TKXH (1)

Mc lc

Chng 1. Tng quan v Thng k v thng k x hi hc . . . . . . . . . . . . . . . . . . . . . . . . 1

1.1. Khi nim thng k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1.2. Mt s khi nim c bn dng trong thng k . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.3. Cc cp bc o lng v thang o. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.4. Khi qut qu trnh nghin cu thng k. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

Chng 2. Thu thp d liu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.1. Xc nh d liu cn thu thp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6

2.2. D liu th cp v d liu s cp . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

2.3. Cc k thut ly mu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

Chng 3. Tm tt v trnh by d liu bng bng v th . . . . . . . . . . . . . . . . . . . . 12

3.1. Tm tt v trnh by d liu bng bng tn s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

3.2. So snh phn phi cc tn s . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16

3.3. Tm tt v trnh by d liu bng biu v th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Chng 4. Tm tt d liu bng cc i lng thng k m t. . . . . . . . . . . . . . . . . . . 26

4.1. Cc s o hng tm ca tp d liu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26

4.2. Cc i lng m t s phn b ca tp d liu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.3. Cc i lng o lng phn tn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29

4.4. Cc i lng m t hnh dng ca tp d liu. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

4.5. Phn phi chun. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

Chng 5. Xc sut cn bn v bin ngu nhin. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

5.1. Xc sut l g?. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

5.2. Bin ngu nhin . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

Chng 6. c lng cc tham s tng th. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

6.1. Tham s tng th v tham s mu . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51

6.2. c lng im . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52

i

6.3. c lng khong . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 54

Chng 7. Kim nh gi thuyt v tham s mt tng th . . . . . . . . . . . . . . . . . . . . . . . 60

7.1. Gii thiu chung v kim nh gi thuyt. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

7.2. Kim nh gi thuyt mt tng th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

Chng 8. Kim nh gi thuyt v tham s hai tng th. . . . . . . . . . . . . . . . . . . . . . . . 73

8.1. Kim nh gi thuyt v s khc bit ca trung bnh hai tng th . . . . . . . . . . . . . . . . . . . . . . 73

8.2. Kim nh gi thuyt cho phng sai ca hai tng th . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

Chng 9. Phn tch phng sai . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

9.1. Phn tch phng sai mt yu t. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

9.2. Phn tch su One-way ANOVA . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 88

Chng 10. Kim nh chi - bnh phng . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

10.1. Kim chng tnh c lp. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

10.2. So snh t l hai tng th. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

10.3. So snh nhiu t l . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

ii

Chng 1

Tng quan v Thng k v thng k xhi hc

1.1. Khi nim thng k

Thng k l ngnh khoa hc lin quan n vic thu thp, tng hp, phn tch, din gii v trnh by

d liu.

1.1.1. Xut pht thut ng Thng k

Thut ng "Thng k" u tin bt ngun t ting Latinh "Statisticum collegium" (hi ng chnh

quyn) v mt t ting "statista" (ngi lm cho chnh quyn hay ngi lm chnh tr).

Trong th k 19, thut ng thng k c hiu mt cch ph bin l thu thp v phn loi d liu

c s dng p ng nhu cu ca chnh ph v cc c quan qun l.

Ngy nay, thng k c s dng rng ri hn nhiu so vi xut pht im u tin l phc v cho

chnh quyn hay chnh ph. Cc t chc v c nhn s dng thng k phn tch d liu v ra

quyt nh.

Thng k c s dng t khoa hc t nhin, cho n khoa hc x hi, y dc hc, kinh doanh v

rt nhiu lnh vc khc.

Thng k x hi l ngnh khoa hc s dng h thng o lng trong thng k nghin cu v hnh

vi ca con ngi trong mi trng x hi.

Chng hn, thng k x hi hc c th s dng cho mt s mc ch nh: c lng cht lng ca

nhng dch v sn c ca t chc, phn tch hnh vi ca mt nhm ngi thng qua mi trng ca

h, d bo xu hng ca mt vn thng qua thng k chn mu . . .

1.1.2. Phn loi thng k

Thng k m t miu t nhng mt quan trng ca tp cc s o. N s dng cc phngphp tm tt, m t mt tp d liu nh: s dng cc loi bng biu, cc i lng thng

k nh trung bnh, mc phn tn, . . .

Thng k suy din s dng cc phng php t cc s o ca mu c th a ra nhngsuy on v nhng mt quan trng ca tng th.

1

2 Chng 1. Tng quan v Thng k v thng k x hi hc

Thng k ng dng bao gm thng k m t v thng k suy din.

Thng k ton l ngnh khoa hc nghin cu c s l thuyt ca thng k.

Trong mn hc ny, chng ta s nghin cu v thng k ng dng c bn (trong iu tra x hi hc).

lm iu ny chng ta s dng cc kt qu nghin cu ca thng k ton.

1.2. Mt s khi nim c bn dng trong thng k

D liu l cc con s, t ng hay hnh nh phn nh thc t ca i tng nghin cu.

Tng th l tp hp ton b cc phn t thuc hin tng cn c nghin cu (thng l ngi,

s vt, s kin). Cc phn t to thnh tng th gi l n v tng th. Ngi ta chia tng th thnh

2 loi: b l v tim n.

Tng th bc l: gm cc n v tng th c th quan st c.

Tng th tim n: gm cc n v tng th khng trc tip quan st c hoc nhn bitc.

V d 1.2.1. Cc tng th sau, u l tng th tim n, u l tng th bc l?

1. Ton b sinh vin Thng Long ra trng trc nm 2014.

2. Ton b cc h gia nh Vit Nam.

3. Ton b cc cng ty trn a bn H Ni.

4. Tp hp ton b cc sinh vin Thng Long yu thch mn Thng k.

5. Tng th sinh vin Thng Long ng h tng hc ph.

Tng th ng cht: gm cc n v tng th ging nhau mt hay mt s c im lin quan

n mc ch nghin cu.

Tng th khng ng cht: gm cc n v tng th khng ging nhau nhng c im lin

quan n mc ch nghin cu.

V d 1.2.2. V d mc ch nghin cu l tm hiu v thi gian t hc nh ca sinh vin th:

tng th sinh vin Thng Long h chnh quy l tng th ng cht.

tng th sinh vin i hc Kinh t Quc dn H Ni khng phi l tng th ng cht v sinhvin i hc KTQD hc nhiu h i hc khc nhau nh: h i hc chnh qui, h i hc

t xa, h i hc ti chc,...

Mu l tp con ca tng th.

Nu ta tin hnh ly s o mi n v tng th th ta gi l tng iu tra.

Nu ta tin hnh ly s o nhng n v ca mu th ta gi l iu tra chn mu.

1.2. Mt s khi nim c bn dng trong thng k 3

V d 1.2.3.

1. Tng iu tra dn s nc ta vo nm 2009 (thng cc nc c 10 nm tin hnh tng iu

tra mt ln, tnh n nm 2015 Vit Nam tin hnh 4 ln nh vy).

2. Ngy 18/9/2014, Cc Qun l Dc (B Y t) c vn bn khn gi Vin kim nghim thuc

Trung ng yu cu ly mu kem nh rng Colgate Total kim tra hm lng triclosan - c

th tng nguy c ung th. y l iu tra chn mu, v hin nhin khng th kim tra c ht

cc tup kem nh rng ca hng ny.

1.2.1. c im thng k, s o v bin

c im ca tng th: l cc tnh cht quan trng ca cc n v tng th lin quan n ni

dung nghin cu.

V d 1.2.4.

1. Khi mun nghin cu v cht lng cuc sng ca ngi dn, cc c im thng k m ta

quan tm, thu thp khi kho st c th l: thu nhp, chi tiu, c nh hay khng, ngh nghip

ca ngi c iu tra, khu vc sng, . . .

2. Nghin cu v s pht trin ca n g khi s dng mt loi thc n mi, ta xt n nhng tnh

cht ca n g nh: trng lng, mu lng, cht lng tht, sc khng vi mi trng,...

Mi c im ca tng th ta gi l mt bin. Mi n v tng th ta tin hnh ly mt s o v

gn cho gi tr ca bin.

V d 1.2.5.

Bin thu nhp th s o l thu nhp mi thng ca ngi c iu tra tnh n trm nghnng gn nht;

Bin ngh nghip th "s o" l cng nhn, nng dn, bc s, cn b, . . .

Khi ni n tng th v mu, cn phn bit bin m ta quan tm ca mt phn t v bn thn

phn t . Trong trng hp s nhm ln, ta c th nhn mnh tng th quan st, tng th s

o phn bit vi tng th v tng t mu quan st, mu s o phn bit vi khi nim

mu. Chng hn, ta cn phn bit tng th sinh vin i hc Thng long tt nghip vi tng th

im trung bnh tt nghip ca cc sinh vin i hc Thng Long.

C g khc nhau gia s o ca bin thu nhp, chi tiu vi s o ca nhng bin nh ngh nghip,

khu vc? Ta thy rng thu nhp v chi tiu c o bi nhng con s thc s, cn khu vc v ngh

nghip khng c o trc tip bng cc con s, nu c nh s cc ngnh ngh bi cc con s th

chng qua ch l k hiu, khng mang ngha v mt lng. Ta phn loi bin thnh hai loi:

nh lng v nh tnh.

Bin nh lng: l bin c s o c th biu hin trc tip bng cc con s. V d bin thunhp, chiu cao, trng lng, tui, . . .

4 Chng 1. Tng quan v Thng k v thng k x hi hc

Bin nh tnh: l bin c s o khng th biu hin trc tip bng con s. V d bin giitnh, khu vc, ngh nghip, . . .

D liu theo nh ngha chnh l s o ca cc bin. Chnh v th ta cng c th ni d liu nh

tnh hay l d liu nh lng.

1.3. Cc cp bc o lng v thang o

Mi mt s o ca bin u nm trn mt "thang o" no . Ty mc tt ca thang o, ta

cp n bn thang o sau:

1.3.1. Thang o nh danh

Thang o nh danh dng cho cc bin nh tnh. S o ca cc bin ny l cc m s phn loi

i tng. Gia cc m s y khng c quan h hn km, ch dng m tn s xut hin ca

cc biu hin.

S o ca bin gii tnh (nam, n), bin mu sc (xanh, , tm, ...), bin khu vc sng,... thuc

thang o nh danh

1.3.2. Thang o th bc

Thang o th bc thng dng cho cc bin nh tnh, i khi dng cho c bin nh lng. Trong

thang o ny gia cc s o ca cc bin c quan h th bc hn km. Tuy nhin, s chnh lch gia

cc s o khng nht thit bng nhau.

Kt qu ca cc cu tr li sau thuc thang o th bc:

1. Bn nh gi th no v vic tip thu ca mnh i vi vic hc cc mn t nhin:

1. Tt 2. Bnh thng 3. Km

2. kin ca bn v vic cht cy xanh m bo hnh lang giao thng tu in trn cao:

1. Rt khng ng h 2. Khng ng h 3. Khng kin 4. ng h 5. Rt ng h

1.3.3. Thang o khong

Thang o khong thng dng cho cc bin nh lng. Thang o khong l thang o th bc c

cc khong cch u nhau. Cc php tnh cng tr u c ngha nhng khng c gi tr khng xc

nh mt cch chnh xc v khng th ly t l gia cc s o.

S o nhit , ch s IQ,... thuc thang o khong.

1.3.4. Thang o t l

Thang o t l dng cho cc bin nh lng. Thang o t l l thang o khong, hn na thang o

ny c gi tr 0 xc nh mt cch chnh xc v c th ly t l gia cc s o.

1.4. Khi qut qu trnh nghin cu thng k 5

n v o tin t (VND, dollar, pound, yen, . . . ); n v o chiu di (cm, m, km, . . . ); n v o

khi lng (kg, tn, t, yn, . . . ), . . . thuc thang o t l

Thang o t l v thang o th bc thuc vo thang o bc thp. Thang o khong v thang o t l

thuc vo thang o bc cao. S o thuc vo thang o bc cao c th chuyn v thang o bc thp

nhng s o thuc thang o bc thp khng th chuyn v thang o bc cao.

Chng hn, vi bin thu nhp thuc vo thang o t l c th chuyn v thang o th bc bng cch

phn lng thnh ba mc: Cao (lng trn 9 triu ng/thng), Trung bnh (4-9 triu ng/thng),

Thp (di 4 triu ng/thng).

1.4. Khi qut qu trnh nghin cu thng k

Qu trnh nghin cu thng k c khi qut qua cc bc sau:

1. Xc nh vn nghin cu, mc tiu, ni dung, i tng nghin cu.

2. Thu thp d liu thng k.

3. X l s liu.

4. Phn tch v gii thch kt qu.

5. Bo co v truyn t kt qu nghin cu.

1.4.1. Mt s cu hi n tp

1. Nu li c im ca tng loi thang o v cho v d.

2. Cc s o sau thuc thang o no?

a. M sinh vin ca mt sinh vin.

b. im thi mt mn hc ca sinh vin.

c. Chiu cao ca mt sinh vin.

Chng 2

Thu thp d liu

2.1. Xc nh d liu cn thu thp

Khi thu thp d liu, trc tin da vo vn nghin cu v mc tiu nghin cu m ta xc nh

d liu cn thu thp, th t yu tin ca cc d liu ny, xc nh r gii hn, phm vi d liu. Vic

ny gip tit kim thi gian v cng sc thu thp v x l s liu sau ny.

Chng hn, nghin cu v vn iu kin sinh hot c nh hng n kt qu hc tp ca sinh

vin hay khng, ta c hai nhm d liu chnh cn phi thu thp l: iu kin sinh hot v kt qu

hc tp. Nhm th hai, kt qu hc tp th d liu thu thp c v r rng. Vi nhm iu kin

sinh hot, c th thu thp nhng d liu sau:

nh vi cha m, ngi thn, tr hay k tc?

C phng ring hay khng?

Nu chung c bao nhiu ngi cng? H c hay gy n khng? H c l sinh vin haykhng?

