Thứ Ba, 21 tháng 1, 2014

Lý thuyết xác suất thống kê - Chương 4

Ch ’u ’ong 4

U
´

OC L

U
.

ONG THAM S
´
ˆ
O C

UA D
¯
A
.
I L

U
.

ONG
NG
˜
ˆ
AU NHI
ˆ
EN
Gi

a s


u ¯da
.
i l

u

o
.
ng ng
˜
ˆau nhiˆen X c´o tham s
´
ˆo θ ch

ua bi
´
ˆet.

U
´

oc l

u

o
.
ng tham s
´
ˆo θ l`a d

u
.
a
v`ao m
˜
ˆau ng
˜
ˆau nhiˆen W
x
= (X
1
, X
2
, . . . , X
n
) ta ¯d

ua ra th
´
ˆong kˆe
ˆ
θ =
ˆ
θ(X
1
, X
2
, . . . , X
n
)
¯d

ˆe

u
´

oc l

u

o
.
ng (d

u
.
¯do´an) θ.
C´o 2 ph

u

ong ph´ap

u
´

oc l

u

o
.
ng:
i)

U
´

oc l

u

o
.
ng ¯di

ˆem: ch

i ra θ = θ
0
n`ao ¯d´o ¯d

ˆe

u
´

oc l

u

o
.
ng θ.
ii)

U
´

oc l

u

o
.
ng kho

ang: ch

i ra mˆo
.
t kho

ang (θ
1
, θ
2
) ch
´

ua θ sao cho P (θ
1
< θ < θ
2
) =
1 − α cho tr

u
´

oc (1 − α go
.
i l`a ¯dˆo
.
tin cˆa
.
y c

ua

u
´

oc l

u

o
.
ng).
1. C
´
AC PH

U

ONG PH
´
AP

U
´

OC L

U
.

ONG D
¯
I

ˆ
EM
1.1 Ph

u

ong ph´ap h`am

u
´

oc l

u

o
.
ng
• Mˆo t

a ph

u

ong ph´ap
Gi

a s


u c
`
ˆan

u
´

oc l

u

o
.
ng tham s
´
ˆo θ c

ua ¯da
.
i l

u

o
.
ng ng
˜
ˆau nhiˆen X. T
`

u X ta lˆa
.
p m
˜
ˆau ng
˜
ˆau
nhiˆen W
X
= (X
1
, X
2
, . . . , X
n
).
Cho
.
n th
´
ˆong kˆe
ˆ
θ =
ˆ
θ(X
1
, X
2
, . . . , X
n
). Ta go
.
i
ˆ
θ l`a h`am

u
´

oc l

u

o
.
ng c

ua X.
Th

u
.
c hiˆe
.
n ph´ep th


u ta ¯d

u

o
.
c m
˜
ˆau cu
.
th

ˆe w
x
= (x
1
, x
2
, . . . , x
n
). Khi ¯d´o

u
´

oc l

u

o
.
ng
¯di

ˆem c

ua θ l`a gi´a tri
.
θ
0
=
ˆ
θ(x
1
, x
2
, . . . , x
n
).
a)

U
´

oc l

u

o
.
ng khˆong chˆe
.
ch
✷ D
¯
i
.
nh ngh
˜
ia 1 Th
´
ˆong kˆe
ˆ
θ =
ˆ
θ(X
1
, X
2
, . . . , X
n
) ¯d

u

o
.
c go
.
i l`a

u
´

oc l

u

o
.
ng khˆong chˆe
.
ch
c

ua tham s
´
ˆo θ n
´
ˆeu E(
ˆ
θ) = θ.

´
Y ngh
˜
ia
Gi

a s


u
ˆ
θ l`a

u
´

oc l

u

o
.
ng khˆong chˆe
.
ch c

ua tham s
´
ˆo θ. Ta c´o
E(
ˆ
θ − θ) = E(
ˆ
θ) − E(θ) = θ − θ = 0
69
70 Ch ’u ’ong 4.

U
´

oc l

u

ong tham s
´
ˆo c

ua ¯da
.
i l

u

ong ng
˜
ˆau nhiˆen
Vˆa
.
u

u
´

oc l

u

o
.
ng khˆong chˆe
.
ch l`a

u
´

oc l

u

o
.
ng c´o sai s
´
ˆo trung b`ınh b
`
˘
ang 0.
⊕ Nhˆa
.
n x´et
i) Trung b`ınh c

ua m
˜
ˆau ng
˜
ˆau nhiˆen X l`a

u
´

oc l

u

o
.
ng khˆong chˆe
.
ch c

ua trung b`ınh c

ua
t

ˆong th

ˆe θ = E(X) = m v`ı E(X) = m.
ii) Ph

u

ong sai ¯di
`
ˆeu ch

inh c

ua m
˜
ˆau ng
˜
ˆau nhiˆen S

2
l`a

u
´

oc l

u

o
.
ng khˆong chˆe
.
ch c

ua
ph

u

ong sai c

ua t

ˆong th

ˆe σ
2
v`ı E(S

2
) = σ
2
.
• V´ı du
.
1 Chi
`
ˆeu cao c

ua 50 cˆay lim ¯d

u

o
.
c cho b


oi
Kho

ang chi
`
ˆeu cao (m´et) s
´
ˆo cˆay lim x
0
i
u
i
n
i
u
i
n
i
u
2
i
[6, 25 − 6, 75) 1 6,5 -4 -4 16
[6, 75 − 7, 25) 2 7,0 -3 -6 18
[7, 25 − 7, 75) 5 7,5 -2 -10 20
[7, 75 − 8, 25) 11 8 -1 -11 11
[8, 25 − 8, 75) 18 8,5 0 0 0
[8, 75 − 9, 25) 9 9 1 9 9
[9, 25 − 9, 75) 3 9,5 2 6 12
[9, 75 − 10, 2) 1 10 3 3 9

