GIÁO TRÌNH XÁC SUẤT THÔNG KÊ VÀ ỨNG DỤNG

Chifdng 6: Lfdc lifdng dac tnfnf dim dong.. Lftic lifting hi§u qua Trong tap cac ifdc lifdng khong chech, ifdc lifdng nao co phUdng sai nho nhat thi goi la iftic lifting hieu qua.. Chifd

Trifdc tie n ta x e t m ot vi du sau:

trifcrng D ai hoc Cong nghiep Tp.HCM noi chung 111 to t hay

b in h thifcrng h ay kem , va m uon co k e t qua tro n g thori gian

n g an (m ot ngay ch an g han) The th i ta kho n g th e k h am stic

khoe cho tr e n to a n bo cho 50.000 hoc sin h sin h vien to&n

trifdng dtfcrc C hung ta d a n h chon n g iu n h ie n m o t vki bo

p h a n sin h vien nao do, roi k iem tr a sufc khoe ho Dtfa vko ket

qua do nhif vay, chung ta co th e k e t lu a n ve tin h h in h sufc

khoe chung cho to a n bo hoc sin h sin h vien tro n g trtfcrng

O day n ay sin h ra v an de: k e t lu an cua chung ta chinh xac d en mtic nao, ro ra n g neu chung ta chon ra n h ieu hoc

sin h va ra i deu tr e n cac ldrp, cac he k h ac n hau th i k e t qua

cang tin cay hon Nhifng nhieu la bao nhieu? N eu n hieu qu&

th i khong xue N eu it qua th i k e t qua it tin cay V an de cua

th o n g ke la dtfa vao nhting phiforng p h ap chon ho p ly, vifa dd

to n kem vifa co k e t qua d an g tin cay

Ve sau n ay ta se difa ra nhufng cong thtic to a n hoc cho

phep chi ro so ltfong co th e can p h ai dieu tr a tro n g m6i b&i

to a n th o n g ke so lifcfng

Vay k h i ta can n g h ien ctiu m ot ta p ltin gom N (hufu h a n

ltin, hay vo h a n dem difOc) p h a n tti (goi la d a m d o n g ), N la

kich thiftic dam dong, do m ot ly do nao do ta k h o n g th e

n g h ien ctiu difcfc t a t ca moi p h an tti cua chung T a lay ra m ot

ta p nho gom n p h a n tti (goi la m au), n goi la kich thiftic m lu ,

va n g h ie n ctiu chung th a t ky b ang phifong p h ap khoa hoc de

Chifcfng 5: Lv thuvet m au 199

roi tif d6 ru t ra diforc k e t lu in khach quan ve toan bo ta p ldn

d im dong N p han tur d tren, phifcfng p h ip nay con goi la

p h U ctn g p h a p m d u

+ Cac dac trung cua X la cic dac trung cua dam dong

+ Xet ve liforng ta quan tam den 2 d#c tn m g sau

Trung binh d im dong fi = E ( X )

Phifcmg sai dam dong cr2 = var(X)

+ Xet ve ch at, ta quan t i m den ti le p cua cac p han tuf co tin h chat A (=1) n io do va khong co tin h chat A (=0),

+ Tien h an h quan s i t ttf m lu n g lu nhien d tren , ta diforc cic

g ii tri cu the X j = th i (xi,x2, ,xn) l i m lu cu th e , ndtfcrc goi la kich thiftic m lu cu the

Chu v:

+ Ta chi xet c ic k e t qua quan s i t doc lap

+ Khi xet ly th u y et ta dung m lu tdng q u it, con lam to in th i

Gia sti d im dong l i m ot binh difng mifcri vien bi, trong do co

3 vien nang lOgr, 5 vien nang 20gr, 2 vien n in g 30gr Goi X

l i BNN d ie tru n g cho trong liforng (x6t ve liforng) th i X c6 luat

p h an pho'i:

200 Chitong 5: Lv thuvet m au.

Tufc Xi co cung p h a n pho'i vdri X

Tiep tuc goi X2 la tro n g liicfng vien bi dugc lay r a la n thuf hai,

Tufc X2 co cung p h a n pho'i vofi X T iep tuc .cho d en la n thuf n Tom la i (Xi,X2, ,Xn) 1& n BNN doc l&p co cung p M n p h6'i vdi BNN X