Ch c n o, cht tri, nng bc khng?

C bn hc ring hay khng?

C phi lm thm, ph gip b m khng? Nu c thi gian lm khong bao nhiu gi mingy?

Ni cch ch hc bao xa?

...

Tuy nhin nhng thng tin nh sau l khng cn thit:

Bn hc mu g? C hay mi?

Nh c xy nm no?

Nh v sinh c p hay khng?

...

6

2.2. D liu th cp v d liu s cp 7

2.2. D liu th cp v d liu s cp

D liu th cp: l d liu thu thp t nhng ngun c sn, thng l qua tng hp, x l.

D liu th cp c th ly t ngun: ni b ca cc doanh nghip nh bo co v sn xut, ti chnh,

nhn s,...; T cc c quan thng k nh nc (Tng cc Thng k, http://www.gso.gov.vn); T

bo, tp ch; Cc t chc, hip hi, vin nghin cu, cc cng ty v t chc nghin cu v cung cp

thng tin theo yu cu.

u im ca vic thu thp d liu th cp l: thu thp nhanh, t tn km chi ph v thi gian. Tuy

nhin n t chi tit, nhiu khi khng p ng nhu cu nghin cu.

D liu s cp: d liu thu thp trc tip t i tng nghin cu.

D liu s cp thng c thu thp theo mt qui trnh bi bn ty theo nghin cu thng k l

nghin cu th nghim hay nghin cu quan st.

Trong nghin cu th nghim, ngi nghin cu o c v thu thp d liu trn cc bin ktqu trong cc iu kin khc nhau ca bin nguyn nhn ang cn nghin cu.

Trong nghin cu quan st th cc d liu c th thu thp t:

Nhiu ngi cung cp thng tin khc nhau nh: ch h gia nh, khch hng, ch doanh

nghip,... Ngi thu thp d liu cn kho st trc tip vi i tng cung cp thng tin

qua kho st dng vit, qua phng vn, ...

Ngun ni b nh phng kinh doanh ca cng ty, cc b phn c chc nng ghi chp s

liu ca c quan,...

Ngun bn ngoi nh cng ty nghin cu th trng, cc t chc chuyn nghip (cc t

chc phi chnh ph, ngn hng th gii, cc cng ty dch v cung cp d liu, ...),...

D liu s cp c u im l chi tit, p ng tt nhu cu nghin cu. Tuy nhin vic thu thp tn

km chi ph v thi gian.

2.3. Cc k thut ly mu

Khi nghin cu mt tng th ln, chng hn ta nghin cu v lng khi im ca cc sinh vin

Thng Long sau khi ra trng, vic thu thp c thng tin lng ca tt c cc sinh vin tt

nghip l iu khng kh thi. Do vy, ta ch thu thp c mt mu ca tng th. T mu ny s

dng cc phng php thng k ta suy din cho tng th. Tuy nhin, nhng suy din c chnh

xc th vic chn mu phi ng.

Ngoi l do tng th qu ln, cn rt nhiu l do khc buc ta phi chn mu nh: qu trnh nghin

cu phi ph hy (chng hn o hm lng ca mt loi cht bo qun trong cc chai nc mm),

khi nghin cu tng th tim n (khi ny chn mu l s la chn duy nht)

Khi vic chn mu khng i din c cho tng th hoc ngi iu tra khng nh gi cao vic

chn mu m ch tin tng vo vic iu tra tng th th ta phi kho st c tng th.

8 Chng 2. Thu thp d liu

2.3.1. Mt s khi nim v chn mu

Chn mu ngu nhin l qu trnh la chn sao cho trong mi ln chn mi n v tng th c c

hi c chn vo mu nh nhau. Qu trnh chn mu khng p ng c yu cu trn th gi l

chn mu khng ngu nhin (chng hn, cc n v mu c chn ra da trn s thun tin hoc

do s phn on ca ngi iu tra . . . ).

Thng tin cha ng trong mu mun phn nh chnh xc tng th ang nghin cu, tc l i din

c cho tng th, th mu phi c chn mt cch ngu nhin t tng th. Trong cc phn sau

ny, mi mu c chn ra u gi s c c t mt qu trnh chn mu ngu nhin.

Chn mu c hon li l cch chn m khi mi n v tng th c chn ra li t tr li tng th.

Cn chn mu khng hon li l cch chn m khi mi n v tng th c chn ra khng c

t tr li tng th.

Chn mu khng hon li cho ta ci nhn y hn v tng th so vi chn mu c hon li v

chn mu khng hon li m bo rng cc phn t trong mu l khc nhau, t cho ta nghin

cu c nhiu phn t khc nhau nht c th trong tng th.

2.3.2. Chn mu ngu nhin

2.3.2.1. Chn mu ngu nhin n gin

Nu s lng n v tng th t (vi trc hoc vi trm), ta c th chn cc n v ca mubng cch bc thm hay quay s.

Chng hn, bn mun phng vn quan im ca ging vin TLU v vic hc gia s ca sinh

vin. Gi s bn mun phng vn 5 ging vin c chn ngu nhin ca khoa kinh t. Bn c

th ghi s th t ca cc ging vin trong danh sch ging vin ca khoa ny (c th s dng

danh b in thoi ca nh trng) ra nhng mnh giy ri bc thm ly 5.

Nu s lng n v tng th l ln (vi trm, vi nghn), vic chn mu bng bc thm tr nnmt thi gian, trong trng hp ny ta c th chn mu bng cch dng bng s ngu nhin.

33276 70997 79936 56865 05859 90106 31595 01547 85590 9161078188 03427 49626 69445 18663 72695 52180 20847 12234 9051133703 90322 92737 88974 33488 36320 17617 30015 08272 8411527156 30613 74952 85689 48237 52267 67689 93394 01511 2635885140 20285 29975 89868 08178 77233 13916 47564 81056 9773585977 29372 74461 28551 90707 51259 77452 16308 60756 9214449442 53900 70960 63990 75601 40719 60268 89368 19885 5532244819 01188 65255 64835 44919 05944 55157 94904 31273 0414618594 29852 71585 85030 51132 01915 92747 64951 58586 2321614513 83149 98736 23495 64350 94738 17752 35156 35749 0999842698 06691 76988 13602 51851 46104 88916 19509 25625 5810414346 09172 30168 90229 04734 59193 22178 30421 61666 9990432812 74103 47070 25306 76468 26384 58151 06646 21524 1522796909 44592 24200 13363 38005 94342 28728 35806 06912 1701264161 18296 22851 87308 58731 00256 45834 15398 46557 4113510367 07684 36188 18510 07351 19731 92420 60952 61280 5000167658 32586 86679 50720 94953 96423 24878 82651 66566 1477876797 14780 13300 87074 79666 95725 26432 46901 20849 8976881536 86645 12659 92259 57102 80428 25280 66432 84673 4002732832 61362 98947 96067 64760 64584 96096 98253 26422 4440744048 37937 63904 45766 66134 75470 66520 34693 90449 9430526766 25940 39972 22209 71500 64568 91402 42416 07844 69618

Bng s ngu nhin l mt dy s c chia thnh cc nhm gm nm ch s d c, trong

mi s gm mt ch s trong bng c gi s l chn ngu nhin t cc ch s t 0 n

2.3. Cc k thut ly mu 9

9, mi s gm hai ch s trong bng c gi s l chn ngu nhin t cc ch s t 00 n

99, mi s gm ba ch s trong bng c gi s l chn ngu nhin t cc ch s t 000 n

999,...

V d 2.3.1. xem xt s ngy thanh ton chm ca cc ha n gi n mt cng ty trong

vng mt nm, ngi ta chn ra 60 ha n t 3250 ha n c gi n ( c nh s t

1 n 3250). S dng bng s ngu nhin, chn 60 ha n ny nh sau:

V tng s ha n l 3250 gm bn ch s, nn ta s ngu nhin chn mt nhm gm bn

ch s t bng, chng hn 2234, y l s ca ha n u tin c chn.

T s ny, ta di chuyn sang tri (phi, ln hoc xung) thm c cc nhm bn ch

s na, chng hn ta i xung chn thm c by nhm gm bn ch s na l: 2234,

8272, 1511, 1056, 0756, 9885, 1273. Do ch c 3250 ha n nn ta loi nhm s 8272 v

9885. Nh vy nm nhm s c chn u tin l: 2234, 1511, 1056, 0756, 1273. y

chnh l s th t ca nhng ha n s c chn o ngy.

C tip tc qu trnh nh vy, ta s thu c s th t ca 60 ha n c chn o

s ngy thanh ton chm.

2.3.3. Ly mu h thng

Khi s n v mu kh nhiu hay rt nhiu (vi trm, vi nghn n v mu), ta c th s dng

phng php ly mu h thng. Theo cch ny ta nh s tt c cc phn t ca tng th.

Gi s ta cn ly mu gm m phn t t tng th c N phn t c nh s. Chia N n v

tng th thnh k nhm theo cng thc k =N

m(nu k khng phi s nguyn dng th k c lm

trn n s nguyn dng gn nht).

Trong k n v u tin, chn ngu nhin ra mt phn t, y l n v mu u tin, cc nv mu tip theo c ly cch n v ny mt khong k, 2k, 3k, . . .

Nu n ht danh sch N n v cha m n v mu ta quay tr li u danh sch vi quic phn t th N + 1 chnh l phn t c nh s 1, phn t N + 2 chnh l phn t c

nh s 2, . . .

V d 2.3.2. Gi s ta cn chn mu gm m = 10 phn t t tng th c N phn t:

Tng th gm N = 63 phn t:

Tnh k =63

10= 6.3, ly k bng 6.

Trong 6 n v u tin, chn ngu nhin ra mt phn t chng hn l 4, khi cc n

v mu c ly ra l: 4, 10, 16, 22, 28, 34, 40, 46, 52, 58.

Tng th gm N = 67 phn t:

Tnh k =67

10= 6.7, ly k bng 7.

Trong 7 n v u tin, chn ngu nhin ra mt phn t chng hn l 5, khi cc n

v mu c ly ra l: 5, 12, 19, 26, 33, 40, 47, 54, 61, 1 (61 + 7 - 67).

10 Chng 2. Thu thp d liu

2.3.3.1. Ly mu theo khi v ly mu nhiu giai on

Trong 3 cch chn mu ni trn, ta thy rng ta u c danh sch cc phn t ca tng th (t

mi c th t ca phn t v chn mu t th t ny). Khi ta khng c danh sch ny (c th

do tng th qu ln, vic ly ton b danh sch l tn km chng hn) th ta khng s dng c

cc phng php chn mu ni trn. Khi ta cn chn mu theo khi, v c th, ly mu nhiu

giai on.

Chn mu theo khi l cch chn lin quan n vic chia tng th thnh nhiu khi, mi khi xem

nh mt tng th con, ly ngu nhin n gin ra m khi, sau kho st ht cc i tng trong

m khi mu c ly ra.

Trong trng hp mi khi c chn c c qu ln, cc phn t trong khi c xu hng ging nhau

c im nghin cu th ta c th chia cc khi ny thnh cc khi nh hn v chn ngu nhin

vi khi nh ny trong mi khi ln kho st. Cch lm nh vy gi l chn mu theo giai on.

V d 2.3.3. Chng hn, ta cn kho st thu nhp ca cc h gia nh ni thnh H Ni. Ta c

th chia thnh cc khi, mi khi l mt phng. Ta chn ngu nhin 10 phng chng hn. Nhng

nh ta bit, mi phng c qu nhiu h gia nh v kho st ht mt phng qu thc tn km.

Ta li chia cc phng thnh cc t dn ph. V trong mi phng c chn trn ta chn ngu

nhin 3 t dn ph. Tip theo ta c th kho st ton b cc h gia nh cc t dn ph c chn.

Nu vn qu ng, ta c th chia cc t dn ph thnh cc khi nh hn, nh ngch, hm, ...

2.3.3.2. Ly mu phn tng

Chn mu phn tng s dng khi cc n v tng th khc nhau v tnh cht lin quan n vn

nghin cu, kho st. Khi tng th c chia thnh cc tng (lp), mc tiu l cc gi tr ca

cc i tng tng th ta quan tm thuc cng mt tng cng t khc nhau cng tt. Sau cc

n v mu c chn t cc tng ny theo cc phng php ly mu thng thng nh ly mu

ngu nhin n gin hay ly mu h thng.

Vic phn b c ca mu iu tra mi tng ty thuc vo mc ch nghin cu. Khi nghin cu

tp trung vo c tng th v khng cn tp trung thng tin ca cc tng ring r th s lng phn

b s c ly theo t l chim ca tng trong tng th. Khi nghin cu t trng tm vo cc

tng hoc so snh cc tng khc nhau, ngi ta c th phn b tng i u s lng cn thu thp

vo cc tng. iu ny m bo cho lng quan st ly ra mi tng l khng qu t. T kt qu

thu thp c, nu mun tnh chung cho c tng th ta s nhn thm trng s. Trng s ca mt

tng c tnh nh sau:

T l phn trm ca tng trong tng th / (c mu rt ra t tng ny/c mu chung)

V d 2.3.4. tm hiu im hi lng ca sinh vin vi chng trnh hc ca trng ta c th

phn tng theo cc ngnh hc, v mi ngnh hc c chng trnh dy ring, ging vin khc nhau v

mc yu cu khc nhau.

Gi d trng ch c 3 ngnh, ngnh A chim c c thy 6000 sinh vin theo hc, ngnh B c 3000,

ngnh C c 1000.

Gi d chng ta ch c kinh ph kho st vi 200 sinh vin

2.3. Cc k thut ly mu 11

1. Khi ta khng cn so snh xem ngnh no sinh vin hi lng hn m ch quan tm n mc hi

lng chung ca c trng th ta s phn b s lng mu ly ra cc tng nh sau: ngnh A

chim 60% nn ta s iu tra 120 sinh vin, ngnh B chim 30% nn ta s iu tra vi 60 sinh

vin, cn ngnh C l 20 sinh vin. Vic chn 200 sinh vin ngnh A, 60 ngnh B, 20 nghnh

C c chn theo cch ly mu h thng hoc bng ngu nhin nh thng thng. Mc hi lng

chung s c tnh trn mu 200 sinh vin c chn nh trn.