50 -13 95
Go
.
i X l`a chi
`
ˆeu cao c

ua cˆay lim
a) H˜ay ch

i ra

u
´

oc l

u

o
.
ng ¯di

ˆem cho chi
`
ˆeu cao trung b`ınh c

ua c´ac cˆay lim.
b) H˜ay ch

i ra

u
´

oc l

u

o
.
ng ¯di

ˆem cho ¯dˆo
.
t

an m´at c

ua c´ac chi
`
ˆeu cao cˆay lim so v
´

oi chi
`
ˆeu
cao trung b`ınh.
c) Go
.
i p = P (7, 75 ≤ X ≤ 8, 75). H˜ay ch

i ra

u
´

oc l

u

o
.
ng ¯di

ˆem cho p.
Gi

ai
Ta lˆa
.
p b

ang t´ınh cho x v`a s
2
.
Th

u
.
c hiˆe
.
n ph´ep ¯d

ˆoi bi
´
ˆen u
i
=
x
0
i
− 8, 5
0, 5
(x
0
= 8, 5; h = 0, 5)
Ta c´o u = −
13
50
= −0, 26. Suy ra
x = 8, 5 + 0, 5.(−0, 26) = 8, 37
s
2
= (0, 5)
2
.

95
50
− (−0, 26)
2

= 0, 4581 ∼ (0, 68)
2
.
a) Chi
`
ˆeu cao trung b`ınh ¯d

u

o
.
c

u
´

oc l

u

o
.
ng l`a 8,37 m´et.
b) D
¯
ˆo
.
t

an m´at ¯d

u

o
.
c

u
´

oc l

u

o
.
ng l`a s = 0, 68 m´et ho
˘
a
.
c ˆs =

50
50−1
0, 4581 ∼ 0, 684
c) Trong 50 quan s´at ¯d˜a cho c´o 11+18 = 29 quan s´at cho chi
`
ˆeu cao lim thuˆo
.
c kho

ang
[7, 5 − 8, 5)
Vˆa
.
y

u
´

oc l

u

o
.
ng ¯di

ˆem cho p l`a p

=
29
50
= 0, 58.
1. C´ac ph

u

ong ph´ap

u
´

oc l

u

ong ¯di

ˆem 71
b)

U
´

oc l

u

o
.
ng hiˆe
.
u qu

a
⊕ Nhˆa
.
n x´et Gi

a s


u
ˆ
θ l`a

u
´

oc l

u

o
.
ng khˆong chˆe
.
ch c

ua tham s
´
ˆo θ. Theo b
´
ˆat ¯d

˘
ang th
´

uc
Tchebychev ta c´o
P (|
ˆ
θ − E(
ˆ
θ)| < ε) > 1 −
V ar(
ˆ
θ)
ε
2
V`ı E(
ˆ
θ) = θ nˆen P (|
ˆ
θ − θ| < ε) > 1 −
V ar(
ˆ
θ)
ε
2
.
Ta th
´
ˆay n
´
ˆeu V ar(
ˆ
θ) c`ang nh

o th`ı P (|
ˆ
θ − θ| < ε) c`ang g
`
ˆan 1. Do ¯d´o ta s˜e cho
.
n
ˆ
θ v
´

oi
V ar(
ˆ
θ) nh

o nh
´
ˆat.
✷ D
¯
i
.
nh ngh
˜
ia 2

U
´

oc l

u

o
.
ng khˆong chˆe
.
ch
ˆ
θ ¯d

u

o
.
c go
.
i l`a

u
´

oc l

u

o
.
ng c´o hiˆe
.
u qu

a c

ua tham
s
´
ˆo θ n
´
ˆeu V ar(
ˆ
θ) nh

o nh
´
ˆat trong c´ac

u
´

oc l

u

o
.
ng c

ua θ.
 Ch´u ´y Ng

u
`

oi ta ch
´

ung minh ¯d

u

o
.
c r
`
˘
ang n
´
ˆeu
ˆ
θ l`a

u
´

oc l

u

o
.
ng hiˆe
.
u qu

a c

ua θ th`ı ph

u

ong
sai c

ua n´o l`a
V ar(
ˆ
θ) =
1
n.E(
∂lnf (x,θ)
∂θ
)
2
(4.1)
trong ¯d´o f(x, θ) l`a h`am mˆa
.
t ¯dˆo
.
x´ac su
´
ˆat c

