lei.;];) Hop Ihu nh4t eo 2 bi trang vii lObi
den; h¢p thu hai eo 8 bi
tr~rig va 4 bi den.tir m6i hQP Illy ng~u nhien ra I bi. Tinh xac suat ca
2 bi I~y ra Iii: a) d~u trAng; b) d~u den; c) I trAng I den?
Giiii
a) GOi A Iii bi~n e6 I~y dU'Q'e2 bi trAng:
Cl
Ci2
CI
Ci2
2 8
12 12
16
144
P(A) =_2x_8=-x-=-=_
Ta co: B = C + 0 + E. Do C. 0, E Iii bi~n c6 XKTD
PCB)= p(e + 0 + E) = P(C) + prO) + prE)
+ P(C)=?
Ta co: C =A,.A2
I
9
P(C)=P(AI·A2
b) GOi B la bi~n e6 lAydl1qe 2 bi den:
c; C~ 10 4 40
P(B)=-x-=-x-=_=_
Ci2
Ci2
12 12
144
= P(A,.A2
5
18
= P(AI
-
--
-
-
.A3+ A, .A2.Ad
.Ad
A, .A2 .A)
-
-
AI .A2·AJ + AI
.A3) + p(AI .A2.AJ)
.Ai .AJ)
+ p(AI .A2 .A3)
+ peAl )P(A2 )P(~) + peAl )P(A2 )P(A)
).P(A2)·P(A3)
c) Goi C Iii bi~n e6 lAyduoc I bi tring I bi den:
P(C)=
=Q7~Q2xQl+Q,27x0,8xQl +Q2~Q2xQ9=QO 151-0.02+0.045=0.0
G C! Go G
2 4 10 8 8 80 88 11
12 12 12 12 144 144 144 18
- x-+-!::x-=-x- +-x- =_+-'-=_ =_
G2
<:;2
G2 G2
-t-
~!)p
thu nh~t co 2 bi trAng va lObi den; hQp thu hai co 8 bi
trAn-gvii 4 bi den.tu m6i hQp liy ngdu nhien ra I bi. Tinh xac suAt ca
2 bi liy ra Iii:a) I tring I den; b) nhi~u nhAt m9t bi trang?
Giai
a) GOi A IIIbi~n e6 liy duoc I bi Iring I bi den:
P(A)=
C~ C~
~IO
-X-+-x-
C~
2
4
10 8
8
80
88
II
=-x- +-x- =_+_ =_ =_
C~'2 Cl11 C;'2 C;'2
12 12 12 12 144 144 144 18
P(O) = peAl .A2 .A3+ AI.A,
-
--,
..
=
1'8
=9 9
+ prE) '" ?
Ta co: prE) = PtA) .".0,54
V;iy: P(B)= P(C)+ P(O)+P(E)=O.08-0.375"'O.5~ = 0.995
-
= -128 = -8 /'
9
0,75x 0,8xO,9
= 0,54
b) Cach I:
Goi B la bi~n c6 c6 it nhc1t I x~ thu ban trung m\1etieu.
Goi C la bi~n c6 co 1 x~ thu bin trung ml,lCtieu.
GQi 0 Iii bi~n c6 co 2 x~ thu bAntrung m\lC lieu.
G9i E hi bi~n cd co 3 x~ thli bAntrung m\lC tieu.
<;:.
_
=-)
t;
-::)(:H0
G»."')jV \\J'~
/ \"'')\)\f)\I~l!)~
.
\.(:0)
~S2
)
:r~
P !'\ I).'" A - 1~IfJ).
=?
P(I)=p(AI
A2 A3)=p(AI).p(Ac).p(A)
= 0,25 x 0,2 x 0,1 = 0,005
V~y: P(F) = P(G) + P(H) +P(I) = 0,375 + 0.08 ~ 0,005 = 0.46
Caeh 2:
G9i F Iii bc eo it I I XII thu b~n kh6ng trung l11\1e tieu.
=>
j: Iii bi~n c6 co ccl3 X~ thli b~n trung m\le tieu.
Ta co: P (F)
=
PtA)
V~y: P(F) = 1- peA)
= I-
0,54
= 0,46
,
,
IJ..
Cau 4: Trong I hOp e6 I00 t~m the du'Q'cdanh s6 tir I den 100. RUI
ngdu nhien 2 the r6i d~t theo thu' t\f. Tinh xac suAt de: a) 2 the I~p
thllnh S6 c6 2 chu- 56: b) 2 the I~p thllnh s6 chia h~l eho 5: 2 the I~p
thimh s6 ehia h~t eho 2?
2.
.I"
0 " l'J : (Jr,
XII thu bAnkh6ng trung ITI\lC liell.
G9i G Iii bi~n c6 c6 1 x~ thti bAn kh6ng trung m\lC tieu.
Goi H Iii bi~n e6 e6 2 Xl;! thu ban kh6ng trllng m\le lieu,
GQi 1111 bi~n e6 e6 3 Xl;! thu bAn kh6ng trllOg m\le lieu.
Ta eo: F'" G + H + l. 00 G, H, I Iii bi~n c6 XKTD.
=> P(F) = peG + H -+- I) = P(G) + (PH) + P(I)
• P(G) '" ?
Ta co: peG} = P(D) = 0,375
• P(H) = ?
Ta c6: P(H) ~ P(C) .: 0,08
• P(I}
VelO
__ .
~ I - 0,25 x 0.2 x
"
C) each
I:
GQi F 1<\ bc co it I I
mQI m\le tieu. Xac suit trung m\le lieu eua
~~
thu nh~t Iii0,75; eua X~ thu thu hai hi 0,8; eua X~ thu thu ba Iii
0,9. Tinh xac suit dS: a) ea 3 x~ thu d€u bin trung m\lC tieu; b) c6 il
nhAt mQt x~ thu bin trung m\le tieu; e) e6 it nhit mQt x~ thu bin
kh6ng trung m\le tieu?
Giai
a) Goi AI Iii bi~n e6 x~ thu thu I bAntrung m\lClieu.
Goi A2 la bi~n c6 x~ thu thu 2 bin trung m\1clieu.
Goi AJ Iiibi~n c6 x~ thu thu ba bin trung m\lCtieu.
A,ia be x~ thu thu i bin trung m\lC tieu (i = I, 2, 3).
Ta co: A = AIA1A,. Do A" A2, AJ Iii bi~n e6 dQCI~p.
00 do: PtA) = P(AIA2AJ) = P(AI).P(A2).P(AJ)
=
.
'" 1._ p(AI ).p(A2 ).p(k)
1- 0 005 = 0,995
Ba x~ thu bin
thu bAntrung ml,lClieu.
=
.
&)
I Xl;!
=>P(B)=I-P(B)=I-P(AIA:A,I
Do do: P(C) = PtA} "'-
144
A.,)
=> B Ii; bi~n c6 ca 3 xa thu d~u b~n khang trung mllC tie~.
Cach 2: -fu~' 41~, Coo!.
Goi B Iiibi~n co I§y dl1qc nhifu nh~t I bi tring.
Goi C Iiibi~n c6 liy du9'C I bi trAng.
Goi 0 IIIbi~n c6 liy kh6ng .duQ'cbi tring.
Ta co: B = C + O. Do C, 0 hi bi~n e6 xung khic.
88
144
cio c~ -x-=10 4 40
P(O)=-x-=
CI') Ci2
12 12 144
88 40
V~y: P(B) = P(C)+ P(O) = +144 144
2'
)P(A2)P(A1) + P(AI )P(A2)P(A»
@_]_:
Goi B Iii bi~n c6 co it nh~t
1- -
-
.A3 + AI.A
+ P(AI )P(A) )P(A,)
=Q2S
=QI8t-UI35t-U06=:U375
= p(AI
b} Goi B la bi~n c6 I~y dU"),cnhi~u nhit I bi Iring:
Cach I: PIB} = I - P( B)
prO) =?
~.r"
a) GQi A Iii bi~n cd 2 ther'l?p thanh 56 co 2 che 56:
Cach 2: G9i B la be chon ducc it nh~t I 6ng thu6c t6t
B Iii be chon duoc kh6ng 6ng Ihu6c 161
P A - m _ A~ _ 8x9 _ 72 ~ ~
( )-;A~oo - 99xl00 - 9.900 - .,l:K
Ta co: PCB)
b) G9i B Iii bi~n cd 2 the I~p thanh s6 chi~ h~t cho 5:
P(B) = A~o~A~o =80x20 =~
Aloo
99x 100 99
c) G9i C Iii bi~n c6 2 the I~p thanh s6 chia h~t cho 2:
P(C) =
A~oxA~o 50x50 25
2
=
=A 100
99x 100 99
Cau 5: Trong I h¢p co chira 7 bi trAng va 3 bi den. LAy ngAu nhien
cung luc 4 bi. Tim xac suit d~ trong 4 bi liy ra: a) co 2 bi den; b) it
nhit 2 bi den; c) it nhit 2 bi tring?
Giili
a) GQi A Iii bi~c c6 4 bi liy ra co 2 bi den:
C~5 C~4 C~3 C~, C~I
210 2 10 210
P(A) = C~~cg = 3x84 = 252 = 84 = 28
C12
495 495 165 55
Cau 9: MQI lOp hQc c.o 4 bong den, moi bong co xac Sll~t bi,cha}£;
0,25. Lap hQc du as nell co it I 3 bong den sang. Tinh xac suat de Iflr
hQc khong au anh sang.
Giai
Ap dJ,mgeong thuc Bernoulli
35
210
=C,~.pK .4"-K
GQi A la lap h9c kh6ng au anh sang.
Pn(K)
G9i K Iii 56 bong den sang.