2. Khi ta cn xt xem mc hi lng tng ngnh hoc so snh xem ngnh no hi lng nht th

vic chn nh trn li khng tt v ngnh C c chn vi s lng qu t. Khi chng ta c th

phn b u mi ngnh chn khong 1/3: ngnh A: 68, ngnh B: 66, ngnh C: 66. Sau thu

thp d liu theo c cu s lng . Nu mun dng s liu ny tnh cho mc hi lng chung

th kt qu ca mi mu (mi ngnh lu 1 mu nh trn) cn c nhn vi trng s: ngnh

A trng s 60%/(68/200)=1.8; ngnh B: 30%/(66/200)=0.9; ngnh C: 10 %/(66/200)=0.3.

2.3.4. Li do chn mu v khng do chn mu

Trong qu trnh thu thp d liu c th chng ta gp li kiu nh: mt d liu, ghi chp sai, phn

tch sai, ngi tr li phng vn khng hiu, khng ni hoc phng i vn c hi, . . . Nhng

li ny gi l li khng phi chn mu. Hu nh khng c phng php thng k no o hoc iu

khin c nhng li kiu ny, tuy nhin trong cc k thut thng k sau ny, ta u gi s rng

khng c li kiu khng do chn mu xy ra.

Sai lm do chn mu xy ra khi mu khng i din c cho tng th. Khi mu khng i din

c cho tng th th cc tnh ton trn mu khng th suy din cho tng th c.

Mt v d kinh in cho sai lm do chn mu l cuc bu c tng thng ca M nm 1936. C

hai ng c vin sng gi cho chc v tng thng l: Franklin D.Roosevelt (ng Dn ch) v Alf

Landon (ng Cng ha). C rt nhiu tp ch v t chc tham gia d on kt qu ca cuc bu

c trong c tp ch Literary Digest v nhm thm d d lun George Gallup do George Gallup

sng lp.

Tp ch Literary Digest chn mt mu gm 2.4 triu ngi chn t danh b in thoi v chn

t tn cc thnh vin ca cc cu lc b iu tra. Tp ch ny d on rng Alf Landon s

nh bi Franklin D.Roosevelt vi t l l 57% v 43%.

Nhm George Gallup chn mt mu gm 5000 ngi theo mt phng php chn mu kiu h

thng v d on Roosevelt s dnh chin thng trc i th Alf Landon.

Thc t Roosevelt dnh chin thng ln. Nh vy nhm George Gallup d on ng cn tp

ch Literary Digest d on sai. Nguyn nhn v sao?

Sai lm ca Literary Digest l chn nhng ngi iu tra t danh b in thoi v tn ca

cc thnh vin ca cc cu lc b. Nm 1936, M cha phc hi t cuc i suy thoi nn rt nhiu

ngi tht nghip v c thu nhp thp khng c in thoi v cng khng tham gia vo cu lc b

no. Chng trnh chn mu ca Literary Digest b qua nhng ngi ny, m chnh h b

phiu ng h Roosevelt. Cn nhm George Gallup vi phng php chn mu thch hp d on

chnh xc chin thng ca Roosevelt.

Chng 3

Tm tt v trnh by d liu bng bngv th

Khi thu thp d liu, mt nhu cu tt yu l ta cn tm tt li d liu xem cu trc ca n

gm thnh phn nh th no, nhng biu hin loi no l ch yu, biu hin no him, ... Bng tn

s gip chng ta c ci nhn tng quan v tp d liu. T bng ny ngi ta c th m t bng cc

hnh nh trc quan tng ng m ta gi l biu .

3.1. Tm tt v trnh by d liu bng bng tn s

Bng tn s l mt bng tng hp cc biu hin c th c ca c im quan st, hoc cc khong

gi tr m trong d liu (nh lng) c th ri vo v s quan st (tn s), t l phm trm chim

(tn sut) tng ng vi mi biu hin hoc khong gi tr d liu. Bng tn s thng gm ba ct:

Ct u tin m t cc biu hin hoc cc gi tr hay khong gi tr c xc nh cho d liu;

Ct th hai m t tn s tng ng vi cc biu hin hay gi tr;

Ct th ba m t cc tn sut tng ng.

Ngoi ra, ta c th b sung thm ct tn s tch ly v tn sut tch ly ca cc biu hin hoc gi

tr.

3.1.1. Lp bng tn s cho d liu nh tnh

Bng tn s ca mt d liu nh tnh c 3 ct dng:

Biu hin Tn s Tn sut

Biu hin 1 f1f1n

Biu hin 2 f2f2n

. . . . . . . . .

Biu hin k fkfkn

Ct u tin lit k cc biu hin c th c ca i tng theo cc c im cn nghin cu,chng hn ta c k biu hin;

12

3.1. Tm tt v trnh by d liu bng bng tn s 13

Ct th hai l tn s mi biu hin va lit k, tn s ca biu hin th i k hiu l fi.

Ct th ba m t cc tn sut tng ng vi mi biu hin. Vi cc gi s nh trn, tn sutca biu hin th i s l

fin 100%, y n l s quan st ca tp d liu, n chnh l tng

ca cc tn s, tc l n = f1 + f2 + . . .+ fk, ngi ta hay dng k hiuki=1

fi biu din cho tng

ny;

thm thng tin, ngi ta cn b sung thm ct tn s tch ly v ct tn sut tch ly. Hai ct

ny c tnh t ct tn s v tn sut nh sau:

Tn s tch ly ca biu hin th i l tng cc s t dng dng u n dng th i ca ct tns;

Tn sut tch ly ca biu hin th i l tng cc s t dng dng u n dng th i ca cttn sut.

V d 3.1.1. Dy sau y lit k xp loi tt nghip ca 20 sinh vin, trong "G" l gii, "K"

l kh, "TB" l trung bnh, "XS" l xut sc.

G XS G XS K G K XS XS K TB K XS K XS K G G TB TB

Khi ta c bng tn s nh sau:

Xp loi Tn s Tn sut Tn s tch ly Tn sut tch lyTB 3 0.15 3 0.15K 6 0.30 9 0.45G 5 0.25 14 0.7XS 6 0.3 20 1

Bng ny cho ta thng tin tng quan v dy d liu xp loi tt nghip ca 20 sinh vin ni trn,

chng hn, loi kh c 6 ngi, chim 30%, c 70% xp loi t gii tr xung, . . .

3.1.2. Lp bng tn s cho d liu nh lng

Nu c im quan tm l d liu nh lng c t biu hin th cch lp bng tn s ging nh cch

lp bng tn s cho d liu nh tnh, xem mi gi tr nh l mt biu hin. Nu c im quan tm

l d liu nh lng c qu nhiu biu hin th vic lit k tng biu hin lm bng tn s di, mt

i tc dng tm lc thng tin. Trong trng hp ny ta phi tin hnh phn t d liu v lp

bng tn s cho d liu c phn t.

Vic phn t d liu hiu n gin l vic nhm cc biu hin gn nhau vo mt nhm. Mi nhm

ta gi l mt t, mi t c cn di v cn trn. Cn di l tr s nh nht ca t, cn trn l tr

s ln nht ca t. Hiu gia cn trn v cn di c gi l khong cch t. Nu cc khong cch

ny bng nhau tt c cc t th ta gi l phn t u.

Lu rng, vic phn t phi m bo cc t phi ri nhau, ng thi cc t c phn chia

phi bao qut ht tt c cc gi tr ca tp d liu v khng c t trng.

14 Chng 3. Tm tt v trnh by d liu bng bng v th

Sau y chng ta s quan tm nhiu hn n phn t u. Cc bc ca th tc phn t u nh

sau:

1. Xc nh s t cn chia k.

2. Xc nh khong cch t h: h =xmax xmin

k, trong xmax v xmin tng ng l gi tr nh

nht v gi tr ln nht ca tp d liu.

3. Xc nh cn di v cn trn ca cc t. Lu bc ny t u tin phi cha gi tr nh

nht xmin v t cui cng phi cha gi tr ln nht xmax ca tp d liu.

4. Cn trn ca t trc trng vi cn di ca t lin sau.

5. Phn chia cc quan st vo cc t: quan st c gi tr ph hp vi t no th th xp vo t .

Di y l mt gi trong khi phn chia thnh cc t:

T 1: [xmin, xmin + h]T 2: (xmin + h, xmin + 2h]T 3: (xmin + 2h, xmin + 3h]. . . . . .T k: (xmin + (k 1)h, xmax]

y t (a, b] c ngha l mt phn t thuc t ny khi n ln hn a v b. Chng ta cng c thchn nht lot cc t thnh dng [a, b) vi ngha hiu tng t.

V d 3.1.2. Chng hn, ta c tui ca mt mu quan st nh sau:

48 30 35 31 21 28 34 43 36 45 41 33 47 47 30 47 44 45 32 46 47 23 30 23 49 20 24 20 40 50

Ta thy tui giao ng t 20 n 50, gi s ta mun chia thnh 6 nhm, nh vy khong cch t

s l (50-20)/6=5. Nh vy ta s chia thnh cc t

[20,25] (25,30] (30,35] (35,40] (40,45] (45,50]

v ta lp c bng tn s nh sau:

Khong tui [20, 25] (25,30] (30,35] (35,40] (40,45] (45,50]

Tn s 6 4 5 2 5 8

Tn sut 0.2 0.13 0.17 0.07 0.17 0.27

Tn s tch ly 6 10 15 17 22 30

Tn sut tch ly 0.2 0.33 0.5 0.57 0.74 1

Qua bng trn ta thy tui t trn 45 n 50 chim nhiu nht, 50% nhng ngi trong mu c

tui khng vt qu 35, ...

Lu , nhiu khi cc t l u v p ta nn ly cn di ca t u v cn trn ca nhm cui

ph ra ngoi khong dao ng ca d liu, chng hn trong d liu trn nu c thm 2 quan st: 18

v 51 th s c thm hai t: [15, 20] (khi t [20, 25] thnh (20, 25]) v (50, 55].

Tuy nhin nu trong d liu ni trn c mt ngi 57 tui, mt ngi 66 tui th ta khng to thm

t: (50, 55], (55, 60], (60, 65], (65, 70] v s c hai t trng. Trong nhng trng hp ny ta ch thm

mt t vi tn l "Trn 50 tr ln" (hay vit tt l ">50").

3.1. Tm tt v trnh by d liu bng bng tn s 15

3.1.3. Bng tn s cho

Bng tn s cho l mt bng tng hp phn phi ca hai hoc nhiu bin vi nhau. Trong mt bng

tn s cho:

Tn s ca mt bin c quan st theo mt hoc nhiu bin khc.

Tn sut c th c tnh theo dng, ct hoc ton b.

Tn s c th c tnh theo cc biu hin hoc khong gi tr ty theo cc bin c so snh.

V d 3.1.3. Bng tn s cho gia gii tnh v ngh nghip ca 474 ngi c iu tra c th

c m t nh sau:

Qua bng ny ta thy c 157 ngi c gii tnh nam v l NVVP (nhn vin vn phng), khng c

cng nhn n no, . . .

Bng tn sut tnh trn ton b nh sau:

Qua bng ny ta thy c 33.1% l nam nhn vin vn phng, . . .

Bng tn sut cho theo dng:

Qua bng ny ta thy ngh qun l v cng nhn ch yu l nam, cn NVVP th n chim nhiu

hn. Ni chung bng ny cho bit kt cu gii tnh trong mi ngh.

Bng tn sut cho theo ct:


Qua bng ny ta thy trong s nhng ngi gii tnh nam c 60.9% l NVVP, chim t trng nhiu

nht trong tng quan cc nhm ngh ca nam. Trong nhm n, cng vic NVVP chim nhiu nht.

Ni chung, bng ny cho thy t trng cc ngh trong mi gii tnh.

3.2. So snh phn phi cc tn s

3.2.1. Phn trm thay i

Phn trm thay i Phn trm thay i o t l thay i tng i ca mt bin t giai on thi

gian ny sang giai on thi gian khc.

Phn trm thay i t giai on ny sang giai on khc c tnh theo cng thc:

p =ftime2 ftime1

ftime1,

trong

ftime1 l tn s trong giai on thi gian 1.

ftime2 l tn s trong giai on thi gian 2.

V d: Dn s Vit Nam nm 2012 khong 88.78 triu ngi v nm 2013 khong 90 triu ngi.

Hy tnh phn trm thay i ca dn s Vit Nam t nm 2012 sang nm 2013.

3.2.2. T s

T s l i lng dng so snh hai gi tr ca cng mt bin da vo tn s ca chng. T s

ca hai gi tr c tnh theo cng thc:

ratio =fv1fv2

,

trong

fv1 l tn s ca gi tr th nht.

fv2 l tn s ca gi tr th hai.

V d 3.2.1. Theo cuc tng iu tra dn s Vit Nam nm 2009, Nm 2009 Vit Nam c khong

756192 b s sinh trai v 689602 b s sinh gi. T s b s sinh trai theo s b s sinh gi l

756192/689602 = 1.096563, n cho thy c s lch v t l gii tnh khi sinh: c khong 10 b gi th

c 11 b trai.

3.3. Tm tt v trnh by d liu bng biu v th 17

3.2.3. T sut

T sut l i lng o tn s ca mt s kin tng i theo c ca tng th hoc n v thi gian.

T sut ca mt s kin c tnh theo cng thc:

rate =Tn s ca s kin

S phn t tng th 10000.

Ch : Con s 10000 c nhn vo trnh s thp phn qu nh. Con s ny c th c thay

bng nhng con s khc ph hp trong tng trng hp.