ua ¯da
.
i l

u

o
.
ng ng
˜
ˆau nhiˆen g
´
ˆoc. Mo
.
i

u
´

oc
l

u

o
.
ng khˆong chˆe
.
ch θ luˆon c´o ph

u

ong sai l
´

on h

on V ar(
ˆ
θ) trong (4.1). Ta go
.
i (4.1) l`a gi
´

oi
ha
.
n Crame-Rao.
⊕ Nhˆa
.
n x´et N
´
ˆeu ¯da
.
i l

u

o
.
ng ng
˜
ˆau nhiˆen g
´
ˆoc X ∈ N(µ,
σ
2
n
) th`ı trung b`ınh m
˜
ˆau X l`a

u
´

oc l

u

o
.
ng hiˆe
.
u qu

a c

ua k`y vo
.
ng E(X) = µ.
Thˆa
.
t vˆa
.
y, ta bi
´
ˆet X =
1
n
n

i=1
X
i
∈ N(µ,
σ
2
n
)
M
˘
a
.
t kh´ac do X c´o phˆan ph
´
ˆoi chu

ˆan nˆen n
´
ˆeu f (x, µ) l`a h`am mˆa
.
t ¯dˆo
.
c

ua X
i
th`ı
f(x, µ) =
1
σ


e
−(x−µ)
2
/2σ
2
Ta c´o

∂µ
lnf(x, µ) =
x − µ
σ
2
.
Suy ra nE

∂lnf(x, µ)
∂µ

2
= nE

x − µ
σ
2

2
=
n
σ
2
. Do ¯d´o V ar(X) ch´ınh b
`
˘
ang nghi
.
ch
¯d

ao σ
2
/n.
Vˆa
.
y X l`a

u
´

oc l

u

o
.
ng hiˆe
.
u qu

a c

ua µ.
c)

U
´

oc l

u

o
.
ng v
˜

ung
✷ D
¯
i
.
nh ngh
˜
ia 3 Th
´
ˆong kˆe
ˆ
θ =
ˆ
θ(X
1
, X
2
, . . . , X
n
) ¯d

u

o
.
c go
.
i l`a

u
´

oc l

u

o
.
ng v
˜

ung c

ua tham
s
´
ˆo θ n
´
ˆeu ∀ε > 0 ta c´o
lim
n→∞
P (|
ˆ
θ − θ| < ε) = 1
72 Ch ’u ’ong 4.

U
´

oc l

u

ong tham s
´
ˆo c

ua ¯da
.
i l

u

ong ng
˜
ˆau nhiˆen
 D
¯
i
`
ˆeu kiˆe
.
n ¯d

u c

ua

u
´

oc l

u

o
.
ng v
˜

ung
N
´
ˆeu
ˆ
θ l`a

u
´

oc l

u

o
.
ng khˆong chˆe
.
ch c

ua θ v`a lim
n→∞
V ar(
ˆ
θ) = 0 th`ı
ˆ
θ l`a

u
´

oc l

u

o
.
ng v
˜

ung
c

ua θ.
1.2 Ph

u

ong ph´ap

u
´

oc l

u

o
.
ng h

o
.
p l´y t
´
ˆoi ¯da
Gi

a s


u W
X
= (X
1
, X
2
, . . . , X
n
) l`a m
˜
ˆau ng
˜
ˆau nhiˆen ¯d

u

o
.
c ta
.
o nˆen t
`

u ¯da
.
i l

u

o
.
ng ng
˜
ˆau
nhiˆen X c´o m
˜
ˆau cu
.
th

ˆe w
x
= (x
1
, x
2
, . . . , x
n
) v`a
ˆ
θ =
ˆ
θ(X
1
, X
2
, . . . , X
n
).
X´et h`am h`am h

o
.
p l´y L(x
1
, . . . , x
n
, θ) c

ua ¯d
´
ˆoi s
´
ˆo θ x´ac ¯di
.
nh nh

u sau:
• N
´
ˆeu X r
`

oi ra
.
c:
L(x
1
, . . . , x
n
, θ) = P (X
1
= x
1
/θ, . . . , X
n
= x
n
/θ) (4.2)
=
n

i=1
P (X
i
= x
i
/θ) (4.3)
L(x
1
, . . . , x
n
, θ) l`a x´ac su
´
ˆat ¯d

ˆe ta nhˆa
.
n ¯d

u

o
.
c m
˜
ˆau cu
.
th

ˆe W
x
= (x
1
, . . . , x
n
)
• N
´
ˆeu X liˆen tu
.
c c´o h`am mˆa
.
t ¯dˆo
.
x´ac su
´
ˆat f (x, θ)
L(x
1
, . . . , x
n
, θ) = f(x
1
, θ)f(x
2
, θ) . . . f(x
n
, θ)
L(x
1
, x
2
, . . . , x
n
, θ) l`a mˆa
.
t ¯dˆo
.
c

ua x´ac su
´
ˆat ta
.
i ¯di

ˆem w
x
(x
1
, x
2
, . . . , x
n
)
Gi´a tri
.
θ
0
=
ˆ
θ(x
1
, x
2
, . . . , x
n
) ¯d

