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8/8/2019 Xc sut thng k CTU

1/23

Cao Ho Thi 43

CHNG 5

BIN NGU NHIN V PHN PHI XC SUT(Random Variables and Probability Distributons)

5. NH NGHA BIN NGU NHIN (Random Variable)5.1.1. nh ngha Bin ngu nhin l nhng bin m gi tr ca n c xc nh mt cch ngu nhin. V mt ton hc, nu mi bin c sng A thuc tp hp bin c no y c th

t tng ng vi mt i lng xc nh X = X(A) th X c gi l mt bin cngu nhin. Bin ngu nhin X c th xem nh hm ca bin c A vi min xc nhl .

Cc bin ngu nhin c k hiu bng cc ch ln X, Y, Z, cn cc gi tr cachng c k hiu bng cc ch nh x, y, z...

5.1.2. Phn loiBin ngu nhin c chia lm hai loi: bin ngu nhin ri rc, bin ngu nhin lin tc.

a) Bin ngu nhin ri rc (Discrete Random Variable)Nu gi tr ca bin ngu nhin X c th lp thnh dy ri rc cc s x1, x2, , xn (dyhu hn hay v hn) th X c gi l bin ngu nhin ri rc.

b) 3.1.2.2. Bin ngu nhin lin tc (Continuous Random Variable)Nu gi tr ca bin ngu nhin X c th lp y ton b khong hu hn hay v hn (a,b)ca trc s 0x th bin ngu nhin X c gi l bin ngu nhin lin tc.

Th d

Lng khch hng n ca hng trong ngy l bin ngu nhin ri rc. Nhit trong ngy Si Gn l bin ngu nhin lin tc.5.2. PHN PHI XC SUT I VI BIN NGU NHIN RI RC

(Probability Distribution for Discrete Variable)

5.2.1. Hm xc sut (Probability Function)Hm xc sut Px(x) ca bin ngu nhin ri rc X dng din t xc sut cho bin ngunhin X t gi tr x. PX(x) l hm ca gi tr x

PX(x) = P(X=x)


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Cao Ho Thi 44

Th d

Trong th nghim thy 1 con xc sc, ta c

P(X=1) = P(X=2) = = P(X=6) = 1/6

Hm xc sut l : PX(x) = P(X=x) = 1/6 vi x =1, 2, 3, 4, 5, 6

5.2.2. Phn phi xc sut (Probability Distribution)Phn phi xc sut ca bin ngu nhin X th hin s tng quan gia cc gi tr xi ca Xv cc xc sut ca xi, s tng quan c th trnh by bng bng th hoc bng biuthc.

Th d

Trong th nghim thy 1 con xc sc, phn phi xc sut l:

Trnh by bng bng:

X 1 2 3 4 5 6

PX(x) 1/6 1/6 1/6 1/6 1/6 1/6

Trnh by bng th :

5.2.3. Hm xc sut tch ly (Cumulative Probalility Function).a) nh nghaHm xc sut tch ly FX(xo) ca bin ngu nhin ri rc x th hin xc sut X khngvt qu gii hn xo. FX(xo) l hm ca xo

FX(xo) = P (Xxo)

PX(x)

1/6

0 1 2 3 4 5 6 x


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Cao Ho Thi 45

b) Tnh chtTa c cc tnh cht sau:

a. FX(xo) = xox X )x(P

xox

X )x(P : tng ca tt c cc gi tr c th c ca x vi iu kin xxo

b. 0 FX(xo) 1 xo

c. Nu x1 < x2 th FX(x1) FX(x2)

Th dTrong th nghim thy 1 con xc sc, ta c hm xc sut tch ly nh sau

FX(xo) =

=+


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Cao Ho Thi 46

5.2.4. K vng ca bin ngu nhin ri rc(Expected Value of Discrete Random Variable)

a) K vng ca bin ngu nhin K vng, E(X), ca bin ngu nhin ri rc X c nh ngha nh sau:

E(X) = x

x )x(P.x

x

:Tng tt c cc gi tr c th c ca x

K vng ca bin ngu nhin c gi l s trung bnh (mean) v c k hiu l xE(X) = x

Th d

Gi X l s li c trong 1 trang sch. Hm xc sut ca bin ngu nhin X c cho bi:PX(0) = 0,81, PX(1) = 0,17, PX(2) = 0,02.