V d 3.2.2. Theo cuc tng iu tra dn s Vit Nam nm 2009, s tr s sinh v dn s H

Ni v TPHCM c cho trong bng sau:

Thnh Ph s tr s sinh Tng dn s

H Ni 119991 6451909TPHCM 102406 7162864

T sut sinh H Ni l (119991/6451909) 10000 = 186 cho ta bit trong nm 2009 H Ni c1 vn ngi th c 186 b c sinh ra. Con s ny TP HCM l (102406/7162864) 10000 = 143.iu ny cho thy t l sinh H Ni l cao hn TP HCM.

3.3. Tm tt v trnh by d liu bng biu v th

Mc ny ta s tm hiu mt s biu thng dng tm tt di dng trc quan cho tp d liu.

Ta s s dng cc dng biu khc nhau ty thuc vo d liu l nh lng hay nh tnh.

3.3.1. Mt s biu v th minh ha cho tp d liu nh lng

i vi d liu inh lng, y chng ta s tm hiu cc dng biu phn phi tn s, a gic

tn s, thn v l.

3.3.1.1. Biu phn phi tn s

Biu phn phi tn s gm cc ct ng dng miu t phn phi tn s ca tp d liu nh

lng.

Cch lp biu phn phi tn s:

Phn chia tp d liu thnh cc t;

Xc nh tn s trong mi t;

V cc ct ng t cnh nhau vi rng l khong cch trong mi t v cao tng ngl tn s trong mi t.

V d 3.3.1. Ta tr li vi v d 6.2.1 ta xy dng c bng tn s nh sau:

Khong tui [20, 25] (25,30] (30,35] (35,40] (40,45] (45,50]

Tn s 6 4 5 2 5 8


T bng tn s ny ta dng c biu sau, gi l biu phn phi tn s (histogram).

Da trn hnh dng ca biu phn phi tn s ta c th bit c:

Mc tp trung tng i ca phn phi ca tp d liu;

Hnh dng tng i ca phn phi ca tp d liu l bng phng, lch hay cn i.

Chng hn, cc biu sau l biu phn phi tn s ca bin thu nhp ca 2 tp d liu khc

nhau.

Biu u cho thy mt s cn i, a s nhng ngi c kho st c thu nhp qung 5 n

8. Cn biu th 2 cho thy biu b lch (ta s gi l lch phi) cho thy a s thu nhp tp

trung t 0 n 10, phn thu nhp cao t.


3.3.1.2. a gic tn s

a gic tn s l th gm cc on thng ni cc im vi nhau dng miu t phn phi tn

s ca tp d liu nh lng.

V d 3.3.2. T biu histogram nh v d 3.3.1 ta c th dng c a gic tn s nh sau:

Hnh th nht v cng vi histogram, hnh th 2 v ring a gic tn s.

Tng t nh biu phn phi tn s, hnh dng ca a gic tn s gip ta c th bit c:



3.3.1.3. Biu thn v l

Biu thn v l c xy dng bng cch chia cc ch s trong mi s ca tp d liu nh lng

thnh hai phn: phn thn v phn l miu t phn phi ca tp d liu.

Cch lp biu phn phi tn s:

Xc nh phn thn, phn l trong mi d liu;

Lp phn thn ca biu ;

Xp phn l tng ng vi mi phn thn lp.V d 3.3.3. Gi s ta c im qu trnh mn thng k ca 30 sinh vin nh sau:

6.9 3.2 6.4 7.2 7.2 4.8 6.1 7.4 7.8 8.0 6.0 6.6 6.5 7.2 2.2

8.7 7.3 7.8 4.7 4.8 5.7 6.1 6.5 8.7 5.0 7.8 6.5 5.6 7.9 8.8

Ta chn phn thn l phn nguyn ca im, l l phn l ca im.

Lp ct thn: 2 3 4 5 6 7 8.

in cc l vo thn theo th t t nh n ln.


Ta c biu thn v l nh sau:

Tng t nh biu phn phi tn s v a gic tn s, hnh dng ca biu thn v l gip ta

c th bit c:



Chng hn, vi biu thn v l trn, ta thy d liu tp trung nhiu t 6 n 7.9. Biu cho

thy khng c s i xng lm, lch tri. iu ny cho thy a s sinh vin t im vng cao sao

vi khong giao ng 2 - 8.

Nhn xt: C 3 biu ni trn u cho ngi c nhng thng tin nh nhau v tp d liu. Biu

thn v l c u im l c th khi phc c d liu gc, cn hai biu trn th khng. Tuy

nhin biu thn v l ch thch hp khi d liu l t, tm vi trm quan st, cn hai biu trn

c th dng biu din cho mi kch c mu.

3.3.2. Biu cho d liu nh tnh

3.3.3. Biu hnh trn

Biu hnh trn miu t d liu nh tnh di dng hnh trn, trong din tch ca ton b hnh

trn i din cho 100% cc phn t ca tp d liu v din tch cc hnh qut i din cho phn

trm ca cc tp con cn biu din.

Cch v biu hnh trn:

Xc nh cc biu hin trong tp d liu (nh tnh);

Xc nh t l cho tng biu hin;

Tnh gc ca hnh qut tng ng vi tng biu hin

Da trn s o gc cho hnh qut v phn din tch cho mi biu hin tng ng.V d 3.3.4. Theo c tnh, lng khch quc t n Vit Nam vo thng 2 nm 2015 theo cc

phng tin nh sau: ng bin: 7 021 lt, ng b: 138 146, ng khng: 610 834. T y ta

c th lp biu hnh trn theo cc bc nh bng sau:Phng tin Lng khch T l Gc ()

ng bin 7 021 0.009287 3.34

ng b 138 146 0.182733 65.78

ng khng 610 834 0.80798 290.87

Ta c hnh trn l


Da trn biu hnh trn ta c th bit c:

Mc phn phi tng i ca mi biu hin so vi ton th; Biu hin no c chim nhiu nht, t nht.

Chng hn, biu hnh trn ni trn cho ta thy, lng khch quc t n nc ta ch yu qua

ng hng khng, chim hn 80%, ng bin rt t, c th ni l khng ng k so vi lng n

t ng hng khng.

3.3.3.1. Biu thanh

Biu thanh bao gm thanh ng (hoc ngang) dng miu t phn phi tn s ca tp d liu

nh tnh.

Cch v biu thanh:

Xc nh cc biu hin trong tp d liu; Xc nh gi tr cho tng biu hin trong tp d liu; V cc thanh ng cho tng biu hin vi chiu cao tng ng vi gi tr mi biu hin.

V d 3.3.5. Vi s liu v lng khch quc t n Vit Nam trong thng 2/2015 nh cp

v d 3.3.4. T nhng s liu ny ta c th lp biu thanh minh ha nh hnh sau (n v nghn):


Da trn biu hnh thanh ta c th bit c:

S so snh tng i gia cc biu hin vi nhau;

Biu hin no c chim nhiu nht, t nht.

Chng hn, t biu thanh trong v d trn ta thy rng s lt khch n bng ng hng khng

l chim ln nht, n gp 4 ln s lt khch n t ng b. V s lt khch n t ng bin

th cc t so vi hai phng tin trn.

Gia biu thanh v trn, bn thy thng tin mang li c g khc bit? Ci no tt hn?

Ta thy rng, biu thanh thin v s lng thnh phn mi loi, cn biu trn thin v t

trng mi loi. Tm l hc ch ra rng biu thanh mang li thng tin tt hn so vi biu

trn. Tuy nhin, khi mun nhn mnh vo t trng, th dng biu trn l thch hp hn.

Mt cu hi nh dnh cho ngi hc: bn thy c s khc bit g gia biu thanh v biu phn

phi tn s?

3.3.4. Biu m t mi quan h gia hai bin

3.3.4.1. Biu tn x

Biu tn x l th hai chiu bao gm cc im m t mi quan h gia hai bin nh lng.

Cch lp biu tn x:

1. Xc nh cc biu hin ca bin nh lng th nht: x1, x2, . . . , xn;

2. Xc nh cc biu hin ca bin nh lng th hai: y1, y2, . . . , yn;

3. Xc nh cc cp im (x1, y1), (x2, y2), . . . , (xn, yn);

4. V cc cp im ny trn mt phng cc.

Ni chung biu tn x cho ta nhng thng tin sau:

Biu tn x cho ta ci nhn tng i v mi quan h gia hai bin nh lng;

Biu tn x cho ta s c on rng liu hai bin nh lng c quan h c tnh quy lutvi nhau hay khng, v nu c th quan h dng tuyn tnh hay phi tuyn.

V d 3.3.6. Gi s ta c thu nhp v chi tiu ca 20 ngi nh sau.

Thu nhp 19 17 18 23 14 11 17 26 19 19 8 26 17 10 17 19 11 15 21 8

Chi tiu 12 7 10 10 6 5 10 15 11 9 3 11 8 6 9 8 4 7 12 4

Ta mun bit: liu c mt quan h c quy lut no gia hai bin thu nhp v chi tiu hay khng?

lm iu ny ta c th dng biu tn x nh mt cng c "khm ph" trc tin cho nhng

phn tch mang ngha v sau. Biu tn x ca 20 cp s trn nh sau:


Qua biu ta thy dng nh hai yu t trn c lin h quy lut vi nhau: thu nhp cng cao, chi

tiu cng tng. Ta ni rng hai bin trn c mi quan h thun.

Chng hn nu nh mt c iu tra no cho thy im hi lng i sng v chng ca cc b v

v s nm chung sng to nn biu nh di y (bn tri) th iu c ngha l cng chung

sng lu, cc b v cng km hi lng vi cuc sng v chng.

Cn nu nh kho st no v cn nng v ch s IQ m thu c biu tn x hnh bn phi,

th ta d on rng khng c lin h c tnh quy lut no gia ch s IQ v cn nng ca ngi

c!

Ni thm rng, cc im biu din tp trung quanh mt ng thng cho, ngi ta gi ng thng,

iu cho thy c mt s lin h "tuyn tnh" mnh gia hai bin trn. Mt i lng thng k

c xy dng o mc lin h v mt "s lng" ny c tn gi "h s tng quan tuyn tnh

mu", n c tnh bng cng thc sau:

r =x1y1 + x2y2 + . . . xnyn nxy

[(x21 + x22 + . . .+ x

2n nx2)][(y21 + y22 + . . .+ y2n ny2)]


Trong , x =x1 + x2 + . . .+ xn

n, y =

y1 + y2 + . . .+ ynn

ln lt l trung bnh cng ca cc bin xi

v yi (i = 1, n).

Gi tr ca r nm gia 1 v 1. Gi tr r cng gn 1 th cng c s tng quan tuyn tnh thunmnh gia hai bin nh lng, tc l cc im cng c xu hng tp trung quanh mt ng thng

dc ln, cn cng gn 1 th cng c s tng quan tuyn tnh nghch mnh gia hai bin nhlng, tc l cc im cng c xu hng tp trung quanh mt ng thng dc xung. Nu r cng

gn 0 th cc im khng tp trung theo ng thng no.

3.3.4.2. Biu cho m t phn phi ca bin nh tnh ny theo bin nh tnh khc

y thc ra l mt dng biu din trc quan ca bng tn s cho gia hai bin nh tnh. Theo ,

biu gm cc nhm theo bin nh tnh th nht, mi nhm ny c cu thnh t cc thanh l

biu din cho s lng ca mi biu hin ca bin th 2.

Chng hn, nu mt iu tra cho ta tng hp cho ca gii tnh phn chia theo khu vc nh sau:

Th c biu cho ca gii tnh chia theo nhm khu vc nh sau:

Biu ny cho ta thng tin v mc cn bng gii tnh mi khu vc. Qua biu ta thy trong

tp d liu ny: hi o, min ni, nng thn c s cn bng v gii tnh, cn thnh ph th mt

cn bng gii tnh.

Cn nu nh t bng tn s cho ca khu vc chia nhm theo gii tnh nh sau:


Th c biu cho ca khu vc chia theo nhm gii tnh nh sau:

Biu ny cho ta thy phn b v cc khu vc sng ca mi gii tnh. Qua y ta thy c nam v

n u tp trung thnh ph, sau n min ni, nng thn ri hi o. Nhng mc tp tp

thnh ph ca nam l ln hn so vi n.

Chng 4

Tm tt d liu bng cc i lng thngk m t

Trong chng trc chng ta bit cch tm tt d liu qua bng tn s v qua biu . Cc

phng php rt hu ch cho vic biu din d liu v truyn ti thng tin nhanh chng. Tuy

nhin, chng li khng dng c trong cc suy lun thng k. Chng hn khi ta cn so snh hai

tp d liu vi nhau, nu ta dng biu so snh th gn nh ng nhin ta s thy rng chng

khc nhau. Nhng ta mun ch ra mc khc bit th s phi m t nh th no? R rng l

n s mang tnh cm tnh v khng khch quan. Nhng hn ch s c khc phc bi nhng

thc o m t bng s.

4.1. Cc s o hng tm ca tp d liu

4.1.1. Trung bnh cng, trung v v mode

nh ngha 4.1. Trung bnh cng n gin c tnh bng cch cng tt c cc gi tr quan st

ca tp d liu ri chia cho s quan st ca tp d liu .

Trung bnh cng n gin x ca cc gi tr x1, x2, . . . , xn c cho bi cng thc:

x =x1 + x2 + . . .+ xn

n

V d 4.1.1. V d: Gi s s tin (n v nghn VND) dng cho chi tiu thc phm trong mt

tun l: 120, 150, 125, 100, 180, 140, 200. Khi s tin trung bnh mt ngy trong tun dnh cho

chi tiu thc phm l:

x =120 + 150 + 125 + 100 + 180 + 140 + 200

7= 145.

nh ngha 4.2. Trung bnh cng c trng s c tnh bng cch cng cc tch ca gi tr

quan st ca tp d liu vi trng s tng ng ri chia cho tng cc trng s ca tp d liu .