u

o
.
c go
.
i l`a

u
´

oc l

u

o
.
ng h

o
.
p l´y t
´
ˆoi ¯da n
´
ˆeu
´

ung v
´

oi gi´a
tri
.
n`ay c

ua θ h`am h

o
.
p l´y ¯da
.
t c

u
.
c ¯da
.
i.
 Ph

u

ong ph´ap t`ım
V`ı h`am L v`a lnL ¯da
.
t c

u
.
c ¯da
.
i ta
.
i c`ung mˆo
.
t gi´a tri
.
θ nˆen ta x´et lnL thay v`ı x´et L.
B

u
´

oc 1: T`ım
∂lnL
∂θ
B

u
´

oc 2: Gi

ai ph

u

ong tr`ınh
∂lnL
∂θ
(Ph

u

ong tr`ınh h

o
.
p l´y)
Gi

a s


u ph

u

ong tr`ınh c´o nghiˆe
.
m l`a θ
0
=
ˆ
θ(x
1
, x
2
, . . . , x
n
)
B

u
´

oc 3: T`ım ¯da
.
o h`am c
´
ˆap hai

2
lnL
∂θ
N
´
ˆeu ta
.
i θ
0
m`a

2
lnL
∂θ
< 0 th`ı lnL ¯da
.
t c

u
.
c ¯da
.
i. Khi ¯d´o θ
0
=
ˆ
θ(x
1
, x
2
, . . . , x
n
) l`a

u
´

oc
l

u

o
.
ng ¯di

ˆem h

o
.
p l´y t
´
ˆoi ¯da c

ua θ.
2. Ph

u

ong ph´ap kho

ang tin cˆay 73
2. PH

U

ONG PH
´
AP KHO

ANG TIN C
ˆ
A
.
Y
2.1 Mˆo t

a ph

u

ong ph´ap
Gi

a s


u t

ˆong th

ˆe c´o tham s
´
ˆo θ ch

ua bi
´
ˆet. Ta t`ım kho

ang (θ
1
, θ
2
) ch
´

ua θ sao cho
P (θ
1
< θ < θ
2
) = 1 − α cho tr

u
´

oc.
T
`

u ¯da
.
i l

u

o
.
ng ng
˜
ˆau nhiˆen g
´
ˆoc X lˆa
.
p m
˜
ˆau ng
˜
ˆau nhiˆen W
X
= (X
1
, X
2
, . . . , X
n
). Cho
.
n
th
´
ˆong kˆe
ˆ
θ =
ˆ
θ(X
1
, X
2
, . . . , X
n
) c´o phˆan ph
´
ˆoi x´ac su
´
ˆat x´ac ¯di
.
nh d`u ch

ua bi
´
ˆet θ.
V
´

oi α
1
kh´a b´e (α
1
< α) ta t`ım ¯d

u

o
.
c phˆan vi
.
θ
α
1
c

ua
ˆ
θ (t
´

uc l`a P (
ˆ
θ < θ
α
1
) = α
1
).
V
´

oi α
2
m`a α
1
+ α
2
= α kh´a b´e (th

u
`

ong l
´
ˆay α ≤ 0, 05) ta t`ım ¯d

u

o
.
c phˆan vi
.
θ
1−α
2
c

ua
ˆ
θ (t
´

uc l`a P (
ˆ
θ < θ
1−α
2
) = 1 − α
2
).
Khi ¯d´o
P (θ
α
1

ˆ
θ ≤ θ
1−α
2
) = P (
ˆ
θ < θ
1−α
2
) − P (
ˆ
θ < θ
α
1
) = 1 − α
2
− α
1
= 1 − α (∗)
T
`

u (*) ta gi

ai ra ¯d

u

o
.
c θ. Khi ¯d´o (*) ¯d

u

o
.
c ¯d

ua v
`
ˆe da
.
ng P (
ˆ
θ
1
< θ <
ˆ
θ
2
) = 1 − α.
V`ı x´ac su
´
ˆat 1− α g
`
ˆan b
`
˘
ang 1, nˆen bi
´
ˆen c
´
ˆo (
ˆ
θ
1
< θ <
ˆ
θ
2
) h
`
ˆau nh

u x

ay ra. Th

u
.
c hiˆe
.
n
mˆo
.
t ph´ep th


u ¯d
´
ˆoi v
´

oi m
˜
ˆau ng
˜
ˆau nhiˆen W
X
ta thu ¯d

u

o
.
c m
˜
ˆau cu
.
th

ˆe w
x
= (x
1
, x
2
, . . . , x
n
).
T
`

u m
˜
ˆau cu
.
th

ˆe n`ay ta t´ınh ¯d

u

o
.
c gi´a tri
.
θ
1
=
ˆ
θ
1
(x
1
, x
2
, . . . , x
n
), θ
2
=
ˆ
θ
2
(x
1
, x
2
, . . . , x
n
).
Vˆa
.
y v
´