Tm s li trung bnh c trong 1 trang sch ?

Gii

x = E(X) = x

X )x(P*x = 0 * 0,81 + 1 * 0,17 + 2 * 0,02

= 0,21 li /1 trang

b) K vng ca hm sca bin ngu nhinGi X l bin ngu nhin ri rc vi hm xc sut PX(x)

g(X) l mt hm s ca bin ngu nhin X

K vng ca hm s g(X) c nh ngha nh sau :

E[g(x)] =

x X)x(P)x(g

PX(x)

0,8

0,4

0 1 2 x

x = 0,21


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Cao Ho Thi 47

5.2.5. Phng sai (Variance)Gi X l bin ngu nhin ri rc.

Gi X l s trung bnh ca bin ngu nhin

Phng sai ca bin ngu nhin X chnh l k vng ca (X - x) v c k hiu2X .

2X = E[(X - X)] = ( )

xXX )x(P*x

2

Phung sai 2X c th tnh theo cng thc :2X = E(X) -

2X =

22X

xX )x(Px

Chng minh

2X = )x(P)x( XX

x

2 = +x

XX

x

XX

x

X xPxPxxPx )()(.2)(22

2X =

22X

xX )x(Px

5.2.6. lch chun x (Standard Deviation) lch chun c k hiu x

X =2X

Th d

Cho hm xc sut ca s li X c trong 1 trang sch l

PX(0) = 0,81, PX(1) = 0,17, PX(2) = 0,02

Tm lch chun ca s li c trong 1 trang sch ?

Gii

Trong th d trc ta c X = 0,21

K vng ca XE(X) =

xX )x(Px

2 = 0 * 0,81 + 1 * 0,17 + 2 * 0,02

E(X) = 0,25

Phng sai2X = E(X) -

2X = 0,25 - (0,21) = 0,2059

lch chun


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Cao Ho Thi 48

x = 4538,02059,02 ==X

5.2.7. Momena) Momen gc cp k (Momen of Order k)

mk= E [Xk] = )x(Px X

x

k

k = 1: m1 = E[X] = XXx

)x(Px =

k = 2: m2 = E[X]b) Momen trung tm cp k (Central Momen of Order k)

Mk= E[(X-X)k] = )x(P.)x( X

kX

k = 2: 2X = E[(X - X)] = m2 - 21m M1 = E [(X - )] = 0

M2 = E [(X - ) ] = (Variance)

M3 = E [(X - )] = (Skewness : lch)

M4 = E [(X - )4] = KM2 = K4

K : h s Kurtorsis

5.2.8. Phn phi xc sut nh thc (Binomial Probability Distubutions)a) Hm xc sut ca phn phi nh thc

(Probability Function of Binomial Distribution).

Tin hnh n php thc lp.

Gi p l xc sut thnh cng trong mi php thc lp => q = (1-p) l xc suttht bi trong mi php thc lp.

Xc sut c s ln th thnh cng l x trong nhng php thc lp c chobi hm xc sut nh sau :

Px(x) = [n!/ (x!(n - x)!)].[px(1 - p)n-x ] vi x = 0,1,2,, n

hay

Px(x) =xnC p

xqn-x vi q = 1 - p

Ghi ch

Phn phi ca s ln php th thnh cng l x c gi l phn phi nh thc..