Ta co: peA) = p. (K ~ 2)
c;
=
x(0,75)2X (0,25)2+c~X(0.75)1
x(0,25)'+C; x (0.75)0X (0.25)4
= P.(K= 2)
=
4
7
1I.BIENNGAUNHIENRal ~C:
C x-2.+_J
Cl Cl x_
C5l
Cl8 Cl7 C8l Cl7
5
3
3
5
8
7
8
7
=
20
-+
56
P.(K = I) ~ P.(K = 0)
15
-+
56
_
Cau I: M¢t ki~n hilng co 5 SP t6t vii 3 SP x6u. Chc;>nngau nhien t.l
Ki~n hilng do ra 2 san ph§m (ch9n I I~n).
a) L~p bang phdn ph6i xac 5U~teua 56 SP t61 eh9n dllgc:
b) Tinh ky vQng, phuong sai ctia 56 san ph~m 16t;
c) Tim ham phan ph6i X3C Sll§1etia 56 san ph~m t6l.
Giai
GQi X Iii s6 san pham t61 ch9n aLIqc.
X co th~ nh?n cae gia tri tiI : 0, I, 2.
l
+_5
-x -+ -x -
-+
0,5625 + 0,047 + 0,004 ~ 0,631
P(X = 0)
5
C~9 C~8 C~7 C~6 C~;
Cau 8: M<)I t6 12 sinh vien g6m 3 niI va 9 nam. Chia 10 nily ra.3
nhom b~ng nhau m9t each nh~u nhien, dnh xac su~t de trong m~
nhom a~u co nfr.
Giili
G9i A Iii bi~n c6 trong m6i nhom d~u c6 nii.
b) Gc;>iB la bc chQn duQ'c it nhit 1 6ng thu6e t6t:
Gc;>iC Iii bc ehQn duQ'c I 6ng thu6c t6t
Gc;>iD Iii be eh9n duQ'c 2 6ng thu6c t6t
Ta co: 8 = C + D. Do C, D xung khAe
Do do PCB) = P(C + D) = P(C) = P(D)
8
c~o
__
P(D)
Cl X_4
CI =-x-=5 4 5
P(A) =_5
C8l C7I 8 7 14
=-x -+
Cl7
95 94 93 92 91 90
=0 729
100 99 98 97 96 95 '
Ciiu 6: M9t h<)pthu6c chua 5 6ng thu6e t6t va 3 6ng thu6e kem chit
IU'<;mg.Ch(,ln ngdu nhien I~n lUq! (khong hoiln I~i) tiI hQp ra 2 6ng
thu6e. Tim xac suit d~: a) eli 2 6ng thu6c ehc;>ndU'gc d~u t6t; b) it I co
I 6ng thu6e t6t?
Giai
a) Gc;>iA Iii b 2 6ng thu6c ehc;>ndU'qed~u t6t:
Cl X_4
CI
PCB) =_5
C8l C7l
Cl
__J_
=-X-X-x-x-x-
C) Gc;>iE Iii bi~n cd 4 bi 16yra c6 it nh~t 2 bi trAng.
G9i F Iii bi~n cd liy dU'C;>,c
2 bi tring (2T, 2D)
Gc;>iG Iii bi~n c6 liy dU'qc 3 bi trAng (3T, ID)
Gc;>iH Iii bi~n c6 liy dU'qc 4 bi trAng (4T)
Ta co: E = F + G + H. Do F, G, H Iii bi~n e6 XKTD.
M~t khk do F = A => P(F) = peA) = 63/210
<=> peE) = P(F + G + H) = P(F) + peG) + P(H)
63 C;xC~ c;
63 35x3
=-+
+--=-+---+210
CI~
CI~ 210 210
63 105 35 203
=-+--+-=210 210 210 210
C8l
I . 6/56 = 50/56
cioo
cixc~ =-+-=-=1/3
63
7
70
CI~
=
X
=--X--X--X--'x--x
= 3x21 = 63 =3/10
CI~
210 210
b) GQi B hi bi~n c6 4 bi liy ra co it nhit 2 bi den.
G9i C hi bi~n c6 liy dugc 2 bi den (2D, 2T)
Gc;>iD Iii bi~n c6 liy dugc 3 bi den (3D, IT)
Ta co: B = C + D. Do C,D la bi~n c6 xung khAc.
M~t khae: C = A => peA) = P(C) = 63/210
<=>
PCB)
peA
+
D)
peA)
63
=-+.
210
Cl
_.1
I· P (B) = I •
Cau 7:, M¢I 16 hang C? 100 SP chua 5% ph~ pham. Ki~m tra ngi
nh_ienIan lu~ 6 san ~ham trong 16 hang (truong hop khong hoiln I~i'
Neu co it nhat I phe pharn thi kh6ng mua 16 hang, tinh xac suit ~
hang decc mua?
Giai
Goi A Iii bi~n e6 16hang dircc mU3
GQi A,Ia be SP ki~m tra I~n thir i lit t6t (i = 1,2,3,4,5,6).
Ta co: A = A, A2A) A4A5A6
=>P(A)=P(A, AJ AJ A. A5A6)=P(A1).P(A21 A,).P(A11 A, A2).P(A.iAI•
A2.A).P(Asi AI.A2.AJ.A4).P(AJ A,.A2.A1.A45)
cJ2xci
P(A)=
-
=
15
50
=_
56
56
_
3
= _.C~ =C82 28
P(X = I) =
CIXCI
S
5x3 =_
15
=_
3
ci
28
C2 =_
10
P(X=2)=_s
28
Ci
c;
C; C~
CJ2
C2l
P (X = 3) = -
°
I
3
28
15
-
28
28
I
2. + I x J2. + 2 x J! = ~
28
28
+~
28
=~
28
28
:::
2. ::: 1,25
4
V~y: E(X) = 1,25
Phuong sai Var(X) =?
Theo cong thuc ta co: Var(X)
VOi E(X2) =
I
X,2P,
[E(X)
= E(X2)-
3
= XI2PI
+
xi
P2
28
V~y Var(X)
=
28
28
28
F(x):::O
F (x) ::: p(X
=0
1
X - 4 + I·, X -26 + 2 1 X -24 + 3 1 X - 6 = -44
60
60
V~y Var(X)
= I)
:::0) + p(X
15 18
N~u 2
F(x):::
0
N~u 1 < x S 2
28
I
N~u 2
O;x:5
0
3
28'
V~y ham phan phdi X3Csu~t la: F(x) :::
F(x)
28'
=0
rex) =p(X
f(x)
= 0) =-4
60
= p(X = 0) + p(X = I)
4 26 30
=-+-=60 60 60
f (x) = p(X = 0) + p(X = I).,. P(X
4 26 24 54
=-+-+-=60
-'O
18
-'I
60 15
= 2,93 - (1,53i = 0,5891
c) Ham phan ph6i xac su~t ella X co dang:
F(x) = p(X < x)
3
:::-+-:::28 28
60
') 9"-'
= __
= E(X1)-[E(X)Y
N~u x < 0:5 I
28
2
1=1
N~u x:5
3
F(x) = p(X::: 0) =-
N~u x < 0:5 I
r
4
= E(X")- [E(X) Y = 1,96 - (1,25)2
= 1,96 - 1,56 = 0,3975 = 0,4
~0
1,53
= Ix," P,
=.!2.+ 40=55=1,96
28
=
=?
E(X2) - [E(X)
Vai: E(X")
28
N~u I < x:5
x 4/60+ Ix 26/60+ 2x 24/60+ 3x60=23/15
Var(X)
c) Ham phan ph6i X8CSU~I co dang:
F(x) = p(X(x)
N~u x
/=1
=0
PJ
/=1
=02x_2_+12x.!2.+22x.!..Q
LX, p,
= "
Theo cong thirc ta co: E (X)
Tim phuong sai: Var(X)
Theo cong thirc, ta co:
Y
xi
+
b) Tim ky vong: E(X) = ?
+ X2P2 + XJPJ
= X1PI
X, P,
1=1
28
6 60
10
-
b) Theo cong thuc ta co: E(X)=
x
60 60 60
= -3 X -2 = _6
C52 CliO
6
2
n
=0
x-
c:,
C;8
phan h6i xac su~t cua X IIi:
-
P
C~ C; C~ 60 60 60
CJ C~ C~xc~ C~ =_+_=_
12 12 24
-X-+---x-
P(X=2)=
V~y bang phin phdi X8esu§t ella X la:
X
c~ e
C~.c~
22
C~ :::_+_=_
24 2 26
--x-+--x-
P(X=)=
28
60
60
=
2)
60
N~u 2(x
F(x) = 1
V~y ham phan ph6i xac su~t ella X la:
O;x:5 0
1;2
4
Cau 2: H<)p I co 3 bi Tva 2 bi D, h<)pII co 2 bi Tva 4 bi E> ChQn
ng&unhien tir h<)pI ra 2 bi (chQn I IAn)va tir h<)pII ra I bi.
a) L~p bang PP x8e sUl1t ella s6 bi trAng eh9n duqe;
b) Tinh ky vQng, phlIcmg sai ella 56 bi lrAng chc;mdugc;
c) Tim ham PP x8e sudt ella 56 bi trang eh9n dLIqC.
Giai
a) G9i X hi 56 bi trAng ehQn duqc.
X co th~ nh~n gia trj tl1: 0, I, 2 , 3
fl(X =0)=
C22 C4l
I 4 4
-x=- X -=_
C2s
6
CliO
6
5
60
-'0
60'
. 30
F(x) =
-'1 < x S 2
60 '
54 '2 < x S 3
60 '
1;3 < x
Cau 3: M<)!co quan co 3 xe 616: 1 xe 4 ch6; I xe 50 ch6 va I xe tal,
Xac su~t d~ trong m<)t ngay lam vi~c, cae xe duqc SD la 0.8; 0.4 \t.I.