oi 1− α cho tr

u
´

oc, qua m
˜
ˆau cu
.
th

ˆe w
x
ta t`ım ¯d

u

o
.
c kho

ang (θ
1
, θ
2
) ch
´

ua θ sao
cho P (θ
1
< θ < θ
2
) = 1 − α.
• Kho

ang (θ
1
, θ
2
) ¯d

u

o
.
c go
.
i l`a kho

ang tin cˆa
.
y.
• 1 − α ¯d

u

o
.
c go
.
i l`a ¯dˆo
.
tin cˆa
.
y c

ua

u
´

oc l

u

o
.
ng.
• |θ
2
− θ
1
| ¯d

u

o
.
c go
.
i l`a ¯dˆo
.
d`ai kho

ang tin cˆa
.
y.
2.2

U
´

oc l

u

o
.
ng trung b`ınh
Gi

a s


u trung b`ınh c

ua t

ˆong th

ˆe E(X) = m ch

ua bi
´
ˆet. Ta t`ım kho

ang (m
1
, m
2
) ch
´

ua
m sao cho P (m
1
< m < m
2
) = 1 − α, v
´

oi 1 − α l`a ¯dˆo
.
tin cˆa
.
y cho tr

u
´

oc.
i) Tr

u
`

ong h

o
.
p 1

Bi
´
ˆet V ar(X) = σ
2
n ≥ 30 ho
˘
a
.
c (n < 30 nh

ung X c´o phˆan ph
´
ˆoi chu

ˆan)
Cho
.
n th
´
ˆong kˆe
U =
(X − m)

n
σ
(4.4)
Ta th
´
ˆay U ∈ N(0, 1).
74 Ch ’u ’ong 4.

U
´

oc l

u

ong tham s
´
ˆo c

ua ¯da
.
i l

u

ong ng
˜
ˆau nhiˆen
Cho
.
n c
˘
a
.
p α
1
v`a α
2
sao cho α
1
+ α
2
= α v`a t`ım c´ac phˆan vi
.
P (U < u
α
1
) = α
1
, P (U < u
α
2
) = 1 − α
2
Do phˆan vi
.
chu

ˆan c´o t´ınh ch
´
ˆat u
α
1
= −u
1−α
1
nˆen
P (−u
1−α
1
< U < u
1−α
2
) = 1 − α (4.5)
D

u
.
a v`ao (4.4) v`a gi

ai hˆe
.
b
´
ˆat ph

u

ong tr`ınh trong (4.5) ta ¯d

u

o
.
c
X −
σ

n
u
1−α
2
< m < X +
σ

n
u
1−α
1
D
¯

ˆe ¯d

u

o
.
c kho

ang tin cˆa
.
y ¯d
´
ˆoi x
´

ung ta cho
.
n α
1
= α
2
=
α
2
v`a ¯d
˘
a
.
t γ = 1 −
α
2
th`ı
X −
σ

n
u
γ
< m < X +
σ

n
u
γ
T´om la
.
i, ta t`ım ¯d

u

o
.
c kho

ang tin cˆa
.
y (x − ε, x + ε), trong ¯d´o
* x l`a trung b`ınh c

ua m
˜
ˆau ng
˜
ˆau nhiˆen.
* ε = u
γ
σ

n
(¯dˆo
.
ch´ınh x´ac) v
´

oi u
γ
l`a phˆan vi
.
chu

ˆan m
´

uc γ = 1 −
α
2
• V´ı du
.
2 Kh
´
ˆoi l

u

o
.
ng s

an ph

ˆam l`a ¯da
.
i l

u

o
.
ng ng
˜
ˆau nhiˆen X c´o phˆan ph
´
ˆoi chu

ˆan v
´

oi ¯dˆo
.
lˆe
.
ch tiˆeu chu

ˆan σ = 1. Cˆan th


u 25 s

an ph

ˆam ta thu ¯d

u

o
.
c k
´
ˆet qu

a sau
X (kh
´
ˆoi l

u

o
.
ng) 18 19 20 21
n
i
(s
´
ˆo l

u

o
.
ng 3 5 15 2
H˜ay

u
´

oc l

u

o
.
ng trung b`ınh kh
´
ˆoi l

u

o
.
ng c

ua s

an ph

ˆam v
´

oi ¯dˆo
.
tin cˆa
.
y 95 %.
Gi

ai
x
i
n
i
x
i
n
i
18 3 54
19 5 95
20 15 300
21 2 42

25 491
Ta c´o x =
491
25
= 19, 64kg.
D
¯
ˆo
.
tin cˆa
.
y 1 − α = 0, 95 =⇒ α = 0, 025 =⇒ γ = 1 −
α
2
= 0, 975 Ta t`ım
¯d

u

o
.
c phˆan vi
.
chu

ˆan u
γ
= u
0,975
= 1, 96. Do ¯d´o
ε = u
0,975
1

25
= 1, 96.
1
5
= 0.39
x
1
= x − ε = 19, 6 − 0, 39 = 19, 25
x
2
= x + ε = 19, 6 + 0, 39 = 20, 03
Vˆa
.
y kho