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Cao Ho Thi 49

Hm xc sut PX(x) l hm xc sut ca phn phi nh thc.b) Strung bnh, phng sai v lch chun ca phn phi nh thcGi X l s ln thnh cng trong n php th, mi php th c xc sut thnh cng l p. X

tun theo phn phi nh thc vi s trung bnh, phng sai v lch chun c tnhtheo cc cng thc sau:

S trung bnh

X = E(X) = np

Phng sai

2X = E[(X - x)] = np(1-p)

Hay

2X = npq vi q = 1-p

lch chun

x = npq

Th d

Mt ngi i bn hng i tip xc cho hng vi 5 khch hng. Xc sut bn chng trong mi ln cho hng l 0,4.

a) Tm phn phi xc sut ca s ln bn c hng.b) Tm s trung bnh, phng sai v lch chun ca s ln bn c hng.c) Tm xc sut ca s ln bn c hng trong khong 2 n 4 ln.Gii

a. Xc sut ca s ln bn c hng tun theo phn phi nh thc :

PX(x) =xnC P

x qn-x = xC5 * (0,4)x * (0,6)5-x

PX(x) =)!x(!x

!

55

* (0,4)x * (0,6)5-x

x = 0 => PX(0) = 0,078

(khng bn c)

x = 1 => PX(1) = 0,259

x = 2 => PX(2) = 0,346

x = 3 => PX(3) = 0,230

x = 4 => PX(4) = 0,077

x = 5 => PX(5) = 0,010

(trong 5 ln bn c c 5)

PX(x)

0,4

0,2

0 0 1 2 3 4 5 Xs ln thnh cng


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Cao Ho Thi 50

b. S trung bnh ca s ln bn c hng x = np = 5 * 0,4 = 2

Phng sai 2X = np(1-p) = 5 * 0,4 * 0,6 = 1,2

lch chun x = 12. = 1,10

c. P(2 < X < 4) = PX(2) + PX(3) + PX(4) = 0,653

5.2.9. Phn phi xc sut Poissona) Phn phi PoissonBin ngu nhin X c gi tun theo phn phi Poisson nu hm xc sut ca X c dng

PX(x) =!x

e xvi > 0,

x = 0,1,2,

b) Strung bnh, phng sai v lch chun ca phn phi Poisson S trung bnh ca phn phi Poisson

x = E(x) =

Phng sai.x = E[(x-x)] =

lch chunx =

Th d

Mt trm in thoi tng nhn c trung bnh 300 ln gi trong 1 gi. Hi xc sut

trm nhn c ng 2 ln gi trong 1 pht cho trc.Gii

S ln nhn c trung bnh trong 1 pht

300/60 = 5 ln/1pht => = 5

Xc sut nhn c ng 2 ln trong 1 pht.

PX(2) = (5 * e-5)/2! = 25/2e5 0,09


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Cao Ho Thi 51

5.3. PHN PHI XC SUT I VI BIN NGU NHIN LIN TC(Probability Distributions For Continuous Random Variables)

Phn phi ca bin ngu nhin lin tc c xc nh bi hm mt xc sut.5.3.1. Hm mt xc sut (Probability Density Function)Gi X l bin ngu nhin lin tc, gi x l gi tr bt k nm trong min cc gi tr c thc ca X.

Hm mt xc sut fX(x) ca bin ngu nhin lin tc l hm c nhng tnh cht sau :

fX(x) 0 , x Xc sut P(a


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Cao Ho Thi 52

Th d

Bin ngu nhin X tun theo lut phn phi vi mt fX(x), trong

fX(x) =

>

=1

1111

2

11 a))((aSdx)x(fX

b. Tm xc sut

P(1/2X1) = +=+1

21

121

2

21

/ /]x

xdx)x(

= [-1

21

1 2

21 2

2 2

+ +] [( / )

/ ]

= 1/2-[-1/8+1/2] = 1/8

P(-1/3X1/3) = 2P(0X1/3)=20

1 3

1/

( ) +x dx

= 2 [ -x/2+x ]0

1 3/

= 2 [-1/18+1/3] = 5/9

c. P(X = ) = 021

21= dx)x(fX

/

/

Th d

Cho hm mt xc sut ca bin ngu nhin X c dng:

1

x

f(x)

1

3/40


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Cao Ho Thi 54

Tm

a) P (X 3/4)b) P (X > 1/2)c)