0.9. Hay I~p lu~t phiin ph6i xac su~t cho s6 xe dlIQ'C su dung trong
m¢t ngay cua co quan.
a) L~p bang phan ph5i xac su~t cho s6 xe dlIQ'CSlr dung trong m¢t
ngay cua co quan;
b) Tinh ky vong, phirong sai cho xe dllQ'c SD trong 1 ngay cua co
a) GQi X la 56 IAnthu C.X co the nhan cac gia trj: 1,2,3,4,5.
P(X = I) = I - 0,3 = 0,7
P(X = 2) = 0, 3 x 0, 7 = 0,21
P(X = 3) =(0,3)2 xO, 7 = 0,063
quan;
P(X = 4) =(0,3)3 xO, 7 = 0,0189
c) Tim ham phan ph6i xac su§t cho s6 xe dlIQ'Csu dung trong m¢t
ngay cua co quan.
Giai
a) GQi X la 55 xe Slldung trong J ngay cua co quan.
X eo th~ nh*" cac gia tri: 0, I, 2, 3.
P(X = 5) = (0, 3)4 X 0, 7 + (0, 3)5 = 0,0081
V~y bang phiin ph6i xac sdt cua X la:
P(X = 0) = 0,
P
X
2 x 0,6 x 0, 1= 0, 012
P(X = I) =0,8xO,6x 0, 1+0,2xO,4x
0, J + 0,2x 0, 6x 0,9
= 0, 048+0,008 +0, 108 = 0,164
P(X = 2) =0,8xO,4x 0, I+0,8x 0,6xO, 9+ 0,2x 0,4x 0,9
=0,032+0,432+0,072
= 0,536
P(X = 3) ~0,8xO,4xO, 9 = 0,288
V~y bang phiin ph6i xac sdt cua X la:
b)
1
2
3
0,7
0,21
0,063
4
5
0,0189
0,0081
Tim ky vong E(X) =?
n
Theo cong thuc ta co: E(X) =
I x, PI
= X,P,
X4P4 = I x 0,7+ 2 x 0,21+3 x 0,063+4 x 0,0189+5 x 0,0081
Tim phuong sai: Var(X) = ?
xl
=
1.425;
3
0,288
b) Tim ky vong: E(X) =?
L x, P,
+ X,P, + X,1Pd
1:1
n
Theo cong thuc ta eo: E(X) =
Theo cong thuc, ta co:Var(X) = E(X2) -
[E( X)]2
1:1
= X1P1+ X2P2 + XlP) + X4P4 = 0 x 0,012 + 1 x 0,164 + 2 x 0,536
+3xO,288 = 21110 = 2,1
Tim phU'ong sai: Var(X) = ?
Theo cong thUc, ta eo:
Var(X)= E(X2)-[E(x)t
=
Lx,2 P, = X,2 p., + X;
P2 + xi P1 + xl P4
I:'
=0,164+2,144+2,592
V~y Var(X) = E(X2) - [E(X)
Vai: E(X)2 = LX,P,
r
=4,9
= 4,9 - (2, 1)2= 0,49
c) Ham phiin ph6i xac su§t cua X eo d~ng:
F(x) p(X < x)
=
N~lIx:50
F(x)=O
N~u O
F(x) = p(X = 0) = 0,012
N~u I
N~u2
= 0,712
N~u 3(x
F(x) = I
V~y ham phiin ph6i xac su~t ella X la:
O;x$O
ix 0,21T fx
/x 0,7+
=
1;3 < x
Ciiu. 4: M<)! ngu'oi va.o clla hang th~y co 5 chi~c tivi gi6ng nhau. Anh
ta de nghi dlIQ'Cthu hin 11lQ'ltimg chi~c d~n khi chQn dllQ'ctivi t5t thi
m~a va n~u cci 5 Idn thu d~u x~u thi khang mua. GQi X leis6 I~n thu.
Biet eae tivi dQe I~pv6i nhau va xac suAt I tivi xAula 0,3.
a) qp bang phan ph5i xae 5U~tclla X;
b) Tinh ky vQng, phll'ong 5ai ella X;
c) Tim ham phdn phdi XBCsudt cua X.
Gicii
E(X2)-[E(X)r=2,6119-(I,4251)~=
°
N~u2
N~u 3
0.58
-=
P(X < x)
F(x)=p(X=I)+p(X=2)=0.91
F(x) = p(X = I) + p(X = 2) +P(X ~ J)
= 0,973
N~u 4
F (x) = p(X = I) + p(X
P(X= 4) = 0,9919
N~u 5 < x
F(x)=1
V~y ham phan ph6i xac 5u~t cua X Iii:
o;x $1
= 2) +P(X -= 3)-
0,7;1 < x :5 2
F(x)
=
0,91;2
0,973;3 < x:5 4
0,9919;4 < x::; 5
1:5 < x
0,176;1 < x :5 2
0,712;2 < x $ 3
0,063+ 42x 0,0189+ 5"x 0,0081 = 2,6119
e) Ham phiin ph6i xac 5U~teua X co d~ng: F(:.:)
N~u x::; I
f(x) =
N~u I
F(x) = p(X = I) = 0.7
0,012;0 < x $ I
F(x)
= X'IP,+ X',P,+ X',p, ... X~'Pj+ X',P,:
I:'
V~yVar(X)=
4
V6i: E(X')
5
Cliu 5: M9t x~ thu co 4 vien d~n, anh ta bdn IAn11IQ1tung vien chI
d~n khi trUng m\le m\lC tieu ho~c h~t ca 4 vien d~n thi thai. Xac sua
ban trung m\le lieu cua m6i vien d~n la 0,7.
a) L~p bang phdn ph6i XBCSU~1CLlas6 vien d~n
b) Tinh ky vQng, phu'O'Ogsai cua 56 vien d~n dii b~n;
c) Tim ham phdn ph6i xac su~t CLlas6 vien d?n da b~n:
Giili
a) GQi X la 55 vien d~n dii bAn. X co th~ nh~n cae gia tri: I.2. 3,,.
P(X=I)=0,7
P(X = 2) =0,3x 0, 7
P(X = 3)
= 0,21
=(0,3)2 xO, 7 = 0,063
P(X=4)=(0,WxO,7+(0,3)4=
0,0189 + 0,0081 = 0,027
V~y bang phan phdi xac suit ella X la:
X
1
2
3
4
P
0,7
0,21
0,063
0,027
P(X=4)= C
C
I
5
I
4
= I x 0,7 + 2 x 0,21 + 3 x 0,063 + 4 x 0,027
l
1,417
5
C I
C' I X C I
X _2
-'
X -'
Cl J
C~
ci
;:!
'2
c: C
c: e
P(X5)=_4
II
=
c;
C
x
o, 8xO , 75x~x_!_=0
3 2
b) Tim ky vong: E(X) =? Thea cong thuc ta eo:
E(X) = LX'P'
1
X _3
I
4
2 1
0,8xO, 75x-x-x)
3 2
c/_
ei
C'
C'
X _2
X _I
X
c:3 Cl2
-l-
= 0,2
V~y bang phan ph6i xac SU§lcua X ia:
Tim phuong sai: Var(X) =?
Thea cong thirc, ta eo:
Var(X)= E(X2)-[E(X)Y
Voi: E(X)2 = 12 x 0,7 + z2 x 0,21 + 32 x 0,063 + 42 X 0,027 = 2,539
V~y Var(X)
=
: F(x)=O
N~ul
:F(x)=p(X=I)=0,7
N~u 2
:F(x)=p(X=I)+p(X=2)=0,7+0,21=0,91
= 3)
+ 0, 21+ 0,063 = 0,973
N~u 4 < x
F (x) = 1
v ~y ham phan ph6i xac suit ella X la:
0,7: 1 < x ~ 2
0,91:2
0,973:3
Thea cong thirc ta co: E(X)
=
LX, p,
,
E(X-)
4
,2
= ~x,
p,
= 12xO.2"'2-
= 0,2 +0,8+
1,8+3,2 +5
V~yVar(X)=
E(X2)-[E(X)t
N~u
=-
5
Cl
Cl5
xO,2+3'
,
xO,2+4'
,
xO.2+
= II
~ II -):
'" 2
: F(x)=O
:F(x)=p(X=I)=0.2
2 < x ~ 3: F (x)
= p(X
= I) + p(X = 2) = 0,2 + 0.2 = 0,4
+0.2
N~u 4 < x ~ 5
:F (x) = p( X = I) + p( X = 2) + p( X
= 0,2 + 0,2 + 0,2 + 0.2 = 0.8
N~u 5
:F(x)=1
= 2) + p( X = 3)
= 0,6
= 3) + p( X = 4)
O:X~ I
0.2: I < x ~ 2
V~y hilm phiin ph6i xac 5U~1ella X Iii: F(x) =
°'
= 2
0,4: 2 < x ~ 3
0,6:3 < x~ 4
0.8: 4 < x ~ 5
el Cl 4 1 4
P(X :02):0 _4 X _I = - X - = - =
Cl Cl 5 4 20
5
'
N~u 3 < x ~ 4 : :F (x) = p( X = I) + p(X
NglICrila Iftn IUQ1 ki~m tra tUng ehai eha d~n khi phat hi~n ehai thude
gia Ihi ngimg ki~m tra (giil su cae ehai thu6e phili qua ki~m Ira mai
X3c dinh dll<;yeehai thu6e gia hay ehai thu6e t6t).
a) L~p bang phan ph6i xae SU§Iella s6 ehai thu6e duge ki~m Ira;
b) Tinh ky vQng, phtrcmg sai ella s6 ehai thu6e dUQeki~m Ira;
c) Tim ham phan ph6i X8C su~t ella s6 ehai thu6e dUQcki~m Ira.
Giai
a) GQi X la 56 h~n ehai thu6c dUQ'eki~m tra
X co th~ nh~n cae gia trj: 1,2,3,4,5.
e5l
3
,=1
N~ul
Cau 6: M¢I h¢p dl,fng 5 ehai Ihu6e trang do eo mQI ehai Ihu6e gia.