ang tin cˆa
.
y l`a (19, 25; 20, 03).
2. Ph

u

ong ph´ap kho

ang tin cˆay 75
ii) Tr

u
`

ong h

o
.
p 2

σ
2
ch

ua bi
´
ˆet
n ≥ 30
Tr

u
`

ong h

o
.
p n`ay k´ıch th

u
´

oc m
˜
ˆau l
´

on (n ≥ 30) c´o th

ˆe d`ung

u
´

oc l

u

o
.
ng c

ua S

2
thay
cho σ
2
ch

ua bi
´
ˆet (E(S

2
) = σ
2
), ta t`ım ¯d

u

o
.
c kho

ang tin cˆa
.
y (x − ε, x + ε) trong ¯d´o
* x l`a trung b`ınh c

ua m
˜
ˆau cu
.
th

ˆe.
* ε = u
γ
s


n
v
´

oi u
γ
l`a phˆan vi
.
chu

ˆan m
´

uc γ = 1 −
α
2
v`a s

l`a ¯dˆo
.
lˆe
.
ch tiˆeu chu

ˆan
¯di
`
ˆeu ch

inh c

ua m
˜
ˆau cu
.
th

ˆe.
• V´ı du
.
3 Ng

u
`

oi ta ti
´
ˆen h`anh nghiˆen c
´

uu


o mˆo
.
t tr

u
`

ong ¯da
.
i ho
.
c xem trong mˆo
.
t th´ang
trung b`ınh mˆo
.
t sinh viˆen tiˆeu h
´
ˆet bao nhiˆeu ti
`
ˆen go
.
i ¯diˆe
.
n thoa
.
i. L
´
ˆay mˆo
.
t m
˜
ˆau ng
˜
ˆau nhiˆen
g
`
ˆom 59 sinh viˆen thu ¯d

u

o
.
c k
´
ˆet qu

a sau:
14 18 22 30 36 28 42 79 36 52 15 47
95 16 27 111 37 63 127 23 31 70 27 11
30 147 72 37 25 7 33 29 35 41 48 15
29 73 26 15 26 31 57 40 18 85 28 32
22 36 60 41 35 26 20 58 33 23 35
H˜ay

u
´

oc l

u

o
.
ng kho

ang tin cˆa
.
y 95% cho s
´
ˆo ti
`
ˆen go
.
i ¯diˆe
.
n thoa
.
i trung b`ınh h`ang th´ang
c

ua mˆo
.
t sinh viˆen.
Gi

ai
T
`

u c´ac s
´
ˆo liˆe
.
u ¯d˜a cho, ta c´o
n = 59; x = 41, 05; s

= 27, 99
D
¯
ˆo
.
tin cˆa
.
y 1 − α = 0, 95 =⇒ 1 −
α
2
= 0, 975. Tra b

ang phˆan vi
.
chu

ˆan ta c´o
u
0,975
= 1, 96.
Do ¯d´o ε = 1, 96.
27,99

59
= 7, 13.
x − 7, 13 = 33, 92; x + 7, 13 = 48, 18
Vˆa
.
y kho

ang tin cˆa
.
y c

ua

u
´

oc l

u

o
.
ng l`a (33,92; 48,18).
iii) Tr

u
`

ong h

o
.
p 3

σ
2
ch

ua bi
´
ˆet
n < 30 v`a X c´o phˆan ph
´
ˆoi chu

ˆan
Cho
.
n th
´
ˆong kˆe T =
(X − m)

n
S

∈ T (n− 1).
76 Ch ’u ’ong 4.

U
´

oc l

u

ong tham s
´
ˆo c

ua ¯da
.
i l

u

ong ng
˜
ˆau nhiˆen
Ta t`ım ¯d

u

o
.
c kho

ang tin cˆa
.
y (x − ε, x + ε) trong ¯d´o ε = t
γ
S


n
v
´

oi t
γ
l`a phˆan vi
.
Student m
´

uc γ = 1 −
α
2
v
´

oi n − 1 bˆa
.
c t

u
.
do v`a s

l`a ¯dˆo
.
lˆe
.
ch tiˆeu
chu

ˆan ¯di
`
ˆeu ch

inh c

ua m
˜
ˆau cu
.
th

ˆe.
• V´ı du
.
4 Dioxide Sulfur v`a Oxide Nitrogen l`a c´ac h´oa ch
´
ˆat ¯d

u

o
.
c khai th´ac t
`

u l`ong
¯d
´
ˆat. C´ac ch
´
ˆat n`ay ¯d

u

o
.
c gi´o mang ¯di r
´
ˆat xa, k
´
ˆet h

o
.
p th`anh acid v`a r

oi tr


o la
.
i m
˘
a
.
t ¯d
´
ˆat ta
.
o
th`anh m

ua acid. Ng

u
`

oi ta ¯do ¯dˆo
.
¯dˆa
.
m ¯d
˘
a
.
c c

ua Dioxide Sulfur (µg/m
3
) trong khu r
`

ung
Bavarian c

ua n

u
´

oc D
¯
´

uc. S
´
ˆo liˆe
.
u cho b


oi b

ang d

u
´

oi ¯dˆay:
52,7 43,9 41,7 71,5 47,6 55,1
62,2 56,5 33,4 61,8 54,3 50,0
45,3 63,4 53,9 65,5 66,6 70,0
52,4 38,6 46,1 44,4 60,7 56,4
H˜ay