Cng thc tnh trung bnh cng c trng s

Trung bnh c trng s x ca cc gi tr xi vi trng s tng ng wi, i = 1, . . . , n c tnhbi cng thc.

xw =x1w1 + x2w2 + . . .+ xkwk

w1 + w2 + . . .+ wk,

26

4.1. Cc s o hng tm ca tp d liu 27

V d 4.1.2. im thi kt thc hc k ca mt sinh vin c cho trong bng di y:

Mn hc im S .v tn ch

XSTK v ng dng 7.0 4Nguyn l k ton 5.0 2Ton Qun l 8.0 2M hnh Kinh t 4.0 2Th trng chng khon 6.0 2

Khi im trung bnh cc mn hc ca sinh vin trn trong k ny s l:

x =7 4 + 5 2 + 8 2 + 4 2 + 6 2

4 + 2 + 2 + 2 + 2= 6.17.

u im v nhc im ca trung bnh cng:

u im:

Trung bnh cng l mt s o hng tm ph bin v n tc ng vo mi phn t ca

tp d liu;

Mi tp d liu u c duy nht mt trung bnh cng;

Trung bnh cng l mt khi nim ton hc quen thuc, hn na n c nhiu tnh cht

ton hc gip cho ta c th thc hin c cc suy din trong thng k.

Nhc im:

Trung bnh cng b nh hng bi cc gi tr ngoi bin;

Trung bnh cng ch dng cho d liu o bng thang o nh lng.

Ch Trung bnh cng ca d liu tng th thng c k hiu l , trung bnh cng d liu mu

thng c k hiu l x.

nh ngha 4.3. Trung v ca tp d liu c sp th t l gi tr m c khng qu 50% s

quan st ca tp d liu c gi tr nh hn trung v v khng qu 50% s quan st ca tp d liu

c gi tr ln hn trung v.

tm trung v ca tp d liu c n quan st ta lm nh sau:

Sp xp li tp d liu;

Nu s quan st n l s l th trung v l quan st v tr th (n+ 1)/2.

Nu s quan st n l s chn th trung v gi tr trung bnh cng ca hai quan st v tr chnhgia ca tp d liu, tc l hai quan st v tr th n/2 v (n+ 2)/2.

Nhn xt: Trong nhiu trng hp ta c th ni c khong 50% s quan st ca tp d liu nh

hn hoc bng trung v v khong 50% s quan st ca tp d liu ln hn hoc bng trung v.

V d 4.1.3. Trung v ca cc tp d liu

28 Chng 4. Tm tt d liu bng cc i lng thng k m t

1. 5, 11, 9, 12, 10, 20, 15, 30, 25 l 12.

2. 5, 11, 9, 12, 10, 20, 15, 30 l 11.5.

u im v nhc im ca trung v

u im:

Trung v khng b nh hng bi nhng gi tr ngoi bin;

Mi tp d liu u c duy nht mt trung v;

C th tm trung v cho tp d liu s dng thang o th bc, khong, t l.

Nhc im:

Trung v nh ng gia gi tr vi th t ca gi tr trong tp d liu;

tm trung v ta phi sp xp li th t d liu v vic ny tn rt nhiu thi gian.

V d 4.1.4. Mode (yu v) ca mt tp d liu l gi tr xut hin nhiu nht trong tp d liu.

tm mode ca mt tp d liu ta lm nh sau:

Lp bng tn s cho tp d liu;

Tm gi tr ln nht trong cc tn s;

Tm cc gi tr ca tp d liu tng ng vi tn s ln nht; v mode ca tp d liu l ccgi tr ny.

V d 4.1.5. Tm mode ca cc tp d liu sau v nhn xt.

0, 1, 3, 1, 5, 2, 6, 2, 9, 2.

2, 3, 2, 5, 7, 8, 7, 15.

0, 1, 2, 3, 4, 5, 6.

u im v nhc im ca Mode

u im:

Mode khng b nh hng bi gi tr ngoi bin ca tp d liu;

Mode l i lng thng k m t duy nht c th s dng cho tp d liu c s o thuc

thang o nh danh.

Nhc im:

Mode ch quan tm n cc gi tr xut hin nhiu nht m khng quan tm n nhng

gi tr cn li ca tp d liu;

Mode ca tp d liu c th khng duy nht, c nhng tp d liu c nhiu mode.

4.2. Cc i lng m t s phn b ca tp d liu 29

4.2. Cc i lng m t s phn b ca tp d liu

4.2.1. T phn v v phn v th p

Gi s rng bn c ngi thn thi vo mt trng i hc no . Trng ny s ly 25% th sinh

nguyn vng 1 xp theo im t cao xung thp. Bn c tp d liu im ca cc th sinh d thi

ca trng ny. Lm sao bn bit nhng th sinh t bao nhiu im tr ln l nguyn vng 1?

Con s ta cn tm phn chia tp d liu thnh hai phn: mt phn nh hn con s ny chim 75%,

cn phn ln hn chim 25%. Mt con s kiu nh vy c gi l phn v.

nh ngha 4.4. T phn v chia tp d liu sp xp theo trt t tng dn thnh bn phn c

s quan st bng nhau.

Cch tnh t phn v ca tp d liu c n phn t:

T phn v th nht k hiu l Q1 l gi tr ca quan st ti v tr xc nh bi cng thc25%(n+ 1);

T phn v th hai k hiu l Q2 chnh l trung v; T phn v th ba k hiu l Q3 l gi tr ca quan st ti v tr xc nh bi cng thc

75%(n+ 1);

V d 4.2.1. T phn v ca tp d liu: 10, 8, 5, 13, 15, 25, 35, 20, 25, 40, 60, 70 ln lt l 12.25,

22.50, 36.25.

nh ngha 4.5. Phn v th p ca mt tp d liu c sp th t l gi tr chia tp d liu

thnh hai phn, mt phn khng qu p% s quan st c gi tr nh hn phn v th p, phn cn li

c khng qu (100 p)% s quan st ln hn phn v th p.

Ch , trong nhiu trng hp ta c th ni khong p% s quan st ca tp d liu nh hn hoc

bng phn v th p v khong (100 p)% ln hn hoc bng phn v th pCch tnh phn v th p ca tp d liu c n phn t:

1. Sp xp d liu theo chiu tng dn.

2. Phn v th p l gi tr c v tr c xc nh bi cng thc i =p

100(n+ 1).

V d 4.2.2. Phn v th 60 ca tp d liu c sp xp

15, 32, 42, 65, 87, 92, 100, 105, 110, 130

l 95.2.

4.3. Cc i lng o lng phn tn

4.3.1. Khong bin thin, tri gia, phng sai, lch chun

nh ngha 4.6. Khong bin thin ca mt tp d liu l hiu gia gi tr ln nht v gi tr

nh nht ca tp d liu.


V d 4.3.1. Khong bin thin ca tp d liu: 18, 8, 19, 13, 13, 19, 15, 13, 13 l R = 19 8 = 11;

u im v nhc im: Khong bin thin l s o phn tn n gin v d hiu nhng n ch

ph thuc vo gi tr ln nht v gi tr nh nht ca tp d liu rt nhy cm vi cc gi tr ngoi

bin v b qua cch phn b ni b ca tp d liu.

nh ngha 4.7. tri gia ca mt tp d liu l hiu chnh lch gia t phn v th ba v

t phn v th nht ca tp d liu. Ta k hiu l RQ: RQ = Q3 Q1, trong Q3, Q1 l t phn vth ba v th nht ca tp d liu.

V d 4.3.2. tri gia ca tp d liu: 10, 8, 5, 13, 15, 25, 35,20, 25, 40, 60, 70 l 36.25 -

12.25 = 24.

nh ngha 4.8. lch tuyt i trung bnh ca mt tp d liu gm n quan st c gi tr

x1, x2, . . . , xn vi trung bnh x c cho bi cng thc:ni=1 |xi x|

n.

u im v nhc im ca lch tuyt i trung bnh:

u im: lch tuyt i trung bnh cn c vo mi im ca tp d liu v ch ra mt cchtrung bnh mi im d liu nm cch xa trung bnh bao nhiu.

Nhc im: Cng thc tnh lch tuyt i trung bnh dng n gi tr tuyt i nn khthc hin cc php bin i ton hc.

V d 4.3.3. Cho tp d liu v s tin chi tiu thc phm trong mt tun: 120, 150, 125, 100,

180, 140, 200. Trung bnh ca s tin chi tiu trong mt ngy l x = 145. Khi lch tuyt i

trung bnh cho s tin chi tiu thc phm l:

A =|120 145|+ |150 145|+ |125 145|+ |100 145|

7+|180 145|+ |140 145|+ |200 145|

7= 27.14.

nh ngha 4.9. Phng sai ca mt tp d liu tng th, k hiu l 2, c xc nhbi cng thc: 2 =

Ni=1(xi )2

N, y l trung bnh ca tng th v N l s quan st

trong tng th.

Phng sai ca mt tp d liu mu, k hiu l s2, c xc nh bi cng thc: s2 =ni=1(xi x)2n 1 , y x l trung bnh ca mu v n l s quan st trong mu.

lch chun ca mt tp d liu tng th, k hiu l , l cn bc hai ca phng saica tng th:

=

Ni=1(xi )2

N.

lch chun ca mt tp d liu mu, k hiu l s, l cn bc hai ca phng sai mu:

s =

ni=1(xi x)2n 1 .

4.3. Cc i lng o lng phn tn 31

V d 4.3.4. Cho tp d liu

10, 15, 32, 18, 25, 65, 30, 38.

Tnh phng sai v lch chun trong c hai trng hp tp d liu dng tng th v dng mu.

nh ngha 4.10. Cho mt tp d liu thu gn (cho di dng bng tn s) vi xi l gi tr quan

st th i hoc gi tr i din ca t th i, fi l tn s ca quan st hoc t th i v s phn t ca

tp d liu n = f1 + f2 + . . .+ fk. Khi trung bnh cng ca d liu thu gn c cho bi cng thc:

x =x1f1 + x2f2 + . . .+ xkfk

f1 + f2 + . . .+ fk.

Ch

Khong d liu dng (, a) hoc (, a] chn gi tr i din l a. Khong d liu dng [a, b), [a, b], (a, b] hoc (a, b) chn gi tr i din l (a+ b)/2. Khong d liu dng (b,+) hoc [b,+) chn gi tr i din l b.

V d 4.3.5. Tnh im trung bnh ca cc th sinh thi mn ton khi A da trn bng tn s thu

gn sau:

Khong im im i din (xi ) Tn s (fi)[0.0, 1.5] 0.75 1317(1.5, 3.0] 2.25 967(3.0, 4.5] 3.75 518(4.5, 6.0] 5.25 151(6.0, 7.5] 6.75 18(7.5, 9.0] 8.25 3

4.3.2. Biu hp v ru

Biu hp v ru l mt biu mang hnh nh v thc o hng tm, phn tn v phn

phi ca mt tp d liu nh lng. Biu hp v ru cho ta cc thng tin v: gi tr cc i, gi

tr cc tiu, ba t phn v v cc quan st ngoi l ca tp d liu.

v biu hp v ru ca mt tp d liu, ta phi tnh mt s i lng thng k m t sau:

Trung bnh, trung v ca tp d liu; T phn v th nht (Q1), t phn v th ba (Q3), tri gia RQ = Q3Q1 ca tp d liu; Gi tr nh nht, ln nht v cc gi tr ngoi bin (nu c) ca tp d liu.

V d 4.3.6. tp d liu

10, 8, 5, 13, 15, 25, 35, 20, 25, 40, 60, 70

c gi tr nh nht l 5, ln nht l 70 v t phn v l: 12.25 22.50 36.25.

v tp d liu

10, 8, 5, 13, 15, 25, 35, 20, 25, 40, 60, 70, 100

c c gi tr nh nht l 5, ln nht l 100 v t phn v l: 13, 25, 40.

Ta c th v c biu hp v du ca hai tp d liu trn nh sau:


Ta thy c hai biu c ui (du) ko di sang bn phi, d liu tp trung ch yu phn u.

D liu phn b nh vy ta cn gi l lch phi.

i vi biu th hai ta thy xut hin mt im ring bit, tch ra khi biu . chnh l gi

tr ngoi bin, l gi tr tri hn hn nhng gi tr cn li ca tp d liu.

4.4. Cc i lng m t hnh dng ca tp d liu

Skewness l i lng o lng mc lch ca phn phi, cn c gi l h s bt i xng.

H s Skewness Sk c th tnh bng cng thc: Sk =Md

, trong , ,Md tng ng l trung

bnh, lch chun, trung v ca mt tp d liu.

Skewness v mi lin h vi trung bnh, trung v v mode.

Nu phn phi ca tp d liu l i xng th h s skewness bng khng v khi trung bnh,trung v v mode trng nhau;

Nu phn phi ca tp d liu tp trung bn tri th h s skewness dng v mode < trungv < trung bnh;

Nu phn phi ca tp d liu tp trung bn phi th h s skewness m v trung bnh < trungv < mode.

M t hnh hc h s Skewness

4.5. Phn phi chun 33

Kurtosis l i lng o mc tp trung tng i ca cc quan st quanh trung tm ca n trong

mi quan h so snh vi hai ui.

H s Kurtosis v hnh dng ca tp d liu:

Khi phn phi tp trung mc bnh thng th h s Kurtosis = 3;

Khi phn phi tp trung hn mc bnh thng (hnh dng ca phn phi cao v nhn vi haiui hp) th Kurtosis > 3;

Khi phn phi khng tp trung nh mc bnh thng (hnh dng ca phn phi phng v tridi) th Kurtosis < 3.

M t hnh hc h s Kurtosis

4.5. Phn phi chun

4.5.1. Tp d liu c phn phi chun

Mt tp d liu tng th c gi l tun theo phn phi chun nu phn phi ca n c dng i

xng hnh chung.

Phn phi chun khc trung bnh, cng phng sai:


Phn phi chun vi cng trung bnh, nhng phng sai khc nhau

Phn phi chun chim v tr rt quan trng trong l thuyt xc sut, l v tr trung tm trong cc

kt lun thng k sau ny.