P(X = I) = _I
=
Vai:
= 0.2 +0.2
1
= I X 0.2 ... 2 X 0.2 + 3 J
0,2 + 4 X 0,2 + 5 X 0,2 =0,2'" 0,4 ...0,6 ...0,8 ... I
Tim phU'011gsai: Var(X) = ?
Thea cong thirc, ta co:
N~ux~1
):4
Cl
0,2
c) Ham phan ph6i xac SU~Icua X co d~ng;
F (x) = p(X < x)
O:xSI
F(x)=
0,2
ky vong: E(X) =?
Var(X)=E(X1)-[E(X)f
N~u 3 < x S 4 ; F(x) = p(X = J) + p(X = 2) + p(X
= 0,7
5
n
E(X2)-[E(X)Y
= 2,539-(1,417)2 = 2,539-2,007 = 0,532
c) Ham phan ph6i xac suit ella X eo d\lng:F (x) = p(X < x)
N~ux~1
b) Tim
4
4
°
'
8XO 25 =
Cl
Cl
J
X_I =08X075X-=02
C4l
ClJ
'
,
3
P(X:03)=_4 X_3
g
,
'
°
,
1:5 < x
2
Cau 7: MQI ro m~n co 20 Irai trang do co 6 trai bi hu. ChQn ngi~
nhien Iii r6 do ra 4 Irai. GQi X Iii 56 Irai m~n hu ehl;mpilai.
a) L~p bang phan ph6i xae suftt ella X;
b) Tinh ky vQng, phuong sai ella X;
c) Tim ham phan ph6i xae SU§ICllaX.
Giai
a) GQi X iii s6 trai m~n hu phai chQn.
X co th~ nh~n cac gia Iri: 0, 1,2, 3,4.
10
•
=
P(X = 0) = CJ~
100 I
1)=
P(x=
1001
--:O
4845
6x364 = 2184
C~XCI34 =
c;
4845
1001
2184
--
p
--
4845
b)
Tim
4845
2
3
4
1365
-4845
280
-4845
--
15
4845
ky vong: E(X) = ?
CT ta
co:
E(X)
=
Cau 8: M¢t nguoi co 5 chia khoa b~ ngoai r§t giong nhau, trong d
chi co 2 chia mo duoc cira. Nguai do tim each rno cua bang each In
tung chia m¢t cho d~n khi mo duoc cua thi thoi (I§I nhien, chia nit
khong me duQ'cthi lo~i ra), Goi X Iiis6 chia khoa nguoi do sir dung
a) L~p bang phan ph6i xac su§t ella X;
b) Tinh ky vong, phuong sai cua X;
c) Tim ham phan ph6i xac su§t cua X.
Giili
a) GOi X lit 56 chia khoa nguoi do sir dung.
X co th~ nhlin cac gia tr]: 1,2,3,4.
L x,P, =
P(X
=
I) =
c;C = 5"2 = 0,4
1='
ox 1001+ IX 2 18\2X
4845
4845
<
1365+3X 280 +4X _!2_ = 5814= 1,2
4845
4845
4845 4845
Tim phlrang sai: Var(X) =?
Theo congtht'rc,ta eo:Var(X) = E(X2) -[E(X)
r
(I
P(X
C'5 C~l
P(X
ifp
=dx leoI+f/l84 +lx 1155+Jx 2m +4x~=2,15
,,; , ,
4845
4845
4845
4845
4845
V~y Var(X)
=
C;
= 3) =-'
[' ;
P(X=4)=-'
C;
C5l
=
50
N~lI 0
F(x)
=
("4 [' 1
F(x) = p(X =0) =-
5 4 3
= 0.2
C; Ci C; 3 2 I 2
x-x-x-=-x-x-x-=_=O.I
CJl
[I4
0,4
°
.,
5 4
Vliy bang phan ph6i xac
- (I, 2)2 = 0, 71
c) Ham phan ph6i xae su~t clla X co dllng:
F(X) p(X < X)
,
= .:: x ..::.= 0,3
C~ C; 3 2 2
x-- x-- = -x-x-
E(X2)-[E(X)f
= 2,15
N~u X
(I
= 2) = _J X ---l.
V&i:
f(X)=
s3
4845
l :4 < x
n
Theo
2
4830 ."
<4
--.J
V'IIY b'ang pih'an pIh~"
. X J'a:
OJ xae suat cua
I
=
F(x)
xC214 = __
15x91 = __
1365
C;o
4845
4845
20 x 14 = __280
P(X=3)= C6J x C'14 = __
C;o
4845
4845
C4 = __15
P(X =4) =_6
C;o 4845
0
3]85 I
-:
4845
4550 : 2 < x
4845
4845
P(X =2) = C26
X
s0
O:X
C;o 4845
CI2
I
5 4 3 2
10
cua X lit:
SU~[
2
3
4
0,3
0,2
0.1
b) Tim ky v9ng: E(X)
=?
Theo eong thlrc la c6:
E(X) =
n
1001
1 X 0,4 + 2 X 0,3 + 3 X 0,2 ...4 X 0, I '"'2
Tim phuO'ng sai: Var(X) ~ ?
4845
N~u I < x $ 2 :F(x) = p(X = 0) + p(X = I)
1001 2184 3185
4845 4845 4845
N~u2
1001
2184
1365
4550
= --+--+
--=-4845
4845
4845
4845
N~u3
=--+--=--
F (x) = p(X = 0) + p(X = I) + p(X = 2) + p(X
1001
2184
1365
280
4830
=--+--+--+--=-4845
4845
4845
4845
4845
N~u 4
:F(x)=I
V~y ham phan ph6i xac 5uc1t cua X Iii:
L,_I x, P, =
= 3)
Theocongthu-c,lac6:
Vai:
Var(X)= E(X~)-[E(X)y
4
E(x2)= LX,2p, =12xO,4+22xO,3+3"xO,2+42xO,I=.
V~y Var(X)
Y = 5 _ (2)2 = I
= E(X2) -[E(X)
e) Hitm phan ph6i xac su~t eua X c6 d!lng:
F(x) = p(X < x)
N~u X $ I
N~u l
°
:F (x) =
:F(x)=p(X=I)=O,4
N~u 2
N~u3 < X $ 4 :F(x)
=0,4"'0,3=0,1
= p(X = 1)+ p(X = 2)+ p(X = 3)
= 0.4 + 0.3 + 0. 2 = 0;9
N~u 4
:F(x) = I
Vliy ham phiin ph6i xlic suc11 clla X Iii:
J
<7
0: x:51
O:X :51
0,4: 1 <: x:5 2
F(x) = 0,7: 2 < x:5 3
0,8: 1< x:5 2
0,96: 2 < x:5 3
=
F(x)
0,9:3
0, 992 :3 < X :5 4
1:4
0,9984 : 4 < X :5 5
I: 5
l 'au 9: Me;)tngllai thg sAnco 5 vien dan. Ngllai do di san voi n~uyen
til ~: neu bAn trung muc lieu thi v~ ngay, khong di san niia. Biet xac
SUi'I trung dich cua mdi vien d\ln bAn ra hi 0,8. Goi X Iii d\li IllQ11g
ng~ I nhien chi 55 vie." d~n ng~ai §y sfr dung trong cuoc sin.
a) Li p bang phan phai xac suat cua X;
b) Til h ky vcng, phuong sai ella X;
c) Tin ham phin ph6i X8C suAt ella X.
Giai
a) Goi , . Iii d\li 11IQ11g
ngdu nhien chi s6 vien dan nguoi ~y SO trong
cuoc san X co th~ nh~n cac gia tr]: 1,2,3,4,5.
P(X"'
) =0,8
0,8 = 0,16
·(0,2)2xO,8=0,032
rO,2)3 X 0, 8 = 0,0064
P(X == 2 == 0,2 x
P(X=3)
P(X = 4):
P(X~ 5) == (0,
2)4 X 0,8 + (0, 2)5= 0,0032 + 0, 00 128 = 0,0016
V~y bang pI in ph6i xac su§t clla X hi:
32['
~
2
I
3
,1,16
- ..
b) Tim k5' v<;>ng:I:.(X) == ?
0,032 _/
4
5
0,0064
0,0016
PHAN sAl TAp TH6NG KE
I. U'OC LU'QNG KHOANG:
'.
Cau I: Thee doi 100 SV cua tnrcng A de XD 56 gia II! hoc a nha
IhAyeo 95 SV COIl! hoc voi 56 gia TB 4,01 gia vai 5= I,54 gia.
a) Ucc 111Q11g
s6 gia II! h<;>cClla SV Imang A vCride;)tin c~y 97%.
b) lfac 111Q11g
Ii I~ SV Irllang A khang tl! h<;>cvai de;)tin c~y 90%.
Giai
a) G9i m Iii s6 gia II! hge trung binh ella sinh vien IruOng A.
Ta dn l1ae 111qngm vai d9 tin c~y 1- a = 97%.
Thee d~ beii la co:
=
Var(X)
{ n = 100 >
(J2
= 4,01
x
S == I 54 ~>S'~ _17_
.
I x, P,
1 X 0,8
=
+2
X 0,16
+
3 X 0,032
S
=>
X S'
...n-I
1/
Thee CT ta co: E(X)"
Chua bi~t.
30
=
100
,
99 x (1.54)- ~ 2,39
= 1,55
= ~2,39
/=1
+
4 X 0,0064 + 5 X O,lIO 16 = 0,8 + 0,32 + 0,0256 + 0,008
Tim phuong sai: Var(~:) =?
Thee cong thuc, ta co:Var(X)
==
I, 25
=
E(X2)-[E(X)(
Vai:
Mat khac
.