u
´

oc l

u

o
.
ng ¯dˆo
.
¯dˆa
.
m ¯d
˘
a
.
c trung b`ınh c

ua Dioxide Sulsfur v
´

oi ¯dˆo
.
tin cˆa
.
y 95%.
Gi

ai
Ta t´ınh ¯d

u

o
.
c x = 53, 92µg/m
3
, s

= 10, 07µg/m
3
.
D
¯
ˆo
.
tin cˆa
.
y 1− α = 0, 95 =⇒ α = 0, 025 =⇒ 1−
α
2
= 0, 975. Tra b

ang phˆan
vi
.
student m
´

uc 0,975 bˆa
.
c n − 1 = 23 ta ¯d

u

o
.
c t
23;0,975
= 2, 069.
Do ¯d´o ε = 2, 069
10,07

24
= 4, 25.
x − ε = 53, 92 − 4, 25 = 49, 67, x + ε = 53, 92 + 4, 25 = 58, 17
Vˆa
.
y kho

ang tin cˆa
.
y l`a (49,67; 58,17).
Ng

u
`

oi ta bi
´
ˆet ¯d

u

o
.
c n
´
ˆeu ¯dˆo
.
¯dˆa
.
m ¯d
˘
a
.
c c

ua Dioxide Sulfur trong mˆo
.
t khu v

u
.
c l
´

on h

on
20µg/m
3
th`ı mˆoi tr

u
`

ong trong khu v

u
.
c bi
.
ph´a hoa
.
i b


oi m

ua acid. Qua v´ı du
.
n`ay c´ac
nh`a khoa ho
.
c ¯d˜a t`ım ra ¯d

u

o
.
c nguyˆen nhˆan r
`

ung Bavarian bi
.
ph´a hoa
.
i tr
`
ˆam tro
.
ng n
˘
am
1983 l`a do m

ua acid .
 Ch´u ´y (X´ac ¯di
.
nh k´ıch th

u
´

oc m
~
^au)
N
´
ˆeu mu
´
ˆon ¯dˆo
.
tin cˆa
.
y 1 − α v`a ¯dˆo
.
ch´ınh x´ac ε ¯da
.
t


o m
´

uc cho tr

u
´

oc th`ı ta c
`
ˆan x´ac
¯di
.
nh k´ıch th

u
´

oc n c

ua m
˜
ˆau.
i) Tr

u
`

ong h

o
.
p bi
´
ˆet V ar(X) = σ
2
:
T
`

u cˆong th
´

uc ε = u
2
γ
σ

n
ta suy ra
n = u
2
γ
σ
2
ε
2
ii) Tr

u
`

ong h

o
.
p ch

ua bi
´
ˆet σ
2
:
2. Ph

u

ong ph´ap kho

ang tin cˆay 77
D

u
.
a v`a m
˜
ˆau cu
.
th

ˆe ¯d˜a cho (n
´
ˆeu ch

ua c´o m
˜
ˆau th`ı ta c´o th

ˆe ti
´
ˆen h`anh l
´
ˆay m
˜
ˆau l
`
ˆan
¯d
`
ˆau v
´

oi k´ıch th

u
´

oc n
1
≥ 30) ¯d

ˆe t´ınh s
2
. T
`

u ¯d´o x´ac ¯di
.
nh ¯d

u

o
.
c
n = u
2
γ
s
2
ε
2
K´ıch th

u
´

oc m
˜
ˆau n ph

ai l`a s
´
ˆo nguyˆen. N
´
ˆeu khi t´ınh n theo c´ac cˆong th
´

uc trˆen ¯d

u

o
.
c
gi´a tri
.
khˆong nguyˆen th`ı ta l
´
ˆay ph
`
ˆan nguyˆen c

ua n´o cˆo
.
ng thˆem v
´

oi 1.
T
´

uc l`a n =

u
2
γ
σ
2
ε
2

+ 1 ho
˘
a
.
c n =

u
2
γ
s
2
ε
2

+ 1.
2.3

U
´

oc l

u

o
.
ng t

y lˆe
.
Gi

a s


u t

ˆong th

ˆe ¯d

u

o
.
c chia ra l`am hai loa
.
i ph
`
ˆan t


u. T

y lˆe
.
ph
`
ˆan t


u c´o t´ınh ch
´
ˆat A l`a p
ch

ua bi
´
ˆet.

U
´

oc l

u

o
.
ng t

y lˆe
.
l`a ch

i ra kho

ang (f
1
, f
2
) ch
´

ua p sao cho P (f
1
< p < f
2
) = 1−α.
D
¯

ˆe cho viˆe
.
c gi

ai b`ai to´an ¯d

u

o
.
c ¯d

on gi

an, ta cho
.
n m
˜
ˆau v
´

oi k´ıch th

u
´

oc n kh´a l
´

on.
Go
.
i X l`a s
´
ˆo ph
`
ˆan t


u c´o t´ınh ch
´
ˆat A khi l
´
ˆay ng
˜
ˆau nhiˆen mˆo
.
t ph
`
ˆan t


u t
`

u t

ˆong th

ˆe th`ı
X l`a ¯da
.
i l

u

o
.
ng ng
˜
ˆau nhiˆen c´o phˆan ph
´
ˆoi x´ac su
´
ˆat
X 0 1
P 1-p p
Go
.
i X
i
(i = 1, n) l`a s
´
ˆo ph
`
ˆan t