Trong thc t, nhiu bin ngu nhin, nhiu qui lut tun theo phn phi chun, hoc xp x chun

nh trng lng v chiu cao ca ngi ln, thng minh ca tr em, im thi ca cc th sinh,

kh nng chu lc ca thanh st, sai s o c, sai s quan st, bn do ca my mc, trung bnh

cng ca mt s ln cc i lng ngu nhin c lp...

4.5.2. Quy tc thc nghim

nh l 4.11 (Qui tc thc nghim). i vi mt tp d liu tun theo phn phi chun, qui tc

thc nghim cho ta bit t l phn trm cc gi tr nm trong vng mt s ln lch chun tnh t

trung bnh, c th ta c cc kt lun sau:

C khong 68% s quan st ca tp d liu tp trung trong khong [ , + ].

C khong 95% s quan st ca tp d liu tp trung trong khong [ 2, + 2].

C khong 99.7% s quan st ca tp d liu tp trung trong khong [ 3, + 3].

4.5. Phn phi chun 35

V d 4.5.1. Ch s IQ ca con ngi c coi l tun theo phn phi chun vi trung bnh = 100

v lch chun = 15. Da vo qui tc thc nghim hy a ra nhng nhn xt v ch s IQ ca

con ngi.

Phn phi chun c mt s nhng c trng sau:

Phn phi c dng i xng hnh chung.

Cc s o hng tm bng nhau.

tri gia bng 1.33 ln lch chun.

Khong bin thin (trong thc nghim) bng khong 6 ln lch chun.

Khong 2/3 s quan st tp trung trong vng 1 lch chun tnh t trung bnh.

Khong 4/5 s quan st tp trung trong vng 1.28 lch chun tnh t trung bnh.

Khong 95% s quan st tp trung trong vng 2 lch chun tnh t trung bnh.

Chng 5

Xc sut cn bn v bin ngu nhin

5.1. Xc sut l g?

Trong t nhin cng nh trong x hi xut hin rt nhiu nhng hin tng m khng th ni trc

n xy ra hay khng xy ra khi thc hin mt ln quan st, nhng hin tng ny gi l hin tng

ngu nhin. Tuy nhin, nu tin hnh quan st kh nhiu ln mt hin tng ngu nhin trong

nhng hon cnh nh nhau th trong nhiu trng hp ta c th rt ra c nhng kt lun khoa

hc v hin tng ny.

V d khi tung ng xu mt hoc vi ln th ta khng th bit c tn sut xut hin mt sp l

bao nhiu nhng khi tng dn s ln tung ng xu th tn sut ny ngy cng n nh quanh con

s 1/2 v sau cng tin ti tr s 1/2.

Xc sut l mt b phn ca ton hc nghin cu cc hin tng ngu nhin. L thuyt xc sut

nhm tm ra nhng qui lut trong nhng hin tng "tng chng" nh khng c qui lut.

5.1.1. Nhng khi nim c bn ca xc sut

5.1.1.1. Khng gian mu, bin c s cp v bin c

nh ngha 5.1. Php th l mt qu trnh din bin c xut hin cc kt qu. Php th c gi

l ngu nhin nu ta khng th d bo trc kt qu no s xy ra.

Chng hn: Tung mt ng xu v quan st xem mt no xut hin (mt s hay quc huy), gieo mt

con xc xc v ta quan tn n mt trn cng ca n c bao nhiu chm, st pht 11m v ta quan

tm n vic c vo hay khng vo, chn ngu nhin mt cng ty H Ni v ta quan tm n cng

ty l thuc loi hnh cng ty no, ...

nh ngha 5.2. Mt kt qu ca php th c gi l mt bin c v thng c k hiu l

A,B,C, ... Bin c i khi cn c gi l "s kin".

Bin c s cp l mt bin c m khng th phn tch thnh cc bin c khc, thng c k hiu

l .

Chng hn, vi php th gieo mt con xc xc, kt qu ca php th c th l: "mt trn cng ca

con xc xc l mt 2 chm" - y l bin c s cp, "mt trn cng ca con xc xc c s chm

chn" - y khng phi l bin c s cp; kt qu ca php th chn ngu nhin mt cng ty H

36

5.1. Xc sut l g? 37

Ni c th c cc kt qu l: "cng ty c chn l cng ty lin doanh" - y l bin c s cp, "cng

ty c chn khng phi cng ty 100% vn nh nc" - y khng phi l bin c s cp, ...

nh ngha 5.3. Khng gian mu l tp hp tt c cc bin c s cp, thng c k hiu l .

V d: Gi s mt gia nh d nh sinh hai con. K hiu G l con gi, T l con trai khi khng

gian mu ca php th trn l: = {GG,GT, TG, TT}

5.1.1.2. Kt qu thun li ca mt bin c

nh ngha 5.4. Bin c s cp c gi l kt qu thun li cho bin c A nu kt qu ca php

th l th bin c A xy ra.

Chng hn, trong v d v d nh sinh 2 con ca cp v chng trn, ta gi A l bin c "cp v

chng sinh c con trai" c cc bin c s cp thun li l TT, TG, GT. Khi mt trong 3 bin c

s cp ny xy ra th A s xy ra.

nh ngha 5.5. Bin c khng th l bin c khng bao gi xy ra (tng ng vi ).Bin c chc chn l bin c lun lun xy ra (tng ng vi ).

V d: Xt php th tung hai con xc xc (l loi xc xc 6 mt, s chm trn cc mt t 1 n 6)

v quan st s chm ca mt trn cng. Khi bin c: "Tng s chm mt trn cng ca hai xc

xc l 13" l bin c khng th; cn bin c "Tng s chm mt trn cng ca hai xc xc ln hn

1" l bin c chc chn.

Mt v d khc, xt php th: chn ngu nhin mt sinh vin lp ny bin c "sinh vin c chn

trn 16 tui" l bin c chc chn. Cn bin c "sinh vin c chn trn 140 tui" l bin c khng

th.

5.1.1.3. Quan h gia cc bin c

nh ngha 5.6. Cho hai bin c A v B.

iu khng inh "C t nht mt trong hai s kin A, B xy ra" c th xy ra hoc khng xy ra,

v vy n l mt bin c. Ta gi n l bin c tng ca hai bin c A v B, k hiu l A B hocA+B.

Nh vy, bin c A + B xy ra trong cc trng hp: A xy ra, hoc B xy ra, hoc c A v B xy

ra.

Khng nh "C hai s kin A, B xy ra" cng l mt bin c, n c gi l bin c tch ca hai

bin c A v B, k hiu l A B hoc AB.Nh vy bin c AB xy ra khi c A v B xy ra.

Bin c i ca bin c A, k hiu l A l bin c xy ra khi v ch khi A khng xy ra. Ta c

A = \A.

V d: Xt php th chon ngu nhin mt ngi ph n H Ni. Ta gi A l bin c "ngi ph n

ny c chng giu c", B l bin c "ngi ph n ny hi lng vi cuc hn nhn". Khi A+ B

38 Chng 5. Xc sut cn bn v bin ngu nhin

l bin c: "ngi ph n ny c chng giu hoc c ta hi lng vi cuc hn nhn", bin c AB l

"ngi ph n ny c chng giu c v c ta hi lng vi cuc hn nhn ny".

nh ngha 5.7. Hai bin c A v B c gi l ng kh nng nu nh chng cng kh nng xut

hin, tc l kh nng xut hin ca bin c ny ging kh nng xut hin ca bin c kia.

Chng hn, tung mt con xc xc cn i, khi hai bin c: "mt trn cng l mt 1 chm" v

"mt trn cng l mt 2 chm" l ng kh nng.

nh ngha 5.8. Hai bin c A v B c gi l xung khc khi v ch khi chng khng th cng xy

ra trong mt php th.

H cc bin c A1, A2, . . . , An c gi l xung khc tng i nu nh hai bin c bt k trong cc

bin c trn xung khc vi nhau.

Nh vy ta thy rng hai bin c i ca nhau th xung khc. Tuy nhin ngc l ng nhin khng

ng.

Chng hn, trong php th tung xc xc ni trn, gi A l bin c "mt trn cng ca xc xc c

s chm chn", B l bin c "mt trn cng c 1 chm". Khi , trong 1 ln tung khng th c c

A, B xy ra c, do A v B l xung khc. Tuy vy, bin c i ca A l A: "mt trn cng ca

xc xc c s chm l" l bin c khc vi bin c B.

5.1.2. nh ngha xc sut

5.1.2.1. Khi nim chung v xc sut

o kh nng xut hin ca mt bin c, ngi ta dng mt khi nim gi l xc sut. Xc sut

ca mt bin c A l mt con s thuc on [0; 1]. Xc sut mt bin c cng gn 1 th n cng c

kh nng xy ra, cng gn 0 th cng t kh nng xy ra. Ta k hiu P (A) l xc sut hay s o kh

nng xut hin ca A.

5.1.2.2. nh ngha c in v xc sut

nh ngha 5.9. Trong mt php th c n kt qu ng kh nng v xung khc, trong m kt qu

c li cho bin c A, khi xc sut (theo ngha c in) ca bin c A l t s:

P (A) =m

n=|A|||(=

S kt qu c li cho A

Tng s kt qu xy ra).

Trong trng hp ny, vic tnh xc sut qui v vic m tng s kt qu c th xy ra v s

kt qu thun li. V d, tung mt ng xu cn i 2 hai ln, cc kt qu c th xy ra l =

{SS, SN,NS,NN}, cc kt qu s cp xut hin vi cng kh nng v ng xu l cn i. Khi bin c A: "xut hin mt S trong hai ln tung" c 3 kt qu c li l SS, SN, NS. Do

P (A) = 3/4 = 0.75

Ta thy rng xc sut theo ngha c in c tnh ton ch da trn t duy logic da trn vic m

c s kh nng xy ra. iu c ngha l s kt qu ca php th l hu hn, hn na, cc kt

qu phi l ng kh nng, iu ny khng phi lc no cng c c. Chng hn, ng xu trn

5.1. Xc sut l g? 39

b li mt S, khi php th trong v d trn, kh nng xut hin ca SS khng cn nh NN

na v nh vy bin c A khng tnh c. Trong trng hp ny ta cn xc nh c xc sut

xut hin mt S l bao nhiu. C mt cch lm iu ny l ta tung rt nhiu ln ng xu ny v

tnh t l s ln xut hin S trn tng s ln tung. S ln tung cng ln th t l ny ch quanh mt

s nht nh v c coi l xc sut ca bin c xut hin S trong mt ln tung. Cch xc nh xc

sut nh vy gi l xc sut theo ngha thng k.

5.1.2.3. nh ngha thng k v xc sut

nh ngha 5.10. Tin hnh n php th c tin hnh trong nhng iu kin ging nhau. Gi k

l s ln bin c A xut hin. Khi s php th n ln, t sk

ns dao ng quanh mt con s. Ta

gi con s ny l xc sut ca bin c A theo ngha thng k.

V d: nghin cu kh nng xut hin mt sp (bin c A) khi tung mt ng xu, ngi ta tin

hnh tung mt ng xu n ln, m s ln xut hin mt sp n(A) v thu c kt qu sau y:

Ngi th nghim n kk

nBuffon 4040 2048 0.5069Pearson 12000 6019 0.5016Pearson 24000 12012 0.5005

Qua bng trn ta thy khi s php th tng ln th tn sut xut hin mt sp s dao ng ngy

cng t hn xung quanh gi tr khng i l 0.5. iu cho ta hi vng khi s php th tng ln v

hn, tn sut s hi t v gi tr 0.5, t ta c th kt lun xc sut theo ngha thng k xut

hin mt sp P (A) = 0.5.

Mt v d khc, gi s mt kho st kin ca 1000 ngi dn M cho thy c 560 ngi ng h

vic bnh thng ha quan h vi Cuba. Khi , xt php th chn ngu nhin mt ngi dn M

v phng vn v thi vi vic bnh thng ha quan h vi Cuba, gi A l bin c "ngi c

chon ng h bnh thng ha quan h hai nc" th xc sut ca bin c A l 560/1000 = 0.56.

Ta thy rng nh ngha thng k v xc sut khng i hi cc kt qu phi ng kh nng v xung

khc nhng n i hi phi thc hin php th nhiu ln. Tuy nhin, c nhng bin c l kt qu

ca php th m ta khng th th vi s ln ln, chng hn, bin c "bn thi mn thng k

x hi k ny" khng tnh c theo ngha thng k v php th, tc l vic tin hnh lm bi thi

ca bn cng lm l 3 ln, v n khng c coi l " ln", v gi d c tin hnh c s ln ln

th n cng khng c din ra vi iu kin nh nhau, v rng kinh nghim lm bi v kin thc

mn ny ca bn s tng theo s ln thi. Vy liu bin c ny c th tnh c theo ngha c in?

Cu tr li l khng, v mc d ch c 2 kh nng xy ra ( v trt) nhng hai kt qu khng

phi l ng kh nng (bn hc vng th kh nng cao hn trt v ngc li, hc khng vng

th kh nng sau xy ra ln hn). Kh nng bin c ny xy ra l bao nhiu? iu ph thuc

hon ton vo nhn nh ca bn thn bn v kin thc, k nng tch ly, v ch c bn mi xc

nh c n "khong chng" bao nhiu. Vic gn xc sut thi ca bn nh vy c gi l xc

nh xc sut theo ngha ch quan.


5.1.2.4. nh ngha xc sut theo ngha ch quan

nh ngha 5.11. Theo ngha ch quan, xc sut ca mt bin c c a ra da vo s nhy

cm hoc kh nng phn tch, phn on ca ngi xc nh xc sut.

Mt s v d minh ha cho cch xc nh xc sut theo ngha ch quan:

Bng vic hi vi cu ngu nhin, thy gio dy thng k cho rng kh nng thi mn thngk ca bn l 80%. Tuy nhin, bn rt lo lng v thiu t tin v nghe thin h n mn ny

kh. Bn d rng kh nng thi khong 60%.