<=>
I- a
2
<=>
Ur=U
4
IiP, =1 xO,8+i>{0, 16+32xO,032+4 xO,(D)4+S2
E(X2)=
2
2
xO,OOI6
la
eo: 1-
= I-
= 0,97
a
=U0985
=2,17
= 2.17x
~
vlOO
1--
2
:::,
,
: 0,8 + 0,64 + 0,288 + 0, 1024 + 0,04 = 1,87
V~y Var(X)= E(X2) - [E(X)f = 1,87 - (I ,25i = 0,3075
c) Heim phin ph6i xac sudl ella X eo d\lng:F (x) = p(X < x)
N~ux.sl
F(x)=O
Vh kheang lin e~y ctia uac luqng la:
N~ul
(mpm2)
F(x)=p(X=I)=0,8
2 < x.s 3 :F(x) = p(X
N~u 3 < x .s 4 : F(X) == P(X:
N~u
= 1)+
p(X
=->&
4)
~
= (x-c;x+c)
0.336; 4.0 I + OJ36)
= (3,674 ;4,346)
b) G<;>iP leiIy I~sinh vien Irl10ng A khang t\[ h9C:
Ta dn l10e Jl1Q11gP E (~ ;/2) vai dk I - a = 90%.
Thee d~ bai la co:
m
5
Ta eo: I- a = 0.9
==> a
n=
100· f= - = - = 0 05
'.
n
100 '
'
J3
= OJ36
= (4,0 I -
I) + P(X= 2)+ P(X: 3)
0,8 + 0, 16 + 0,032 : 0,992
N~lI 4 < x.s 5 : F(X) = P(X= I) + P(X= 2)+ P(X= 3) +P (X=
= 0,8 + 0,J6 + 0.032 + 0,0064 == 0,9984
N~u 5 < x
F (x) = I
V?y hilm phan ph6i xac sudl ctia X la:
= UrX
Vn
= 2)= 0,8 + 0,J6 == 0,96
==
= 0,03 => a2 = 0.0 I 5
= 0.985
0.015
a
a
=>
= 0,1 <=>
a = 0.05
2
-
J~
<=> 1- a
2
<=> Ur
= 1_ 0,05 = a 975
=U
I_~
2
'
= U0,95 = 1, 645
<=>1- a
2
t(l
n- f)
= 1.645x
0.0;~.95
= 0.0359
=>e = Ur x r
-m
= 1.96
2
0,05
1,96x rn;
=
v81
= 0,0 II
V~y khoang tin C?y cua uoc hrong la:
= (x-c;x
(ml,m2)
V~y khoang tin eb ella uoc hrong Iii:
U;'/2)
0,025 = 0,975 <" Ur = C,_~. = U;..97<
s
1
., e = Uy x
= 1-
+c)
=(19,996-0,011 ;19,996+0,011)
= (19,985; 20,007)
= (j -&;/ +c)
= (0,05 - 0,0359; 0,05 + 0,0359)
= (0,0141 ;0,0859)
C'au 2: D0 d'uoog klmh d cua
' 100 ch"~
iv d 0 I XN
I tiel may
19,8 19,8 19,9 19,9 20,0 20,0
d
OSOSOS(m
19,8
19,9
19,9
20,0 20,0 20,1
m)
5
0
5
0
5
0
S6
chi
tiet
3
5
16
28
23
14
SX co. so~ I"I~U:
20,1 20,1
OS20,1 20,2
5
0
7
4
, ,
Quy d!nh nhiing chi Ilet may c6 dLIang kinh Iii 19,9mm den 20, Imm
la d\ll chu~n,
a) U'oe hrqng ti I~chi ti~t may d\ll ehu~n voi dQtin c~y 99%,
b) lJ&c 111qngdWJng kinh trung binh clla chi tiet may dill ehu~n vai
dQlin c~y 95%.
,
Giai
.
a) G?i P ili I)' I~ chi tiet may d\lt tieu chuan:
Ta can l1ue lugng P E U;;/2) vai dk I - a
,
i~
\~~~~.J-~
SI~
Thea de bili ta co: n = I00' f = - = = 0 81
,
n 100
'
a
Ta co: I-a = 0,99==>a = 0,01 <=>- = 0,005
<"">1-2 = 1-0,005 = 0,995<=> Ur = UI_C!.= UO.995
.
Ta dn LIoelugng mE (m1 :m2) vai dQlin c~y
Thea d~ blii ta co:
= 2,576
2
Chua bi~t. ~ = 46.443
S'= 2.521
=2576x
n'
V~y khaang tin c~y ella uoc lugng ili:
U; ,f2) = (j- c;f
O,8Ix(I-O,81 =0101
100
'
= (0,709;0,911)
=
var(X) = a2
{ n = 81> 30
Chua bi~t.
x = 19,996
s' = 0,05
M~t khac ta co:
S
=>&=Urx r
a = 0,025
I - a = 0,95 => a = 0,05 =>2"
= 1,96x 2.521
(;-;-; =0,461
vl15
V~y khoang tin c~y ella LIaC lugng la:
-
(m"m1)
b) GQi m hi duang kinh trung binh ella chi tiet may dill ehu~n,
Tacdnuoelugng mE (ml;m2) yaidk I-a
95%.
Thea d~blii la co:
"
2
vn
+c) = (0,81- 0,101; O,SI+0,101)
= 0.025
<=>1- a = 1- 0,025 = 0,975
2
<=> Ur = U a = U097< = 1,96
1--
=>&=UrxJ/(I-f)
1- a = 0,95.
1- a = 0,95 => a = 0,05 =>-a
2
M~t khae ta co:
2
!
-.!
Var(X) = a2
{ n=1l5>30
~
m
a
Cau 3: Nang 5U~1Ilia trong I vung la bi~n ng~u nhien. G~t ngcil
nhien 115 ha ella vung nay, nguci, , ta t hu duoc bang so~I"ieu:
Nang sual
40 42 46 50
48 44
(ta/ha)
42
44
46
48
50
52
DT (ha)
I]
7
25 J 35
j
5
. vm. , dQnn cay]0 95%,
a) Uoc hrong nang suat TB cua vung tren
b) Nhiing thira ruong trong vung tren co nang su~t khong qua 44
ta/ha hi nhiing thua co nang suat th~p (gia sir co phan phci ehual~
U'~e hrong nang 5u~t lua trung binh ella nhung thLIarU9ng co na~
suat thap vai dQ tin c~y 99%.
Giai
a) G9i m la nang 5u~t Ilia trung binh etia yung.
-
= (x - e;X +&)=
(46,443 - 0,461; 46,443 + 0,461)
= (45,982; 46,904)
b), GQi,m la nang 5U~1Ilia trung binh ella nhiing thlra rUQngco nart
suat thap
Ta dn u&clugng
Thea d~ biti ta co:
Var(X) = (72
{ n = 20 < 30
M~t khac ta co:
mE
(m, :m2) v&idQ tin e~y
X
X co phein ph6i ehu~n
1- a
= 0.99.
= 42,3
S'= 0,979
a = 0,005
I - a = 0,99 ~> a = 0,00 I =>2"
<=> J
<=>
a
- -
= 1- 0' 005 = 0,995
2
a) Uoc hrong ti I~ h(lp thit d~t tieu chuan trong kho voi d(l tin c$l
94%,
b) Wi sai s6 cho phep khi ucrc lvong ti I~ h(lp thjt khong d~1li~
chuan trong kho la 1% va dQ tin c~y la 99% thi dn kiem Ira 16i Ihie
bao nhieu h(lp thjt?
Ur = U1--a = Uo'995 = 2,861
2
= Uv X
=> C
Gi
a) GQi P Iiity I~ h¢p thit di!t tieu chudn trong kho:
= 2,861X 0,;;: = 0,6263
~
vn
,,20
Ta dn llOChrong
Thea d~ bai ta co:
V~y khoang tin c~y cua ucrc lugng la:
= (x-e;x
(ml ,m2)
+e)= (42,3 -0,6263; 42,3 + 0,6263)
= (41,6737; 42,9263)
m
= 5000,
(= -
' .
n
'.
Mat khac ta co: 1-
'
a
<=> 1- -2 = J - 0,
<=> U y = U
{1
1--
2
= UO.98l
=
m II
/=-;=100=0,11
Ta cci:
°
a
= 1-0.005 = 0.995
2
<=> Uy = U a = [/0,995 = 2.576
=00043
'
1--
V~y khaang tin c~y cua uoc lugng la:
2
Theo cang thuc, ta CO: =>
(J;;fJ=(f -e;/ +e)
= (0,02 - 0,0043 ; 0,02 + 0,0043)
= (0,0157; 0,0243)
b) Ta co:
I- a = 0,99; e < 0,0 I
Ta cci: 1-
a = 0,99
<=> 1- a
2
=J-
<=> Uy = U
a
1-2
=>
£
°
a = 0, 1 <=> a = 0,005
2
= UyxJfx(l-
n
2,576X ~O,02X(I- 0,02)
---------
0,01
f) =
e
= 3606 ' 4
I~\= 2,576 O,IIX~-0,11)
0,11)
= 80.6
V~y nguai ta dn ki~m Ira 6496 + I = 6497 hQpthit.
= Vo'995 = 2,576
=>Fn
""
=> n = (80,6)2 = 6496
0.. 005 = 0 , 995
e=uyxJfX(I-f)
=urx~
\:,\
"nI = 2,576X ~O.lIX(I0,01
Thea cang thuc, ta co:
=>
a = 0.005
2 .
1- a = 0,99 => a = O. f <=> -
<=> 1--
5000
= 0.0589
n'
100
V~y khoang tin c~y cua llOClugng hi:
(J;;!;) = (j -e;f +e)
= (0,89 - 0,0589 ; 0,89 .;.0,0589)
= (0,83' I ; 0,9489)
b) Thea d~ blii la co: I - a 0,99; G = 1 % , Tim n ::; '1
= 2,17
=2,I7x
=UO.97 =1.881=>
e = Uyx ).1(1- f) = 1881 x 0,89x (1-0,89)
2
=>e=uyx~
a
1--
,
2
'
0,02x(I-O,02)
.
a
a = 0,94 => a = 0.06
. <=> -2 = 0.03
'
<=>Uy=U
= 0.985
.