Bay gicf ta tid'n h&nh 5 quan s£ t sau: (xi,x2,x3,x4,x5), m lu cu

th e cua 5 la n quan s&t ch&ng h a n la: X i=xi=10gr, X2=X2=1 0gr,

X3=x3=2 0gr, X4=X4=30gr, X,5=x5=20gr,

vay (xi,x2,x3,x4,x5) =(10gr,10gr,20gr,30gr,20gr)

VI D u 2

Gia suf dam dong can n g h ien cufu ti le phe p h am cua m ot loai

sa n pham Goi X 1& BNN dac tru n g cho dau h ieu phe pham (x e t ve ch at) cua dam dong T a co b&ng p h a n phoi xac s u it sau:

Neu m lu cu th e (xi,x2, ,xn) co nhieu quan s a t k h ac nhau,

k h oang each giufa cdc quan s a t khong dong deu hoac cac Xj

nhan r a t ft th i ta sap xep chung dudri dang khoang

Khi cd m lu n khd l<Jn th i c£c dac trifng m lu xap xi cac dac

trifng tufcfng ting cua dam dong

Do k e t qua cua h ai lu a t so' lorn d chifdng 3, ta co cac k e t qua

T h a t vay, chu y c&c Xi co p h a n phoi Bernoulli B (l,p))

• Thifcrng th i X * / i , nhifng E ^ x ) = fi, va ta co he thufc lien

he giufa X va f i nhif sau:

Chifdng 5: Lv thuvet m au.

206 Chifdng 5: Lv thuvet m au.

Difa vao Ky vong va phifong sai cac dac trifng m lu or p h an

tr e n vifa tin h , va difa vao d in h ly gidi h a n tru n g ta m or

chifong 3, ta co th e suy ra p h a n phoi cua cac dac tn fn g m lu cotin h p h a n phoi g an chuan nhif sau:

3.1 P h a n p h o i c u a t i le m l u F n

pq ] Vdi n k h a lota th i Fn e N Pi

Ta phan biet 4 tnfdng hdp sau:

TrUdng hop a,b dudi day, Xi khong can la p h a n phoi nhimr

c6 p han pho'i S tu d en t vdi ( n - 1 ) bac tif do (chuong 4)

Cho b ie t a vk n ta tin h dugc f£_1 sao cho:

208 Chifdng 5: Lv thuvet m au.

co p h a n phoi z l - i p h a n pho'i chi b in h phifcfng n-1 bac ttf do

Xem chting m inh ti m en h de sau:

Cho b ie t a va bac tif do k ta tin h dtfcfc Zk (a ) sao c^ o:

p [ z k £ z i ( « ) ] = *

co p h a n phoi x l p h a n phoi chi b in h phifcfng n bac tif do.

Tom lai: T a co t a t ca cac k e t qua tr e n qua m e n h d l sau:

M e n h d e : Gia sti n BNN -X1,X2,A 3, X#l,deu co phan phoi chuan

X / e iv(//,£T2 ), chting minh:

a/ Ket qua nay hien nhien, do dinh ly gidi han tning tam (chifdng 3)

nhifng d day ta muon dung ham dac trifng (trong chifdng 2) de co phong

phu hdn ve cac cong cu chting minh

Chifdng 5: Lv thuvet m au 209

(x~nf Vdi X e 7v|//,CT2j Ta co ham mat d6: /(-*) = 1 —* 2<r2

Cac ket qua trifdc ta cd: j = ^ J , ta suy ra ham dac trifng

(7 c/V i Xj e n [/ x ,(72 j nen

210 Chifdng 5: Ly thuvet m au.

==> V= U - W G Xn-i

Hay ta c6 the ly luan nhif sau:

Vi bien so' V chi phu thuoc vko phifcfng sai m lu S 2 va W chi phu thuoc vao trung binh m lu X nen V vk W doc lap Do d6:

(l-2<)2 (1— 2/)5

=>My(t) = - 5 -^j- ==> Ve z i - i ■

(1- 21 )2

e/ Theo dinh nghia phan phoi Student, th i e/ la h# qua true

tiep cua a/ va d/, th a t vay: X e TV — a 2 '

Chifcftig 5: Lv thuvet m lu 211

Gii sur X ~ z f 0 ■ Tinh P(2.156 < X), Ti'nh P(2.558 < X<3.940)

Giai:

Bang tra phan pho’i Khi binh phifdng cuo'i sach cho ta cd dinh hang thtf

k=10 la m6t day Zio tang theo a , ta tim a: ^j20 (a)=2.156, ta co :

a/ Tinh xac suat de do dai trung binh cac chi tiet nay lay ra khong nho hdn 200mm

b/ Tim xac suat de phifdng sai m lu hieu chinh khong nho hdn 230(mm)2

Giai:

a/ n=25 < 30 va X la p h a n pho'i chuan, g=20 da biet,

vay theo ly thuyet

• Trong m lu cu the kich thifdrc n, co m phan tuf c6 tin h ch at A

ma ta quan tam th i ta n so:

Chifdng 5: LV thuvet mau 213

214 Chifdng 5: Lv thuvet m lu

N eu khong doi bien ta lap b ang nhif sau:

(so lieu tin h todn r a t ldn)

Cach bam may: 1/ stat clear,

Vao dOT li£u:

s=14.282/ Mod = Reg, Lin205

215 il[8jlM ± l Ket qua:

| t l |S -K A fl|k l B xn = 239

[tl|S -V A * ' n - 1 s =14.28

N eu doi b ien ta lap b an g nhif sau: (so' lieu tin h to a n be )

Lay xo =235 (thifcrng la mod(X)), h=10, xt = ./ _ xi ~ x0

Chifcfng 5: Lv thuvet mau 215

1/ Ngifori ta goi bieu do tren la d a g ia c ta n , cu the da giac tan

so la do th i bar th e h ien toa do (x;, ni), hay da giac ta n suat la

do th i bar th e h ie n toa do (xj, fi), no the h ien th on g tin sa bo

216 Chifdng 5: Lv thuvet m lu

ban dau ve d&m dong P h a n m em excel th e h i$ n d i d&ng dieu nay Vdri vi du tre n , xi =1,2,3,4,5,6,7, bi<lu do 7 cot cao 1& da giac ta n so', bieu do 7 cot th a p tifcrng ting 1& d a gi£c t i n suat 2/ N eu x 111 rdi rac th i xo = Xj ting vdri nj ld n n h a t (modX), neu

x dtfcrc chia th a n h ltfp th i khi tin h to a n ta tin h ckc gi& tri d

gitfa cdc ldrp d6 T ham so' h la k hoang c&ch gitfa c&c Xj.

a1 Tinh trung binh, phuong sai mau cua X va Y.

Ve da giac tan so cua X, Y

b/ Nhung san pham co chi tieu Y duoi 92cm la san pham loai A Tinh trung binh, phuong sai mau cua cac san pham loai A

Chifdng 6: Lfdc lifdng dac tnfnf dim dong 217

Chu’oTng 6

l f d c L I/0 N G DAC TR l/N G DAM DONG

£ l/ofc lifofng d ie m

1.1 D in h n g h ia m p t th o n g ke:

M6 t h&m cua m lu tong qudt T = T(Xi, , Xn) la m o t th o n g k e

chi phu th u6c vao m lu n g lu nhien, k h6ng phu thuoc vao th am

s6 n&o cd C h in g han: X n = — (Xx + + X n).

n

1.2 \J dc lifd n g d ie m c u a th a m so

Jjdc lifdng diem cua th a m so 0 la m ot th o n g ke

0 = 0(A ’j, ,A fl) chi phu thuoc vao n quan s£t Xi, ., Xn va

khong phu thuoc v&o th a m so' 0

Vi D u :

+ TI le m lu F„ = - - + —1 + —+ la iftfc lifdng diem cua tl le

n

dam d6ng 0 = p chifa biet.