Bng nhng kin thc v phn tch k thut, phn tch doanh nghip v kinh nghim ca mnh,ti cho rng kh nng tng gi mnh trong nm 2015 ca c phiu HVG l 90%.

R rng rng xc sut theo ngha ch quan khng c nhng tip cn khoa hc, thng da vo vn

kin thc, s hiu bit v kinh nghim tch ly c ca ngi xc nh xc sut.

Mc d vy, xc sut a ra theo ngha ch quan vn c chp nhn nh mt s "tham kho",

chng hn trong nhng tnh hung cc chuyn gia d on mt xu th, . . .

5.1.3. Mt s qui tc tnh xc sut

Qua cc nh ngha ta thy rng, xc sut ca mt bin c nm gia 0 v 1, bin c chc chn c

xc sut l 1, bin c khng th hay cn gi l bin c rng c xc sut l 0. Sau y ta s tm hiu

v cc quy tc tnh xc sut.

5.1.3.1. Qui tc cng xc sut

Mnh 5.12. Cho A v B l hai bin c ca cng mt php th. Khi ta c

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

c bit khi A, B l hai bin c xung khc, tc bin c tch AB l bin c khng th, ta c P (AB) = 0

nn ta c:

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

c bit, ta c A v A l h xung khc v P (A+ A) = 1 nn:

P (A) = 1 P (A).

Mt cch m rng, nu cc bin c A1, . . . , An xung khc tng i th ta c:

P (A1 + . . .+ An) = P (A1) + . . .+ P (An).

V d: Theo s liu thng k mt vng, t l dn truy cp internet nhiu chim 40%, t l dn xem

ti vi nhiu chim 50%, t l dn va truy cp internet nhiu va xem tivi nhiu chim 10%. Chn

ngu nhin mt ngi dn vng trn.

1. Tnh xc sut ngi ny truy cp internet nhiu hoc xem tivi nhiu.

5.1. Xc sut l g? 41

2. Tnh xc sut ngi ny khng truy cp internet nhiu.

tin cho vic trnh by, ta gi A l bin c ngi c chn truy cp internet nhiu, B l bin c

ngi c chn xem tivi nhiu. Khi :

1. Xc sut ngi ny truy cp internet nhiu hoc xem tivi nhiu l P (A + B) = P (A) +

P (B) P (AB) = 0.4 + 0.5 0.1 = 0.8

2. Xc sut ngi ny khng truy cp internet nhiu l P (A) = 1 P (A) = 1 0.4 = 0.6.

5.1.3.2. Qui tc xc sut c iu kin

Xc sut ca mt bin c chu nh hng bi iu kin. iu ny d thy qua v d sau:

Tung mt ng xu hai ln, gi A l bin c c hai ln u c mt sp, B l bin c: ln tung u

tin c mt sp. Khi = {SS,NN, SN,NS}, A = {SS}, do P (A) = 1/4. Nhng nu bitbin c B xy ra, khi ch c hai kh nng l SN v SS. Xc sut A xy ra khi ny l 1/2 v ta

ni: xc sut xy ra A khi B xy ra l 1/2.

nh ngha 5.13. Xc sut ca bin c A c tnh vi iu kin bin c B xy ra gi l xc

sut c iu kin ca A khi B xy ra v k hiu l P (A|B).

Mnh 5.14. Xc sut c iu kin ca A khi B xy ra (P (B) > 0), c tnh bi cng thc:

P (A|B) = P (AB)P (B)

.

Do vy vi A v B l hai bin c ty , ta c

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

Nu P (A|B) = P (A) th c ngha rng vic xy ra B khng lm thay i kh nng xy ra A. Tiu ny ta c th ch ra c P (A|B) = P (A), tc l nu B khng xy ra th cng khng nh hngn kh nng xy ra A. Khi ngi ta ni A, B c lp vi nhau.

nh ngha 5.15. Hai bin c A v B c gi l c lp vi nhau nu nu vic xy ra hay khng

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

Ta thy rng hai bin c A v B l c lp vi nhau th P (A|B) = P (A) hoc P (B|A) = P (B). Nnnu A, B c lp ta s c P (AB) = P (A)P (B).

kim tra xem hai bin c A, B c c lp nhau khng ta kim tra mt trong cc iu kin sau:

P (A|B) = P (A).

P (A|B) = P (A).

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

V d 5.1.1. iu tra mc thng xuyn xem cc chng trnh th thao ca 500 cp v chng,

ta thu c bng s liu sau:


Mc vMc chng Thng xuyn Khng thng xuyn

Thng xuyn 100 150Khng thng xuyn 50 200

Chn ngu nhin mt cp v chng trong 500 cp trn.

1. Tnh xc sut ngi chng trong cp c chn thng xuyn xem chng trnh th thao.

2. Tnh xc sut ngi chng trong cp c chn thng xuyn xem chng trnh th thao bit

rng ngi v thng xem chng trnh th thao.

3. Hai bin c "V xem chng trnh th thao" v "Chng xem chng trnh th thao" c c lp

vi nhau khng?

Ta gi A l bin c: "ngi chng trong cp c chn thng xuyn xem chng trnh th thao".

B l bin c "ngi v trong cp c chn thng xuyn xem chng trnh th thao".

1. Ta c P (A) =100 + 150

500= 0.5.

2. C tt c 150 cp v chng m v thng xem chng trnh th thao. Trong s c 100 cp

l c chng xem chng trnh th thao, do vy P (A|B) = 100150

=2

3.

3. Ta thy rng P (A|B) > P (A), do , vic v thng xuyn xem chng trnh th thao lmtng kh nng chng thng xuyn xem chng trnh th thao. Do vy, A v B l khng c

lp.

5.2. Bin ngu nhin

Bin ngu nhin l i lng nhn gi tr s, xc nh trn tng bin c s cp ca khng gian

mu ca php th. Bin ngu nhin dng m t nhng c im ca cc bin c trong khng

gian mu.

V d 5.2.1. Xt php th: tung hai ng xu.

K hiu S l trng hp ng xu xut hin mt s v N l trng hp mt cn li, ti gi l mt

nga, ta c:

= {SS, SN,NS,NN}Gi X l i lng ch s mt nga xut hin trong php th.

Ta c X(SS) = 0 X(SN) = 1 X(NS) = 1 X(NN) = 2.

Nh vy X c th nhn ba gi tr: 0, 1, 2 ty thuc vo kt qu ca php th.

Mi bin ngu nhin l mt hm s, gn mi phn t ca khng gian mu vi mt s. Tp gi tr

ca mt bin ngu nhin c th l hu hn, c th v hn. Ta c th ly thm mt s bin ngu

nhin, chng hn, X trong mi tnh hung sau l bin ngu nhin:

1. X l s khch hng c mua hng trong ca hng trong s 10 khch hng tip theo.

5.2. Bin ngu nhin 43

2. X l s t mt ca hng bn c trong mt ngy

3. X l im thi mn TKXHH ca mt sinh vin c chn ngu nhin trong lp.

4. X l s t qua cu trong mt ngy.

5. X l chiu cao ca mt ngi trng thnh no c chn ngu nhin.

6. X l s thnh vin ca mt gia nh c chn ngu nhin H Ni.

7. Tung ng xu n khi no xut hin mt s th dng. Gi X l s ln tung.

5.2.1. Phn loi bin ngu nhin

Mt cch n gin ha, y chng ta gii hn xt n hai loi bin ngu nhin:

1. Bin ngu nhin ri rc: ch nhn gi tr: x1, x2, . . .

2. Bin ngu nhin lin tc: nhn gi tr trong mt khong [A,B]

V d 5.2.2. 1. Chn mt phng hc ngu nhin ti HTL. Gi X l s sinh vin ang hc

lp . Khi X l bin ngu nhin ri rc.

2. Chn ngu nhin mt thi im trong ngy hm nay. Gi Y l nhit chnh xc ti thi im

(s o ny khng lm trn bt c con s no). Khi Y c th l mt con s bt k trong

khong (11, 30), v tp hp cc gi tr ca Y c th ly lp y mt khong s no . Do vy,

Y l bin ngu nhin lin tc.

5.2.2. Phn phi xc sut ca bin ngu nhin ri rc

Cho mt bin ngu nhin ri rc X. Khi ta m t tt c cc gi tr X c th nhn cng vi xc sut

tng ng X nhn cc gi tr , ta ni ta c quy lut phn phi xc sut ca X.

C nhiu cch m t phn phi cho bin ngu nhin ri rc thng c cho di dng cng thc

hoc bng nh:

Gi tr ca X x1 x2 x3 . . . xkXc sut p1 p2 p3 . . . pk

V d 5.2.3. Xt php th tung hai ln mt ng xu, gi X l s ln xut hin mt N. Ta c:

Bin c SS SN NS NN

Gi tr ca X 0 1 1 2

Phn phi xc sut ca X:

Gi tri ca X 0 1 2

Xc sut1

4

2

4

1

4

Lu :

Gi s bin ngu nhin X c bng phn phi xc sut sau:



Khi ta c:

1. cc s pi nm gia 0 v 1.

2. tng p1 + p2 + . . .+ pk = 1.

5.2.3. Phn phi xc sut ca bin ngu nhin lin tc

i vi bin ngu nhin lin tc. Chng hn xt php th: o nhit ti mt thi im ngu nhin

trong ngy hm qua. Gi X l s o nhit chnh xc ti thi im (tc l ta khng lm trn

bt c s no). Nh vy mi gi tr ca X c th c v hn s, c th l s v t. Din bin tng

gim ca nhit ngy hm qua l t 15 n 26 , th X c th mang bt c mt gi tr no trong

on [15, 26].

Ta c th hnh dung rng, nhit ch bng 20 ti mt s thi im nht nh (lu rng 20 6=20.000001 , . . . ) trong v hn cc gi tr nhit trong khong [15, 20]. Do P (X = 20) = 0.

Ti bt c gi tr no, xc sut cng nh vy. Do m t cho phn phi xc sut trng hp

ny ngi ta m t xc sut dng P (a < X < a).

nh ngha 5.16. Mt cch tng qut, phn phi xc sut ca bin ngu nhin lin tc c m t

qua cc xc sut dng P (a < X < b). y ta xt trng hp m cc xc sut ny c th c tnh

t mt hm (qua mt cng thc tch phn), gi l hm mt ca X.

Nh hnh di y, ng cong gii hn l th ca hm mt , cn din tch ca phn c t

chnh l P (a < X < b). Tng din tch gii hn bi ng cong v trc honh bng 1.

5.2.4. Trung bnh, phng sai ca bin ngu nhin

Lng thng va qua ca 10 c cho dy sau (n v triu):

5 6 8 3 5 8 8 6 4 5

Chn ngu nhin mt ngi trong s 10 ngi, gi X l lng thng va qua ca ngi . Ta c

bng phn phi xc sut ca X nh sau:


X 3 4 5 6 8

Xc sut 1/10 1/10 3/10 2/10 3/10

Ta thy rng lng trung bnh ca 10 ngi trn l

(5 + 6 + 8 + 3 + 5 + 8 + 8 + 6 + 4 + 5)/10 = 5.8

Ta cng c th tnh gi tr ny t bng phn phi xc sut ca X:

1

10 3 + 1

10 4 + 3

10 5 + 2

10 6 + 3

10 8 = 5.8

Gi tr trung bnh tnh t bng phn phi xc sut ca X ny gi l k vng ca X. Nh vy k vng

ca X chnh l trung bnh ca tng th (cc s o) m ta ang xt.

Mt cch tng qut, nu bin ngu nhin ri rc X c phn phi xc sut


th

k vng ca X lEX = x1 p1 + x2 p2 + . . .+ xk pk

phng sai ca X l

V X = (x1 EX)2 p1 + (x2 EX)2 p2 + . . .+ (xk EX)2 pk

lch chun ca X l X =V (X).

i vi bin ngu nhin lin tc, tnh k vng v phng sai, ta cn dng n tch phn suy rng

v s khng c cp y.

5.2.5. Mt s phn phi quan trng

5.2.5.1. Phn phi nh thc

nh ngha 5.17. Php th nh thc l php th c cc c im sau:

Php th bao gm n th nghim ging ht nhau.

Mi th nghim ny ch c kt qu l "thnh cng", "tht bi".

Xc sut "thnh cng" trong mi php th u l p (v xc sut "tht bi" l q = 1 p).

Cc th nghim l c lp ln nhau (kt qu ca th nghim ny khng nh hng ti kt quca th nghim khc).

V d 5.2.4.

1. Tung mt ng xu 10 ln, trong mi ln tung, xc sut xut hin mt S u l 0.5.


2. Mt th sinh khng hc bi, chn p n ngu nhin tt c cc cu trong mt gm 100 cu

hi trc nghim (mi cu c 4 phng n la chn). Mi ln chn, kh nng ng u l 0.25.

3. Chn 5 sn phm t mt l hng (c rt nhiu sn phm) m t l t tiu chun l 84%. Khi

, xc sut sn phm l t tiu chun trong mi ln chn l 0.84.

nh l 5.18. Gi X = s ln "thnh cng" trong n th nghim ca php th nh thc.

th X c gi l bin ngu nhin c phn phi nh thc (bin ngu nhin nh thc), k hiu X B(n, p). Xc sut c k ln thnh cng trong n th nghim l

P (X = k) = Cknpkqnk =

n!

k!(n k)!pkqnk

V d 5.2.5. Tung mt ng xu 10 ln, trong mi ln tung, xc sut xut hin mt S l 0.5. Gi

X l s mt sp, X B(10, 0.5).