5
100
a
°
J
a = 0.94.
J-
<=> I - - = I - 0 03 = 0 97
a = 0,97 => a = 0,03 <=> -a2 = 0, 15
°
n
Mat
. khac ta co: 1-
= - 100 = 0 02
5000
voi dQtin c~y
:{),l(
I~
m 89,r
n = 100· ( = - = - = 0.89
Cliu 4: NglIai ta x~p 100 trai 6i vao I thung, co nit nhi~u thung
nhu th~. Kiem tra ng~u nhien 50 thung thAyco 100 tnii 6i khang d~t
tieu chu~n,
a) U'crclugng ti I~trai 6i khang d~t tieu chu~n vcrid¢ tin c~y 97%.
b) MU6n llcrClugng ti I~trai 6i khang d~t tieu chu~n vcri dQ chinh
xac nh6 hO'l10,1% va dQ tin c~y 99% thi dn phili ki~m tra t6i thi~u
baa nhieu thung?
Giai
a) GQi P Iii la ty I~ trai 6i khang d~t tieu chu~n:
Ta dn ucrchrc;mg P E U;; J;) vcrid(l tin c~y 1- a = 0,97.
Thea d~ bili ta co:
n
P E (~;.I;)
==
> n=
60
V~y nguoi ta dn ki~m tra 130061 + 1 = 130062 trai 6i.
Cau 5: Nguai ta ki~m tra ng~u nhien 100 h(lp thit trong I kho thi
th~y co II h(lp khang d~t tieu chudn,
Cliu 7: Giam d6c ngan hang A mUdn LIOC11Igngs6 ti~n gLIi c~
m(it khach hang b4ng cach chQn ngdu nhien 30 khach thi th~y: 56
ti~n gui trung binh la 4750$ va d(l I~ch tieu chu~n la 200$,
a) Voi d(l tin c~y 95%. uoc lugng s6 ti~n gui cua mQt khach hilr~
t~i ngan hang A?
,
b) Neu mU6n co d¢ chfnh xac cua uoc lugng trung binh lei 100$tf,z
dam bao d¢ tin c~y Iii bao nhieu?
Giai
a) GQi: m la s6 ti~n gU'itrung binh cua khach hang t~i ngan hang A.
Ta dn llOCIllgng mE (ml; m2) voi d(li tin c~y 1- a = 95%
Thea d~ bai, ta co:
n = 30
_
'x=4750,
{ Var(X) = a2 = (200)2 .
,
l-a=O
,
95
l-a=O,95
Tac6:
U, = Va::
V097S
1-,
=>
I-~=O 975
2
'
= 1,96
<=>
Ur
2
a
200
vn
..;30
=> C = U, X-r = 1,96x
r:;;;
= 71,569
e
= Ur X ,jfOn- f) = 1,96x 0,3(1- 0,3) = 0.0898
100
(J;; /2) = (I
71,569)
=
- &; /
+ s) (0,3 - 0,0898; 0,3 + 0,0898)
(0,2102; 0,3898)
=
= (4678,431;4821,569)
b) C6 e = 100, Tim 1- a = ?
1(£
PH..\N8..\1 TAp TU6NG
Theo cong thuc, ta c6:
a
&xJ;, 100xJjQ
x r => V = =
= 2,7386
'..;n
r
a
200
=> e = V
r = 0,997 <=>
2'
V~y khoang tin c~y cua uae luong Iii:
= (x-&;x+&)
= (4750-71,569;4750+
=>
2
= U1--a = UO.975 = 1,96
2
V~y: khoang tin c~y cua l1ac 111QT1g
Iii:
(lnl; m2)
= 0,9S=:>a=0,05<::>-a = 0,025<::>1--a = 0.975
Tac6: I-a
1. V'OC LU'QNG )(HOANG:
'.
Ciu I: Theo doi 100 sinh vien ella tnrong A de xac dinh 56 gia"
hoc a nhil thi th~y c6 95 sinh vien c6 tv hoc voi s6 gib Irung bin
4,0 I gia voi s= 1,54 gib,
a) U'6c hrong s6 gib nr hoc cua sinh vien tnrong A voi dO tin c~
97%,
I- a2 = 0.997
b) U6c luong ti I~ sinh vien tnrong A khong tv hoc voi dO tin c$
90%.
<=>
a = 0,003 <=> a = 0,006 <=> 1- a = 0,994
Giai
a) G9i m Iiis6 gia t\1"hoc trung binh ella sinh vien truong A
Ta dn trae ILIgngOJ v6i d¢ tin c~y I - a = 97%.
Thea d~ bili ta co:
2
= 99,4%
V~y I-a
Ciiu 8: D~ khao sat trQng lugng X cua m9t IO\li v~t nuoi trong
,
,'
z
nof!£.tr~lf,~UO'I
ta_guan sat." mgt mau
va, co• k~
et qua, sau:
X (k.B)
36
42
48
54
60
66
72
So
15
12
25
18
10
10
10
con
.. d9
"'
"
a) UO'cIlf(;mgtrQng IUQTIgtrung bmh cua 10\11V\l1nUOJIren
" VaJ
tin c~y 96%,
b) Nhfrng con v~t e6 tn;>nglugng tiI 60kg tra len dtrge g9i hi nhCing
con "d\ll tieu ehu~n'". Hay uac lugng ti I~ con din lieu chu~n vai d9
tin e~y 95%.
= a2
{
Var(X)
n = 100>
x
= 4,01
Chua bi~t.
30
"
Giai
al Ggi OJ Iii trgng IUQ'Ilgtrung binh clia 10(liv~t nuoi
Ta dn uae Itrgng
Thee d~ bili ta c6:
m
E (mJ; m2)
Var(X) = a2
{ n = 100> 30
Chua bi~t. x
1- a
= 0,96.
= 51.96
S'= 11,061
2
= U1--a = Vo'98 = 2,054
11-1
100
99
x (1.54)'
1
'" 2.39
= 1,55
j - a = 0,97 => a = 0.03
=> S = J2.39
M~t khac la co:
a =0,02<::>1--a =0,98
= 0.96=> a =0,04<::>-
Tac6: I-a
Vr
v6i d9 tin c~y
S = 1,54 ~>S' = _11_ x S'-
a
=>-
2
= 0,015
= I-
<=>
I- a2
<=>
Ur = V
a
1--·
0.015 = 0.985
.
= UO.985
= 2,17
2
2
= Vr
=>&
X
S'
1.55
.In
= 2,17 JWO = 0,336
X
2
C
=Vrx
~
= 2,054x
I ~I
-.;100
-.;n
V~y khoang tin c~y elJa l1aC lugng Iii:
= 2,272
V~y khoang tin e~y cua uoe Itrgng Iii:
(mJ;m2) = (x -
&;X
= (51,96·2,272;
+ c)
51,96 + 2,272)
= (49,688;
54,232)
bl Ggi P Iiity I~v~t nuoi d(lt chu~n
Ta dn uac lugng
Thee d~ bili ta c6:
n
=
300
{ l=~=~=03
.
P
n
E
(J;;
12)vai <19tin c~y I - a = 0,95.
-
c; X
b) Ggi P Iiity I~ sinh vien truimg A kh6ng tv hge:
Ta
cdn u6c
111gng
P E (J;;f2) vai dQ tin
I -a =90%
Theo d~ bciita c6:
Chua bi~t.
100
-
= (x -
+ c)
= (4,0 I - 0,336; 4.0 I + 0,336)
= (3,674; 4,346)
(ml, m2 )
'
n = 100; f
Ta co:
= -m = - 5 = 0.05
n
100
I - a = 0,9 => a = 0.1<=> -a = 0,05
2
J/}
c;J
°
a
<=> Uy = U
= U09S.
I-~
e
°
<=>J- a
2
J
I)
= Uy x f(l-
= 1,645x 0,05 x 0,95 = 0,0359
100
n
0,0359; 0,05 + 0,0359)
= (0,0 14 I;0,0859)
P E (~ ;j2) vai do tin e~y
1- a = 0,99.
Thea d~ bili ta co:
m
= (19,996 - 0,0 II; 19,996 + 0.0 II)
= (19,985 ;20,007)
Cau 3: NAng su~t lua trong I vung Iii bien ng~u ~hien. G?I ngS
nhien 115 ha ella vung nay, nguoi ta thu duQ'c bang 56 I'I~U:
NAng
suat 40 42 44 46 48 50 (ta/ha)
42
44
46
48
50
52
Di~n tich (ha)
7
13
25
35
30
5
.
. ..
. 95%
a) Uoc hrong nang suat trung binh cua vung tren vO'l.d9.tin cay
b) Nhiing thua ruong trong vu~g Ir,en eo nang Sual khon,g qua. 4.
ta/ha Iii nhiing thira co nang suat thap (gia su eo phan phoi chuan)
lJae 11IQ'Tlgnang SU~I lua trung binh clla nhiing IhlIa ru(lng eo na[J
SU~Ith~p vai d(l lin e~y 99%.
.
Giai
a) G9i m Iii dltong kinh trung binh.
'.
Ia can
Var(X) = a1
{ n = 115 > 30
=.!..!_ = a 81
100
'
= UO.995
f) = 2 576x
u; '/2 ) = (f
0,81x(l- 0,81 = 0,101
100
.
=>c
= U097",
= 1,96
.-
- c;j + c)
b) G9i m Iii dl10ng kinh trung binh.
Ta dn l1aC 111Q'1lgmE (ml; m2)
= 0,95.
= 0-2
{ n=81>30
S
.In
(mpm2)
2,661
= J,96x -= 0.486
J1I5
= (x-c;x+c)
= (46,383 - 0,486; 46,383 + 0,486)
= (45,897; 46,869)
b) G9i m la ducmg kinh lrung binh.