+ Trung b in h m lu X n = — + — + —+ — la iftfc lifdng diem

n

cua tru n g b in h dam dong 0 =E(X) chifa biet.

+ Phifdng sai mau S 2 = — £ (X t - X n j hay Phifdng sai mau co hieu

chinh S 2 = — — Y ( X j - X „ Y la tide lifdng diem cua tru n g binh

difdc goi \k ifdc lifdng diem cua 0 Do gid tri 0 chifa b iet nen

ta khong th e so s£nh 0 vdi 0 de d&nh gid ch at liforng cua 0

Vi vay ngifdi ta difa ra cac tieu chuan lfdc lufdng nhif sau:

i-n

218 Chifdng 6: Lfdc lifdng dac tnfnp d£m done.

II Cac tieu chuan iftic lifting

+ E^Xn) - ji (trung b in h m lu la tide liforng k h o n g chech cua

tru n g b in h dam dong //)

T rong thifc h a n h Khi co m lu cu th e (xi, ., xn) t a lay

2.2 Lftic lifting hi§u qua

Trong tap cac ifdc lifdng khong chech, ifdc lifdng nao co phUdng sai nho nhat thi goi la iftic lifting hieu qua

Vdi moi 0 khong chech cua tham so" 0 va X la bien ngau nhien co ham mat do f ( x , 0 ) phu thuoc vao tham so 0 Ta cd ket qua bat dangthtic Cramer-Rao (C-R):

Chifdng 6 : Lfdc lifdng dac tnfnp <Mm dong i_ v v,

Ta thtfa nhan ket qu5 nay

Bieu thtfc tren cho ta can difdi ciia phifdng so" 0 la ifdc lifdng cua tham

so 0 Ta co the difa vao bat dang thtfc nay de kiem chtfng ifdc lifdng

tham so dang xet co phai la hi#u qua khong Xet qua c5c vi du sau:

Theo BDT C-R, ta co: var [fi) > = — = v a r ( x )

Bat dang thtfc thanh dang thtfc

That vay do dung X - de ifdc lifdng fi thi da biet la ifdc lifdng

n khong chech vi E ( X ) = f i , va

var(■'f ) = T - y a r[ Z A:i' n \ n n n

220 Chifdng 6: Lfdc lifdng dac trtfnp d£m done.

That vay do dung X =

n de ifdc lifdng // thi da biet la ifdc lifdng

khong chech vi E{^x) = f i ^ &

Thong ke 6 ( X x, ,Xn ) la ifdc lifdng vufng cua 0 neu

—— >0 N ghla lli vdi n du ldn ta c6 :0 = 0

Chifdng 6: l/dc lifrtnp dac tnftig dam dong 221

tide liforng vumg, Udc liforng khong chech va la ifdc liforng hieu

qua cua ky cong dam dong, cua ta n so' dam dong, phuorng sai dam dong

01 - 0 \ = 2 e , E la do chinh xac cua ifdc liforng

a nho bang 1% den 5%, diforc goi la mdc y nghla.

Bai to an tim k h o an g tin cay cua 0 la bai toan udc liforng

khoang

TrUdc k h i di vao tim cac khoang Udc liforng cho cac dac trifng th o n g ke, ta can nhac lai kien thufc d chuorng 2: Cho

X n e <72 ) , va cho trUdc do tin cay ( I - a ) , ta tim duac t a

(tra b ang B) sao cho:

P [ \ X n - A z < T J a \ = l - a = l 9 ( t a )

P \ X n - t a c j < n < X n + t a <r] = 1 - a ( * )

Vav k h oang tin cay la [ X „ - t a cr, X n + t a a ] (**)

222 Chifdng 6: iTdc lifdng dac trifng dam d6ng

trong do: fi\ = x n ~ taa ' M l= X n + ta ° va £ ~ ° Ja

TBay gid ta x et cu th e cac bai to a n tim ifdc liforng k h oang sau:

2.2 K h o a n g tin c a y c h o t i 1$ d a m d o n g p

B a i to a n

Vdi ti le p cac p h an tijf co tin h ch a t A cua d&m dong chifa bi£t

va do tin cay \ - a cho trUdc, khoang tin cay cho p la [p\',pi\

Chifdng 6: Lfdc lifdng dac trifng dam dong 223

Trong dot bau cur tong thong, ngifdi ta phong van nglu nhien 1600 ctf tri thi difdc biet co 960 ngifdi se bau ctf tfng vien A Vdi do tin cay 99%,hay xem tfng vien A trung ctf khong? Biet tl le qua ban la th a n g ctf

De Udc liforng so" h ai cau tre n 1 hon dao ngifdi ta deo vong cho

2000 con Sau m ot thdi gian b at lai 400 con th ay co 80 con

deo vong H ay Udc liforng so hai cau vdi do tin cay la 95%.

224 Chifdng 6: Lfdc lifdng dac tnftip d im dong.

Gia suf dam dong co tru n g b in h n chifa b iet C an cuf vko

(Xi, ,Xn), tim m ot khoang

X n - t a ~r= - f ^ - X n + i oc r~

yin V «

= l - o r =2f>(fa )

= 1- a

+ n < 30, cr2 chifa biet, X c6 p h an phoi chuan.

Khi d6 ^ c6 p h a n phoi Student (n - 1) bac tu do, biet

% 3 Tn

I - a , ta tim diforc f"_1 (bang C p h an pho'i Student) sao cho:

P

Suy ra: —X n — t £ * — j=, M 2 ~ ^ n + *a /—

Xn ~ fl cr Phan tich: t = - — = —r===- = - f = = »

4/ w < 30,AT e A^//,<r2 ^ c r2 chifa b iet

Khi co m§u cu th e , ta th a y X n bdi x n , S2 bori s2

VI D u l : C an 100 tr a i cay, cho b iet jcjqo = 500#r, g 2 =

Lfdc liforng tro n g liforng tru n g b in h // vdi do tin cay 95%

Chifdng 6: l/dc lifdng dac trifng dam dong 227

cdc bao bot la X co p han phoi chuan Lfdc liforng tro n g lifong

trung binh fi vdi do tin cay 95%.

Ta p h an b ie t hai trUcfng hop: n chufa b iet va [i da biet.

• Gia suf dam dong co p han phoi chuan vori ca hai o'2 va n

228 Chifdng 6: Lfdc lifdng dac trifng d im done.

Chitong 6: Lfdc lifdng dac tnfne dim dong 229

vdi n la so" bac tti do (xem m enh de cua ly thuyet mau, p han

bai ta p chticrng 4) Ta co: P

Vi D ul: Mtic hao phi nguyen lieu cho mot loai san pham X tren mot

ddn vi san pham tuan theo qui luat chuan Ngifdi ta can thti mot mau

Vdi do tin cay 95% hay tim khoang tin c3y cho do lech tieu chuan cua mtic hao phi nguyen vat lieu tren Trong ba trtidng hdp

a/ Da biet E(X) =fi= 19

b/ Khong biet E(X)=n

d Da biet E(X) =n=20.

230 Chifdng 6: Lfdc lifdng dac tnftig dam dong.

a/ Da biet E(X) =|i.=19 Phan phoi chi binh phifdng n bac tif do.

Chifdng 6: Lfdc lifdng dac trifng dim dong 231

N h a n x e t: vi x = 19.94 v& ^=19 nam ngoai bang phan pho'i 3 tren, nen khoang tin cay gitfa hai cau a/ va b/ khong xap xi nhau Trong khi do k e t q u i cau d r a t s i t vdi cau b/, vi x = 19.94 ~ ji = 20

VI Du2: Tien hanh 15 quan sat ve chi tieu X cua mot loai san pham, ta

difdc s = 14.574 hay Udc lifdng phifdng sai cua X vdi do tin cay 95%, gia thiet X co phan phoi chuan.