Phn phi ca X c minh ha nh hnh sau y:

nh l 5.19. Nu X B(n, p) th trung bnh ca X l EX = np, phng sai ca X l V X =np(1 p).V d 5.2.6. Mt th sinh khng hc bi, chn p n ngu nhin tt c cc cu trong mt gm

100 cu hi trc nghim (mi cu c 4 phng n la chn). Mi ln chn, kh nng ng u l

0.25. Khi gi X l s cu tr li ng th X B(100, 0.25).

Khi :

1. Xc sut n tr li ng 45 cu hi l

P (X = 45) = C45100 0.2545 0.7555 = 6.67 106

2. Xc sut tr li ng di 50 cu l P (X < 50) = C0100 0.2545 0.75100 + C1100 0.251 0.7595 + . . .+ C49100 0.2545 0.7551 = 0.9999999

3. S cu tr li ng trung bnh l EX = 100 0.25 = 25 cu.

Trong SPSS, ta c th tnh cc xc sut lin quan n phn phi ca bin ngu nhin nh thc

X B(n, p) nh sau:

Vo Transform trong thanh menu, chn Compute Variable trong danh sch s xung v g vo cc

cu lnh tng ng nh sau:


tnh P(X=k) g PDF.BINOM(k,n,p)

tnh P(X k) g CDF.BINOM(k,n,p)

tnh P(X>k) g 1-CDF.BINOM(k,n,p)

Lu rng X mang cc gi tr nguyn nn

P (X < k) = P (X k 1)

P (X k) = P (X > k 1)

5.2.5.2. Phn phi chun

Phn phi chun l tn gi ch mt h phn phi c dng hnh chung (cn gi l phn phi

Gauss). L loi phn phi thng gp trong thc t.

Bin ngu nhin X c phn phi chun c trng bi 2 tham s: trung bnh , phng sai 2. Ta

vit l X N(, 2). Ta c:

1. X l bin ngu nhin lin tc, c hm mt f(x) =1

2pie

(x )22 .

2. Khi trung bnh = 0, phng sai 2 = 1 th ta ni X c phn phi chun tc.

3. Nu X c phn phi N(, 2) th bin ngu nhin Z xc nh bi Z =X

c phn phi

chun tc.

V d 5.2.7. 1. Ch s IQ ca con ngi tun theo phn phi chun.

2. im thi Ielts ca tng th nhng ngi thi tun theo phn phi chun.

3. Trng lng ca n c (cng mt ging) trong mt h tun theo phn phi chun.

4. . . .

th ca hm f(x) =1

2pie

(x)22 ca bin ngu nhin X no c phn phi chun th hin

di dng mt ng cong, cng vi trc honh to nn mt hnh chung i xng, vi im cc

i c honh .

Khi phng sai cng ln, chung cng b "b" rng. Hnh sau minh ha cho cc trng hp c phng

sai khc nhau, nhng cng trung bnh:


Cc bin ngu nhin c phng sai nh nhau th c hnh dng chung ging nhau, ch x dch v tr

ty thuc mc khc nhau v trung bnh, nh minh ha sau:

Bin ngu nhin X phn phi chun vi tham s trung bnh v phng sai 2 th Xc sut X

nm gia 2 s a, b vit l P (a X b) v bng s o phn din tch c t mu nh hnh diy:

Trong SPSS ta c th tnh c xc sut ca phn phi chun X N(, 2) trong mt khong no, theo cch sau:

Trn thanh menu chn Transform, danh sch s xung chn Compute Variable, t tn cho bin v

g vo khung biu thc tnh cu lnh tng ng vi xc sut cn tnh:

tnh P (X a) hoc P (X < a) ta g CDF.NORMAL(a, , ).

tnh P (X a) hoc P (X > a) ta g 1 CDF.NORMAL(a, , ).


tnh P (a X b), hoc P (b < X < a), hoc P (a X < b), hoc P (a < X b) ta g:

CDF.NORMAL(b, , ) CDF.NORMAL(a, , )

i khi ta cn tm gi tr ngc li so ca xc sut ni trn. Tc l cho X N(, 2). Ta bit mts a thuc (0,1) v ta mun tm x0 P (X x0) = a. lm iu ny, ta vo mc tng t nh tnh xc sut ni trn. Nhng lnh g vo l IDF.NORMAL(a, , ).

Nu cn tm x1 P (X x1) = a th ta bin i v P (X < x1) = 1 a.

5.2.5.3. Quy tc thc nghim

nh l 5.20. i vi mt tp d liu tun theo phn phi chun, qui tc thc nghim cho ta bit

t l phn trm cc gi tr nm trong vng mt s ln lch chun tnh t trung bnh, c th ta c

cc kt lun sau:

C khong 68% s quan st ca tp d liu tp trung trong khong [ , + ].

C khong 95% s quan st ca tp d liu tp trung trong khong [ 2, + 2].

C khong 99.7% s quan st ca tp d liu tp trung trong khong [ 3, + 3].

Ta c th tm tt quy tc trn qua hnh sau:

5.2.6. Mt s phn phi khc

Ngoi hai phn phi trn, cn nhiu phn phi khc c coi l ph bin v ng vai tr quan trng

trong l thuyt thng k, nhng y chng ta khng xt n m ch gii thiu. Chng ta c th

tham kho trong nhiu ti liu khc su hn.

5.2.6.1. Phn phi t

Phn phi Student hay cn gi l phn phi t, k hiu l tn, trong n l tham s ca phn phi

gi l bc t do. Vi bc t do n khc nhau th ta c cc ng mt khc nhau tng ng. n

cng ln th ng mt cng gn ng mt ca phn phi chun ha. Nh hnh di y,

khi n = 100, hai ng (phn phi t100 v N(0,1)) gn nh l mt.


5.2.6.2. Phn phi Chi bnh phng

Phn phi Chi bnh phng, k hiu l 2n, trong n l tham s ca phn phi gi l bc t do.

5.2.6.3. Phn phi F

Phn phi F k hiu l Fn,m trong n,m l hai tham s, n c gi l bc t do t s, m c

gi l bc t do mu s.

Chng 6

c lng cc tham s tng th

6.1. Tham s tng th v tham s mu

6.1.1. Tham s tng th v tham s mu

Ta nhc li rng, mt tng th l tp hp tt c cc i tng m ta ang quan tm. Trong tng

th ny ta quan tm n mt t im no c o bi mt thang o. Mi phn t ca tng th

cho ta mt gi tr trn thang o ni trn. Tp hp tt c cc gi tr ny cho ta tng th s o v c

im m ta quan tm. Sau y ta s ch ni tng th, rng l tng th cc s o m ta quan tm

n. Chng hn, xt tng th ngi dn H Ni. Ta quan tm n thu nhp ca mi ngi dn,

mi ngi dn s cho ta mt con s th hin thu nhp ca ngi trong nm qua. Tp hp tt c

cc s thu nhp cho ta mt tng th thu nhp ca ngi dn H Ni. Khi ta ang cp n

thu nhp ca ngi dn H Ni ta s ni n t tng th, ta hiu ngm vi nhau l tng th cc s

o th hin thu nhp ca mi ngi dn.

Gi s ta c mt tng th c N phn t, ta ang quan tm n mt c im (bin) nh lng ca

n, mi mt phn t ta mt s o v bin ny, k hiu cc con s ny l x1, x2, . . . , xN . T nhng

con s ny ta tnh c trung bnh cng ca ca chng bng cch cng li cc phn t s ni trn

ri chia cho N, ta c s gi l trung bnh ca tng th, k hiu l , ta c:

=x1 + x2 + . . .+ xN

N

Phng sai ca tng th ni trn c tnh bi cng thc

2 =(x1 )2 + (x2 )2 + . . .+ (xN )2

N

lch chun ca tng th l =2.

Gi s ta quan tm n mt du hiu no trong tng th trn. Nu c m phn t m c du hiu

trn th t l tng th c du hiu trn l P =m

N.

Nhng i lng nh , 2, P, . . . ni trn gi l tham s ca tng th ang xt.

Tham s tng th l c trng ca tng th dng m t nhng c tnh ca tng th nh:

trung bnh, trung v, mode, phng sai, lch chun, . . .

Bng phng php chn mu, ta bit rng t tng th c th rt ra c mt tp con c gi l

51

52 Chng 6. c lng cc tham s tng th

mu.

Tng t nh tng th, ta cng c khi nim tham s mu, n l c trng ca mu dng m

t nhng c tnh ca mu nh: trung bnh mu, phng sai mu, lch chun mu, . . .

Vic xc nh cc tham s ca mt tng th ni chung l kh thc hin v a phn l khng thc

hin c. Chng hn, xc nh chiu cao trung bnh ca ngi dn tui 20 Vit Nam l khng

thc hin c, v tng th ny ln v ta s tn rt nhiu thi gian v cng sc. Mt v d khc l

ta xc nh nng cht c no trong cc chai nc tng bn trn th trng H Ni, vic ny

cng khng th thc hin c, v rng ta khng th m ht cc chai nc tng trn th trng

o, . . .

Do vy ta s c lng cc tham s ca tng th ch khng tm chnh xc c chng. Trong chng

ny ta s tm hiu cc phng php thng k t mt mu c chn c th suy din c lng

cho cc tham s ca tng th. Ta s tm hiu hai khi nim: c lng im v c lng khong.

6.2. c lng im

6.2.1. Hm c lng v c lng

Sau y, tin cho vic xy dng cc khi nim, ta s khng ni ring ti tham s c th no: trung

bnh, phng sai hay t l, . . .m ta gi chung cho chng mt ci tn, chng hn, . V sau ny khi

tin hnh c lng cho cc tham s c th ta s xt ring tng tham s.

Gi s cn c lng tham s ca tng th, ta chn mu ngu nhin gm n phn t X1, X2, . . . , Xn

v xy dng hm c tnh bng cng thc no t X1, X2, . . . , Xn, ta s k hiu cng thc

bi f(X1, X2, . . . , Xn), n biu din tham s mu tng ng vi tham s ca tng th cn c lng.

Khi hm c gi l hm c lng ca .

Chng hn, nu y l trung bnh ca tng th. Khi t mu X1, X2, . . . , Xn ta tnh trung

bnh mu X =X1 +X2 + . . .+Xn

nv ta gi l mt hm c lng ca .

Nu chn mt mu c th ta o c X1 = x1, X2 = x2, . . . , Xn = xn v tnh cc gi tr =

f(x1, x2, . . . , xn) ca hm tham s mu th mt gi tr tnh trn mu c th ny c gi l mt

c lng ca .

Nu vi mi mt mu c th hm c lng = f(x1, x2, . . . , xn) cho mt gi tr duy nht th

c gi l hm c lng im ca tham s tng th v gi tr tm c ny gi l c lng

im ca tham s tng th . Ta thy rng vi mu c th khc nhau th ta s c c lng im

khc nhau.

Trong l thuyt thng k, ngi ta xy dng cc tiu chun nh gi mc tt mt hm thng

k nh tnh "khng chch", "hiu qu", c lng "vng", c lng "", . . .

Gi s ta cn c lng tham s . Ta xy dng hm trn mu ngu nhin c chn. Ta hnh

dung rng do mu c chn l ngu nhin nn l khc nhau nhng mu khc nhau. Nu mi

mu ta thu c mt v ta tnh trung bnh ca cc con s ny m li ra c s bng ca tng

th th c gi l c lng khng chch ca

Tc l l c lng khng chch ca khi v ch khi E() =

6.2. c lng im 53

Gi s c l mt c lng khng chch ca . Tp hp cc gi tr m thu c cc mu khc

nhau ni trn cng c dao ng, ta ly phng sai ca chng lm thc o cho dao ng ny, chng

hn trong v d trn phng sai ca l xp x 0.361. Nu ta dng mt c lng khng chch khc

c lng cho th ta cng c mt i lng phng sai o mc dao ng ca cc gi tr

tnh trn cc mu khc nhau. Phng sai t v thng l khc nhau, ci no nh hn th c

lng ng vi n cho c coi l hiu qu hn (tt hn).

Nu nh ta tm v chng minh c c mt cng thc khng chch no c lng cho m c

c phng sai nh nht trong tt c cc cng thc c lng khng chch c th c th ta gi

l mt c lng "hiu qu nht" (hay tt nht) cho .

Chng ta tha hng kt qu sau, cc hm c lng sau y l khng chch v tt nht cho cc

tham s tng ng:

Tham s tng th Hm c lng c lng qua mu c th

=x1 + x2 + . . .+ xN

NX =

X1 +X2 + . . .+Xnn

x =x1 + x2 + . . .+ xn

n

2 =(x1 )2 + . . .+ (xN )2

NS2 =

ni=1(Xi X)2n 1 s

2 =

ni=1(xi x)2n 1

S =

ni=1(Xi X)2n 1 s =

ni=1(xi x)2n 1

P =m

NP =

K

np =

k

n

Trong K k hiu cho s phn t trong mu ngu nhin c du hiu m ta quan tm cn tnh t

l, k l gi tr ca K trong mu c th ta chn. Nh vy, ta c th tm gn li bng li:

Trung bnh tng th c c lng t trung bnh mu;

Phng sai tng th c c lng t phng sai mu;

T l tng th c c lng t t l mu.

tin cho vic minh ha, ta xt mt tng th gm t phn t.

V d 6.2.1. Chng hn, Lan mun tnh thu nhp trung bnh ca nhm bn thn hi cp 2 gm 8

ngi c thy. Ta s gi 10 ny l tng th bn thn cp 2 ca Lan. Gi s rng lng (n v triu

ng) mi thng ca 10 ngi h nh sau:

10 9 11 9 9 5 15 12

Nhng do iu kin no , Lan khng th thu thp c thng tin v lng ca h. Lan nh mt

thm t lm gip vic ny. Do qu bn bu ng thm t ny ch chn iu tra ngu nhin 3 ngi

trong danh sch m Lan a.

Chng hn rng ng thu thp c 3 trong s 8 ngi c lng nh sau:

15 10 9

T d liu ny ng ta c lng lng trung bnh ca tng th bn thn cp 2 ca Lan l X =15 + 10 + 9

3=

34

3, tc l xp x 11.3, y l c lng im cho lng ca 8 ngi trn.

54 Chng 6. c lng cc tham s tng th

Nu ng ta quan tm n phng sai ca tng th, ng ta s c

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