. Vi mc ngha 5%, hy kim nh li bo co trn ?

VD 10. Kho st ngu nhin 400 sinh vin v mc nghim tc trong gi hc th thy 13 sinh vin tha nhn c ng trong gi hc. Trong kim nh gi thuyt H : c 2% sinh vin ng trong gi hc, mc ngha ti a l bao nhiu H c chp nhn ?

VD 11. kim tra 1 loi sng th thao, ngi ta cho bn 1.000 vin n vo 1 tm bia th c 670 vin trng mc tiu. Sau , ngi ta ci tin k thut v kim tra li th thy t l trng ca sng lc ny l 70%.





VD 12. Cng ty A tuyn b rng c 40% ngi tiu dng a thch sn phm ca mnh. Mt cuc iu tra 400 ngi tiu dng thy c 179 ngi a thch sn phm ca cng ty A. Trong kim nh gi thuyt H : c 40% ngi tiu dng thch sn phm ca cng ty A, mc ngha ti a H c chp nhn l: A. 7,86%; B. 6,48%; C. 5,24%; D. 4,32%.

Trong kim nh gi thuyt H : t l bn trng ca loi sng th thao ny trc ci tin l 70%, vi mc ngha 3% c gi tr thng k v kt lun l:

A. 2, 0702z = v ci tin k thut l tt. B. 2, 0702z = v ci tin k thut l cha tt. C. 2, 0176z = v ci tin k thut l tt. D. 2, 0176z = v ci tin k thut l cha tt.


1) Da vo mu trn, nu mun c lng lng ko trung bnh c bn ra hng ngy ca hng ny c chnh xc l 1,2309 kg th m bo tin cy l bao nhiu?

2) Hy c lng t l nhng ngy ca hng bn c nhiu hn 40 kg vi tin cy 90%?

3) Bng cch h gi bn, ca hng bn c lng ko trung bnh hng ngy l 40,5 kg. Vi mc ngha 5% hy cho kt lun thc t v vic h gi ny?

BI TP TNG HP Bi 1. Kim tra lng ko X c bn ra hng ngy

ti mt ca hng, c kt qu X (kg) 25 30 35 40 45 50 55

S ngy 9 23 27 30 25 20 5

ChngChng 6.6. KiKimm nhnh GiGi thuythuytt ThThngng kk Bi 2. Lng cht m ca g ta th vn t chun l cao hn hay bng 252g mi con. Kho st lng cht m X (g) ca loi g ny ti mt nng tri, c kt qu

X (g) 248 250 252 254 256 258 S con 5 17 31 45 19 3

1) c lng lng cht m trung bnh c trong mi con g ca nng tri trn vi tin cy 95% ?

2) c th ni t l g t chun v cht m ti nng tri trn l 87% th mc ngha ti a l bao nhiu ?

3) Ngi ta th nui nht loi g ny cho n lc g c cng trng lng vi cch nui nh trn th thy lng m trung bnh l 249g. Vi mc ngha 5% hy cho kt lun thc t ca cch nui nht ?


3. KIM NH SO SNH HAI C TRNG CA HAI TNG TH

3.1. So snh hai trung bnh ca hai tng th

Xt hai c tnh ,X Y ca hai tng th. Gi s ta cn so snh hai trung bnh tng ng l

x v

y , ta t

gi thuyt :x y

H = . C 4 trng hp v vic chp nhn hay bc b H ta u lm nh kim nh so snh trung bnh vi 1 s.

Trng hp 1. C mu 30, 30x yn n> > v

phng sai 2 2, x y bit.


Ta tnh thng k 22

| |

yx

x y

x yz

n n

=

+

v so snh vi /2

z .

Trng hp 2. , 30x yn n > v 2 2,

x y cha bit.

Ta thay 2 2, x y trong trng hp 1 bi 2 2,

x ys s .

Trng hp 3. , 30x yn n v 2 2,

x y bit

X , Y c phn phi chun. Ta lm nh trng hp 1.


Trng hp 4. , 30x yn n

v 2 2, x y cha bit

xstkdh-sv14.pdf

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