Ta can lIae IUQ'ng m E (nil; m2)
vai do tin c~y
1-
Theo d~ bai ta co:
= Uy x -
V~y khming tin caY clia lrae itr9'l1gIii:
= (0,8 1- 0,10I; 0,81 + 0,101)
= (0,709;0,91 I)
X
= U I-!?
2
YaY khoang tin c~y cLialfcYeIlf9'l1gIii:
Var(X)
Chua biet.
2
<=> Uy
n'
I-a
e~'
<=>1- a = J- 0.025 = 0.975
= 2,576
=>
Jf(l-
"oi dc;i tin
a
=>-=0025
2
'
2
°
c = Uyx
mE(ml;m,)
J- a = 0,95 => a = 0.05
M?I khac la co:
a
<=>I - -2 = 1- a ' 005 = ,995
a
luvng
UlJc
]-a=0,95.
a
1- a = 0,99=>a = 0,01 <=>- = 0,005
1-2
,,81
Theo d~bili la co:
n
<=> Uy = U
"n
= (x-c;x+c)
(mpm2)
.
Ta co:
0,05
r;;;=0,011
r=I,96x
V~y khoang lin c~y cua uerc 11IQ'IlgIii:
Cau 2: Do duong kinh d cua 100 chi ti~t may do I xi nghiep san xu~t
co s61i~u'
19,8 19,8 19,9 19,9 20,0 20,0 20,1 20,1
d
s- o- soOS05(m
19,8 19,9 19,9 20,0 20,0 20,1 20,1 20,2
m)
0
5
5
0
5
0
5
0
S6
chi
3
5
16
28
23
14
7
4
tiet
- chi. tret" may co duong klnh tu 19,9mm den 20,1 mm
Quy dinh nhtmg
Iii dat chuan.
a) U'ac ILr9'l1gti I~ chi ti~t may dill chuan voi d(l ~in c~y 99%.
.
b) Vac ILl9'l1gduc'mg kinh trung binh cLla chi tiet may dilt chuan VcYl
do tin cay 95%.
Giai
a) G9i P Iii ty I~ chi ti~t may d~t tieu chu~n:
=
S'
=>c=Uyx
= (0,05 -
n = I00; j'
=U097S. =1,96
I-~
2
U;
lfcYC IU9'l1g
= 1- 0.025
. = 0.975
<=> Uy=U
V~y khoang tin c~y cua Lrac hrong la:
,/2) = (j -c;f + c)
Ta dn
= 0,95 => a = 0,05
a
=>-2 = ' 025
= 1,645
2
=>
1- a
M?t khac ta co:
<=>1- -2 = 1- 005
' = ,95
vai d(l tin e~,
a = 0,99.
Theo d~ bili ta eo:
Chua bi~t.
= 19,996
Var(X) = (72
{ n = 20 < 30
M~t khac ta co:
S' = 0,05
21.
Chua bi~t. X - pp Chu~n
1- a
= 0,99 => a = 0,001
a
=>
= 0005
'
=>-
2
J
~ e = U r x f x (1- f) => Fn = U r x
a
<=> I -
-2 = I - 0' 005 = 0,995
=
= 2,861
<=>
Ur = U1--a = U0995
.
=>&
S·
0,979
= Uyx r = 2,861x ~ = 0,626
-in
...;20
2
11)
0,001
= 360,64
=
C~u 5: Ngucri ta kiem tra ng~u nhien 100 hQp Ihil trong I kho
thay co 11 hc)pkh6ng dill lieu chuan,
a) lf6c IUQ1lgIi I~ hc)pIhil d\lt tieu chuan trong kho voi dc)lin~,
= (x-&;x+&)
M~
th
~~
b) Vai sai 56 cho phep khi uoe IUQ1lgIi I~ hc)p thil kh6ng d\lt liCA
chu~n trong kho la I% va dc)lin c~y la 99% thi cdn kiern Ira 16i Ihie,
Oiu 4: NglIai ta x~p 100 trai 6i van 1 thung, co .it nhi~u thung nhu
th~. Kiern Ira ng~u nhien 50 thung th~y co 100 trai 6i kh6ng dilt lieu
chuan.
a,' lfac Illl!l1gti I~ tnii 6i kh6ng dilt tieu ehu~n vai dc)tin c~y 97%.
b) Mu6n llac IUQ1lgIi I~trai 6i kh6ng d\lt lieu ehu~n vai dc) ehinh xac
nhtl hon 0, I% va dc)tin c~y 99% Ihi dn phili ki~m tra t6i thi~u bao
nhi~u thung?
Giai
a) GQi P Iii Iii ty I~ trai 6i kh6ng d\ll tieu ehuftn:
Ta dn lrae 1U'c,mg PE(J..;f2)v6i
dc) lin c~y
I-a = 0.97.
Theo d~ bili ta co:
baa nhieu hc)pthit?
Gilii
a) GQi Phi uoc hrong Iy I~ hOp thit dat lieu chuan trong kho
Ta dn U'ae 1U'c,mg P E U;; ,f;) veri dO tin c&
'(
1-a
= 0,94.
Theo d~ bili ta co:
n = 100;
m
II
= 0 II
100 '
M~I khac ta co: I - a = 0,94
f =- =-
.
n
7>
a = O. 06 ~
a =0 03
2
m
100
= 5000; f = -;; = 5000 = 0,02
1- a = 0,97
khac ta co:
M(L
JO, 11x (1-0,
2,576x
n=(360,64)2
130061,2
V~y ngllcri ta dn ki~m Ira 130061 + I = 130062 trai 6i,
= (42,3 - 0,626 ;42,3 + 0,626)
= (41,674;42,926)
Ii
C
=>
V~y khoang tin e~y ella uae luong hi:
(m"m2)
Jf x (I - f)
n
'
1- a2 = I - 0 ' 03 = 0.97
,
<=> Uy=U a =[/09' =1,881
J-'
<=>
=>
a = 0,03
<=>
2
a = 0,015
=>
2
<=>
a
I---2
<=>
u:· = U = U0,985 = 2' 17
c=UyxJfX(nl-f)
= I- 0' 015 = 0, 985
c = Uy X f X (1-- f) = 2 17X 0,02 X (1- 0, 02) = 0. 0043
'
5000
.
= (/ - c;j
+c)
= (0,02 - 0,0043 ; 0,02 + 0,0043)
= (0,0157; i ,,0243)
b) Ta co:
= 0,99;&
I-a
Ta co:
< \001
I - a = 0,99
<=>
a
1- -2
<=>
Uy=U
>
= I- 0 ' 005
l-!:
°
°
: , 995
=U 0,995 ,,,2,576
2
Thea cong thfrc, ta eo:
°
a = , 01 <=> a2' = 005
+E)
= (0, II - 0,0589 ; 0,11 + 0,0589)
=(0,0511 ;0,1689)
b) Ta co:
1- a
= 0,99; c = 1%
Ta co:
1- a = 0, 99
V~y khoang ti n qiy clla uac IUl!l1ghi:
U;;J;)
= (j -c;f
(.~;f2)
=>
n
=0.058~
V~y khoang tin c~y cua lraC 1U'c,mg
Iii:
2
J
O,lfx(l-O.II)
100
a
1--
I
=1,88Ix
~>
a = 0,01
<~>
a = 0,005
2
= 1- 0' 005 = 0 , 995
<=>
1- a2
<=>
U r = U I-!:
= U 0,995 = 2 ' 576
2
Theo cong thLi'c,ta co:
=>
J
c = U y x f x (1 - f) => Fn
= U y x ~f x (I -
n
C
2,576 x ~O, 02 x (1- 0, 02)
0,01
=> n = (80,6)2 = 6496
=
V~y ngU'ai la dn ki~m tra 6496
f)
+
= 806
'
1 ~ 6497 h9P thit.
.
1- a = 0,95
Ta co:
<=>
a
1- - = 0.975
2
Cau 7: Giam, ddc ngan hang A mUdn llcre hrong sd ti~n giri cua m<)t
khach hang bang each chon ng~u nhien 30 khach thi th4y: Sd ti~n gui
trung blnh la 4750$ va d<)l~ch tieu chuin la 200$.
a) Va; d9 tin c~y 95%, lIac 11Iqngs6 tien gfri cua m¢t khach hang t~i
ngan hang A?
b) Neu mu6n eo dQ chinh xac cua llOCIllc;mgtrung binh la 100$ thi
dam baa dQ tin c~y la baa nhieu?
Gilli
a) G9i: m la sd ti~n gfri lrung binh eua khach hang tiii ngan
himg A.
Ta dn lloe luc;mg mE (ml; m2) vai d<)i tin e~y
II. KIEM DINH CIA THIET TH6NC KE:
Ca~1: Ti l~ph~pham do. cong ty A SX la 5%. Nham giam u I¢ phc
pham, cong ty A da eli; tien ky thu~t. Sau cai I;~n. nguei ta kiem-ire
ngiu nhien 400 san pham th~t co 18 phe pham. Voi mire y nghia 5%
hay eho ket lu~n v~ hi~u qua cua vi~e cai tien ky thu~t cua cong ty A.