P (1/4 X 4

1

1 )Gii

a. P (X 3/4) = dx)x(fdx)x(fdx)x(f X/

X

/

X

/

+=43

0

21

0

43

0

= 1/2(1/2 *1) + 1 (1(3/4 - 1/2) = 0,5

b. P (X > 1/2) = dx)x(fx21

1

2

1

= 1(1-1/2) + 1/2 (1)(1

1

2 1 ) = 0,75

c. P (1/4 X11

4) = dx)x(fx4

11

4

1

= 1-2 [1/2 * 1/4 * 1/2] = 7/8

5.3.2. Hm phn phi tch ly (Cumulative Distribution Function)Hm phn phi tch lycn c gi l hm phn tch hay hm phn phi xc sut

a)

nh nghaHm phn phi tch ly, FX(x) ca bin ngu nhin lin tc X th hin xc sut Xkhng vt qu gi tr x. FX(x) l hm ca x.

Fx(x) = P(X x)

b) 3.3.2.2. Tnh cht Fx(x) = x X dx)x(f vi fX(x) l hm mt xc sut. FX(x)dx = f X(x) = dFX(x)/dx FX(x) l hm khng gim => FX(x + x) FX(x) 0 FX(x) 1 F(- ) = 0 F(+ ) =1 P (a < X < b) = FX(b) FX(a)


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Cao Ho Thi 55

x

y

FX(x)1

-1

FX(x)

Th d

Bin ngu nhin X c cho bi hm phn phi

FX(x) =

0

1 2

1

( ) /x

Tnh xc sut bin ngu nhin X nm trong khong (1.5, 2.5) v khong (2.5, 3.5)

Gii

P(1,5 < X < 2,5) = F(2,5) - F(1,5)

= (2,5 - 1)/2 - (1,5 -1)/2 = 0,5

P(2,5 < X < 3,5) = F(3,5) - F(2,5)

= 1 - (2,5 -1)/2 = 0,25

5.3.3. K vng ca bin ngu nhin lin tca) K vng ca bin ngu nhin

K vng E(X) ca bin ngu nhin lin tc X c nh ngha nh sau :

E(X) = dx)x(xfx

K vng ca bin ngu nhin c gi l s trung bnh k hiu l x

E(X) = x

b) K vng ca hm sca bin ngu nhin

dx)x(f)x(g)]x(g[E X

=

Nu x 1


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Cao Ho Thi 56

5.3.4. Phng sai = E[X - x)]

=

[x - x)]fX(x)dx

hay

= E(X) - x

5.3.5. lch chun : = 2x

5.3.6. Hm phn phi chun (The Normal Distribution)a) Hm mt xc sut ca phn phi chun

Nu hm mt xc sut ca bin ngu nhin X c dng

fX (x) =2

2

2

22

1

)x(

e

Vi - < < + v 0 < < +

Th bin ngu nhin X c gi l tun theo lut phn phi chun.

b) Tnh cht ca phn phi chunGi X l bin ngu nhin tun theo lut phn phi chun vi cc tham s v . Ta ccc tnh cht sau

a. S trung bnh ca bin ngu nhin X tun theo lut phn phi chun l .

E(X) =

b. Phng sai ca bin ngu nhin X l

Var(X) = E[(X - )] =

c. ng cong ca hm mt xc sut c dng hnh chung i xng qua tr s trungbnh v c gi l ng cong chun (normal curve)


15/23


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Cao Ho Thi 58

c) Hm phn phi tch ly ca phn phi chun (Cumulative Distribution Function ofNormal Distribution)

nh ngha

Cho X ~ N (,). Hm phn phi tch ly ca bin ngu nhin X tun theo phn phichun c nh ngha nh sau :

FX(x) = P(X


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Cao Ho Thi 59

ng cong ca hm mt xc sut ca phn phi chun chun ha gi l ngcong chun chun ha (standard normal variable)

x

= 0

2 = 1

f(x)

Tung ca 1 im bt k trn ng cong chun sc xc nh t phng trnh cahm mt xc sut ca phn phi chun.

fX(x) = 2

2

222

1

)x(

e

Vi = 0 , = 1 v x = z 22

20

2

1)(

z

exf

=

Gi tr ca hm phn phi tch ly ca phn phi chun chun ha (cng bng dintch nm di ng cong chun) c lp thnh bng v c cho sn trong cc phlc ca sch thng k. Cc bng ny cho gi tr ca

FZ(z) = P (Z z) =

z

Z dz)z(f

Z

0

f(x)

Mt s bng lp sn, ch cho ta din tch nm di ng cong chun t 0 n z.