Giai:

Lfdc liforng khoang cho phuorng sai a 1 cua dam dong co p h an

pho'i chuan v l (I khong noi gi nghla la chifa biet

^dA ,, - 0\ < e ) = 1 - a Trong do ta goi 3 chi tieu sau:

• e: Do chinh xac cua ifdc lifdng

• (1-a) : Do tin cay cua Udc lifdng

• n : Cd m lu cua Udc lifdng

Trong p h a n nay ta chi quan tam tin h 3 chi tieu d tren , va d

cac p h a n tr e n ta da b iet E = 0’'xta , trong do a dUdc th a y bang

I— hay J — , con ta ta tr a d bang B (tich p h an Laplace)

- 2

hay b ang C, p h a n pho'i Student, khi cr chifa biet

Vay qua cac quan he tre n ta co the tim diforc chi tieu thuf ba neu ta b iet trUdc difdc 2 chi tieu con lai

232 Chifdng 6: Lfdc lifdng dac tnltap dam dong.

Vi D ul:

Chu cufa h&ng cung cap san muon ifdc lifang lifang sa n chtfa trong 1 th u n g duac sa n xuat ttf cong ty A B iet do lech tieu chuan cua liforng san cong ty la 0,08 thung Dieu tr a 50 thung duac lifang san tru n g b in h la 0,97 thung Vdi do tin cay 99% hay ifdc luang: Luang san tru n g b in h chtfa tro n g m ot thung

tr a sa bo cho ta x = l 1.8kg, vdi do lech tieu chuan la s=0.7kg

Hoi can p h ai lay kich thifdc m S tfto i th ieu la bao n h ieu de dat duac sai so" kho n g vUOt qua 0.2kg?

Chifdng 6: Lfdc lifting dac trifng dim dong 233

Mot may san xuat hang loat Ngtfdi ta tien hanh mot so m lu kiem tra Ket qua cho trong bang sau :

Ngtfofi kiem tra Ngay, Thcfi

gian

Kich thtftfc miu

So" san pham loai A

a/ Tim tfdc ltfdng cho ty le loai A vdi do tin cay 95%.

b/ Gia suf muon tfdc ltfdng ti le loai A vdi do chinh xac 9% va do tin cay 95% thi can kiem ra them it nhat may mau.

c/ Muon do chinh xac 9% thi do tin cay dam bao dtfdc bao nhieu ? d/ Muon do tin cay 90% thi do chinh xac dam bao dtfdc bao nhieu ?

Muon tim n ta cong cot 3 lai Con muon tim m ta cong cot 4 lai

Khoang ifdc lifdng can tim la (0,1346 ; 0,1854)

Ket luan : Bang phifdng phap ifdc lifcfng khoang vdi do tin cay 95%, vdi

m lu cu the nay, ta dif doan ty le san pham loai A cua may la tif 13,46% den 18,54%

VG m6t cufa hang che" bien thuy sdn, theo doi nhu cau cua mat hang

Chufcfnp 6 : Lfdc lifdng dac trifng dam dong.

So ban ra (lit) So" ngay

c/ Hay ifdc lifdng phifdng sai cua nifdc mam ban trong ngay vdi d$ tin cay 95% (gia thiet dai lifdng nay co phan phoi chuan).

/Vdi

a) So" dong ho trung binh ban difdc trong ngay

b) Ty le nhffng ngay e" hang

II Vdi so" lieu cho trong bang, ifdc lifdng:

so lieu cho trong ban g, ifdc lifdng:

236 Chifdng 6: Lfdc ltfdng dac trtfnp d im dong.

a) Lifdng mu trung binh mot cay cho trong ngay

b) Do bien dong cua lifdng mu do

c) Ty le cay loai II

4/ Can thuf 100 trai cay cua mot nong tnfdng, ta co ket qua nhif sau:

trai hong

Chufdng 6: Lfdc lifdng dac trifng dam dong 237

p) Tim ifdc luang cho ty le trai cay hong trong nong tnfdng

b) Tim ifdc lifdng cho ty le trai cay hong trung binh d moi sot

c) Tim ifdc lifdng khong ch6ch cho do bien dong ty le trai cay hong

d moi sot.

Ifd mot cufa hang che' bien thuy san, theo d5i nhu cau cua mat hang

nifdc mam trong mot so ngay, ta co ket qua :

c) Hay ifdc ltfdng phUdng sai cua ltfdng nifdc mam ban trong ngay vdi do tin cay 95% (gia thiet dai lifdng nay co phan phoi chuan).