Giai
L~p gia thuyet thdng ke:
Ho: P = Po = 0,05
HI: P < Po = 0,05
Theo d~ bai, ta eo:
I-a =95%
n = 400; / = m = ~
Theo d~ bili, ta co:
n =30
{ Var( X) = (]'2 = (200)
Ta co:
Uy
=>
= 0,95
I-a
<=>
= V1--a = V
O•97S
2;x=4750;I-a=0,95
• Ti!mi~n
2
= 1,96
(]'
i=I,96x
"n
(/ - Po) x Fn = (0,045 -0,05) x J400 = -0.459
JO.05 xO.95
Vo (i!: Wa => Ch~p nh~n Ho
Uo=
200
~=71,569
,,30
= (-00; -I, 645)
Wa
* Tinh gia tri quan sal:
2
=>&=Vyx
= (-oo~-UI~~~
a = 0,05 => 1- a = 0,95 => UI_a = Uo 9l = I. 645
Ta eo:
=> MBB:
a
1-- = 0,975
W(/
bac bo:
= 0,045 ; a = 5%
=>
J Po x qo
K~t lu~n: Phuong phap eai ti~n rncri khang hi~u qua_
V~y: khaimg tin c~y eua LIoeiLrqngIii:
(ml; m2)
= (x-&';x+&')
= (4750-71,569;4750+ 71,569)
= (4678,431; 4821,569)
b) Co &' = 100. Tim I-a =?
Thea eong thfrc, ta co:
£=u
x_!!_=>V
y
Fn
=>
?
=&,x';;;=IOOxJ30=27386
(]'
200
'
r=0,997
<=>
I - a = 0 997
2
Ciiu 2: M<)t cang ty di~n tho~i noi rAng se lap d~t di¢n thO\ii ehc
khach hang trong thanh ph6 eh~m nh~t 1<\30 ngay k~ tif khi eli yeL
duo Ki~m tra ng~u nhien 30 khaeh hang th~y thai gian Irung biJlf
cha l:1p di~n tho\li Iii 34,5 ngay vai ~Q I~ch mau hi~u c~lOh Iii 3,3
ngay. Vai mile y nghia 3%, co the chap nh~n lai tuyen bci eua ecine
ty duqc khong?
'
Giai
L~p gia thuy~t th6ng ke:
Ho: m = mo = 30
HI: m > mo = 30
Theo d~ bili, ta co:
n=30~30
'
a
0,003
<=> a =---0,006
----r=------- -----------------_-------
<=> -
I - a = 0,994
V~y I-a = 99,4%
<=>
{ Var(X)
_,,._I;
2
= (]' : Chuabiel
x = 34,5; .~ = 3. 3 : a = 0.03
* 'Finrnribr bac-bo~- --W;-=ib'I--;~+ce) --- - -- Ta co: a
= 0,03 => 1- a = 0,97
=>
UI_II = U097 = I. 881
MBB: W(/ = (1,881; +(0)
* Trnh gia tri quan sat:
=>
-
U = (x- mo)
sf
o
x';;; = (3,45 -30) x50
3.3
= 7 469
'
=> Vo e Wa => Bac b6 Ho, ch~p nh~n HI.
K~t lu~n: Uri tuyen b6 cua cong ty khong dung 51! th~l.
I
Giai
G9i m Iii tr9ng 11Igngtrung binh cua lo~i v~t nuoi
Ta dn lIoe luc;mgm (mi. m2) va dQ tin c~y 1-& = 0,96
Theo d~ bili, ta co:
n = 30
{ Var(X) = 0-2 = (200)2
-
; x=4750;
l-a=0,95
C8u 3: Nguai ta ki~m tra ngau nhien 100 chai nuoe ng9t IOlili2 Ih
do nhit may A san xu~t thi th~y luc;mg nuoc ngQt trung binh tront
ehai hi 1,99 lit vii 5=0,05 lit. Voi mlic ~ nghia I%. hay cho bi6t luqn~
mrcrc ngQt trong ehai IO\linay co bi thieu khang"
'&
Giai
L~p gia thuy~t thbng ke:
Ho:
m = mo
HI: In
=2
< Ino = 2
Theo d~ bai, ta co:
n = 100 > 30
K~I lu~n: Phuong phap cai ti~n moi khdng hi~u qua.
_
{ Var(X) = (7 2 : Chuabiet ; X = 1,99 ; S = 0,05
n
'2
Cliu 6: Truce b~u cu nguai ta tham do 10000 cu lri lhi thav cC
3?00 nguai noi ra~g se be)phj~u cho ong A. MQt tu~n sau (v~n ~hlf(l
b:u cu), ng~aj .ta to chit,e. I cu¢c lh~~ do k~ac va thAy co 6890 Iro'!j
so ]5000 cu rn dllQ'ChOI se bo phleu cho ong A, Va; rmrc y nQhlu..
5%, Ii I~ cli tri se bo phi~u cho ong A co thay d6i khong?
~
100
=>5 =-xS2=-x(005)2=00025
n-I
99'
,
s = .JO,0025 = 0,0502
=>
wa = (-ClJ'-V
,
I-a )
* Tim mi~n bac bo:
Ta co:
a = 0,0 I => 1- a = 0,99
Giai
L~p gici thuy~t th6ng ke:
= VO•99 = 2,326
=> VI_a
=> MBS: Wa = (-CIJ ;-2,326 )
* Tinh gia tri quan sat:
HI :
-
V = (X-~10)xfn=
(l,99-2)xMo=_199
0,0502
s
o
Theo d~ bai, ta co:
'
{n
Vo e Wa => ChAp nh~n Ho.
==>
3900
= 0,39
10000
P ":t Po = 0,39
= Po = --
P
Ho:
6890
r. = -mn = --15000
= ° 459;
'
= 15000;
* Tim mi~n bac be:
K~t lu~n: LlI(;mg mrac ngQt trong chai 2 lit khong bi thi~u,
= (-oo;-U i--a)
Wa
a
= 5%
a ;+(0)
U (U
1 __
2
,a
Ta co: a = 0,05 => -
~au 4: M(>t con~ ty tuye~ bd rang 75% khach hang IIU ,thich san
pham cua minh. Dieu tTa ngau nhien 400 khach himg thi thay co 260
nguai LTathich san ph~m cong ty, Vai muc y nghia 3%, hay cho y
ki~n v~ 10; tuyen b6 tren?
Gi3i
L~p gia lhuy~t th6ng ke:
2
=> U
de bili, ta co:
Ta co:
"'>
Wa
=>
Cliu 5: Mc)l 16 hang dut;1c
xem la du tieu chu~n ~~ xuAt kh~u n~u ti
I~ ph~ ph~m khong Vltt;YI qU,aO,_3%. ~i~m; tra n~~u ~hie.n 12~O soan
pham cua 10 hang miy thi thay co 4 phe p~am. ym mue y nghla 5 Vo,
hay eho biet 10 hilng tren co du(,Ycphep xuat khau khong?
Giai
L~p gia thuy~t thdng ke:
Ho : P = Po = 0,05
HI : P < Po = 0,05
m 18
Theo de biti ta co: n = 400; 1=- = = 0,045 ; a
'
n 400
.
* Tim mi~n b,k bo:
= 5%
Wa = (-ClJ; -VI-a)
MBB: Wa = (-00; - I, 645)
Tinh gili trj quan sat:
=>
co Ihay doL
JO,05xO,95
=> Ch~p nhan
Ho
.
= mo = 13
HI : 111 > 1110 = 13
Ho: m
Thea d~ bili, ta co:
n
{
.
= i5 < 30
=
Var(X)
(72 :
Chuabiet
/
0
; x=3,62; S =0,405; a = 51'0
* Tim mi~n bac bo:
Wa = (lI(~'(~I);+oo)
Ta co: a = 0,05 => I - a = 0,95
VI.O = UOQ,) = 2,32 6
W a = (1,761
* Tinh gicitri quan sat:
=> MBB:
= (~-
I
uo= U - Po) x In = (0,045 -0,05) x ~400 = -0,459
Q
Uo ~ W" => Bac b6 He, ch~p nh~n HI
K~t lu~n: Ty I~ cu' tri bo phi~lI cha ong A
=>
a = 0,05 => I-a = 0,95 => UI_a = UO,9S = 1,645
Uo '"W
~
.
3.3
K~t lu~n: Phu'ang phap cai ti~n mai khong hi~u qua,
=>
J0,39 x 0,61
Cau 8' Trong luang trung binh khi Xll~t chu6ng a mot tr~i ch~n
nuoi tru6c Iii kl!ica~. Na~ nay. nguoi ta SlI d\lOg mOl 101lithlfe iLn
mai, can thti 15 co~ khi xuAIchu6ng dugc cae 56 li~lI sau: 3.25; 2.50j
4,00; 3,75; 3,80; 3.90; 4,02; 3,60; 3.80; 3,20: 3.82; 3.40: 3.75; 4.00;
3,50. Gia thi~t trQng Ilfgng gil Iii DLNN co phan ph6i ehuan. v6i
mlie y nghia 5%, hay eho k~t 11I~nv~ lac d\mg clla 101lithlJ'Can nily?
Gicii
L~p gia thuy~t Ih6ng ke:
U-Po)xf,;'=(0,045-0,05)xJ400=_0,459
()- JPoxqo
JO,05xO,95
=> {j o ~ W/1 => Ch§p nhan
Ho
'
J Po x qo
= 17.33
= (-00; -I, 645)
u-
*
1
~ Po x qo
= (-oo;-UI_a)
* Tinh gia trj quan sat:
Ta co:
= V0975 = 1,96
II - Pol x fn = 0.459 - 0.391 x JI5000
Uo =
a = 0,05 => l'iJh = 0,95 => VI_a = VO,9~= 1,645
MBS: WI>
= 0.975
=
n = 400'; f = m = ~ = 0,045 ; a = 5%
n 400
* Tim mi~n bac be:
2
=>MSB: Wa (-oc·;-i,96) u (1,96;+00)
...Tinh gia trj quan sat:
!fa ' P = Po = 0,05
HI : P < Po == 0,05
Thea
0
1-2
2
a
= 0,025 => 1- -
o
mo) x
S·
=-=>
(II-il
II-a
= 10.95 = 1. 761
i~
; +00 )
fn = (3,62 - 3,3) x Ji5 = 3.06
0,405
=> 10 E Wa => Bac bo Ho.ch~p nh~n HI.
K~I lu~n: Thuc an mai co tac dl,mg.