Z

0

f(x)

Da vo bng ny ta c th tnh c xc sut cho bin ngu nhin Z nm trongkhong no . C th.


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Cao Ho Thi 60

P(Z < a) P(a Z b) P(Z > b)

b) Chun ha bin ngu nhin (Standardization of Variable)Nu bin ngu nhin X c s trung bnh l v phng sai l , th bin ngu

nhin Z = (X-)/ s c s trung trung bnh l 0 v phng sai l 1.Z c gi l bin ngu nhin c chun ha (standardized).

Nu X tun theo phn phi chun th Z tun theo phn phi chun chun ha v Zc gi l bin ngu nhin chun chun ha (Standard normal variable). Khi :

P(a < X < b) = P[(a-)/ < Z < (b-)/

X

+ +33

3 0 1 2 32 1

2 +2

Th d

Cho Z ~N(0,1). Tm xc sut gi tr ca Za) Nh hn - 1,25

b) Nm trong khong (-0,50 , 0,75)c) Ln hn 1Gii

a. P(Z - 1,25) = FZ (-1,25)

= 1 - FZ(1,25)

= 1 - 0,8944

= 1 - 0,1056

Ghi ch

FZ(-zo) = 1 FZ (zo)

b. P(-0,50 Z 0,75)

= FZ (0,75) FZ(-0,50)

= FZ(0,75) - [1 FZ(0,50)]

= 0,7734 - [1 - 0,6915)]

0

f(x)

1.251.25

Z

0

f(x)

0.750.5


19/23

Cao Ho Thi 61

= 0,4649

c. P(Z > 1) = 1 - P(Z 1)

= 1 FZ(1)

= 1 0,8413

= 0,1587

Th d

Cho X ~ N(15,16). Tm xc sut X c gi tr ln hn 18

Gii

P (X >18) = P(Z> [(18 -)/] = P(Z> [(18 - 15)/4] = P(Z> 0,75)

= 1 - P(Z18) = 0,2266

Th du

Nu X l bin ngu nhin tun theo phn phi chun c s trung bnh l 3 v lchchun l 2. Tm P(4


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Cao Ho Thi 63

P(a X b) P )5,05,0

(

+

bZ

a

P(X=a) P (a-0,5 X a+0,5)

Th d

Mt ngi bn hng i cho hng vi 100 khch hng. Theo kinh nghim hy vng bnc hng cho mi mt khch hng l 40%. Tm xc sut s khch hng s mua hngnm trong khong 45 n 50.

Gii

Gi X l s khch hng s mua hng X tun theo lut phn phi nh thc vi

= np = 100 * 4 = 40, lch = 246,0*4,0*100)1( ==pnp

Dng cch tnh gn ng ca phn phi chun

P(45 X 50) = P[24

4050

24

4045

Z ]

= P[1,02 Z 2,04]

= FZ(2,04) FZ(1,02)

= 0,9793 - 0,8461

= 0,1332

Nu kn s hiu chnh lin tc.

P(45 X 50) = P[24

40550

24

4045

.Z ]

= P[0,92 Z 2,14]

= FZ(2,04) FZ(1,02)

= 0,9838 - 0,8212

= 0,1626Ghi ch : Nu tnh trc tip bng phn phi nh thc.

P(45X50) = PX(45) + PX(46) + PX(46) + PX(47) + PX(48) + PX(49) + PX(50)

Cc bng nh thc ng vi n 20.

Th d

Mt nh my sn xut th mt loi sn phm mi. Mi sn phm sn xut c xc sut bh l 0,16. Tm xc sut c ng 20 sn phm b h trong 80 sn phm.

Gii

Gi X l s sn phm b h. X tun th theo lut phn phi nh thc vi.


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Cao Ho Thi 64

= np = 80 * 0,16 = 12,8

= 279,384,0*16,0*80)1( == pnp

P(X = 20 ) = P(19,5


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