u c´o t´ınh ch
´
ˆat A trong l
`
ˆan l
´
ˆay th
´

u i.
Ta c´o X =
1
n
n

i=1
X
i
ch´ınh l`a t
`
ˆan su
´
ˆat

u
´

oc l

u

o
.
ng ¯di

ˆem c

ua p = E(X). M
˘
a
.
t kh´ac, theo
ch

u

ong 2, nX c´o phˆan ph
´
ˆoi nhi
.
th
´

uc B(n, p). T
`

u ¯d´o E(X) = p v`a V ar(X) =
p(1 − p)
n
.
Cho
.
n th
´
ˆong kˆe U =
(f − p)

n

p(1 − p)
, trong ¯d´o f l`a t

y lˆe
.
c´ac ph
`
ˆan t


u c

ua m
˜
ˆau c´o t´ınh
ch
´
ˆat A.
Khi n kh´a l
´

on th`ı U ∈ N(0, 1). Gi

ai quy
´
ˆet b`ai to´an t

u

ong t

u
.
nh

u


o

u
´

oc l

u

o
.
ng trung
b`ınh, thay X b


oi f, σ
2
b


oi f (1− f) ta ¯d

u

o
.
c
f − u
γ

f(1 − f)
n
< p < f + u
γ

f(1 − f)
n
T´om la
.
i, ta x´ac ¯di
.
nh ¯d

u

o
.
c kho

ang tin cˆa
.
y (f
1
, f
2
) = (f − ε, f + ε), trong ¯d´o
f l`a t

y lˆe
.
c´ac ph
`
ˆan t


u c

ua m
˜
ˆau c´o t´ınh ch
´
ˆat A
ε = u
γ

f(1 − f)
n
(¯dˆo
.
ch´ınh x´ac) (4.6)
78 Ch ’u ’ong 4.

U
´

oc l

u

ong tham s
´
ˆo c

ua ¯da
.
i l

u

ong ng
˜
ˆau nhiˆen
v
´

oi u
γ
l`a phˆan vi
.
chu

ˆan m
´

uc 1 −
α
2
.
T
`

u (4.6) ta c´o
u
γ
=
ε

n

f(1 − f)
n = u
2
1−
α
2
f(1 − f)
ε
2
 Ch´u ´y Ta c´o th

ˆe t`ım kho

ang tin cˆa
.
y c

ua p b
`
˘
ang c´ach kh´ac nh

u sau:
T
`

u kho

ang tin cˆa
.
y c

ua p:


f − u
γ

p(1 − p)
n
< p < f + u
γ

p(1 − p)
n


hay


|f − p| < u
γ

p(1 − p)
n


Gi

ai b
´
ˆat ph

u

ong tr`ınhn`ay ta t`ım ¯d

u

o
.
c
p
1
=
nf + 0, 5u
2
γ


0, 25u
2
γ
− nf(1 − f)
n + u
2
γ
, p
2
=
nf + 0, 5u
2
γ
+

0, 25u
2
γ
− nf(1 − f)
n + u
2
γ
Khi ¯d´o (p
1
, p
2
) l`a kho

ang tin cˆa
.
y c

ua p v
´

oi ¯dˆo
.
tin cˆa
.
y 1 − α.
• V´ı du
.
5 Ki

ˆem tra 100 s

an ph

ˆam trong lˆo h`ang th
´
ˆay c´o 20 ph
´
ˆe ph

ˆam.
i) H˜ay

u
´

oc l

u

o
.
ng t

y lˆe
.
ph
´
ˆe ph

ˆam c´o ¯dˆo
.
tin cˆa
.
y 99 %.
ii) N
´
ˆeu ¯dˆo
.
ch´ınh x´ac ε = 0, 04 th`ı ¯dˆo
.
tin cˆa
.
y c

ua

u
´

oc l

u

o
.
ng l`a bao nhiˆeu?
iii) N
´
ˆeu mu
´
ˆon c´o ¯dˆo
.
tin cˆa
.
y 99% v`a ¯dˆo
.
ch´ınh x´ac 0,04 th`ı ph

ai ki

ˆem tra bao nhiˆeu
s

an ph

ˆam?
Gi

ai
i) n = 100, f =
20
100
= 0.2
X´et U =
(f−p)

100

pq
∈ N(0, 1).
Ta c´o
1 − α = 0, 99 =⇒ α = 0, 01 =⇒ 1 −
α
2
= 1 − 0, 005 = 0, 995
ε = u
0,995

0, 2.0, 8

100
= 2, 58.
0, 4
10
= 0, 1
f
1
= f − ε = 0, 2 − 0, 1 = 0, 1
f
2
= f + ε = 0, 2 + 0, 1 = 0, 3
2. Ph

u

ong ph´ap kho

ang tin cˆay 79
Vˆa
.
y kho

ang tin cˆa
.
y l`a (0, 1; 0, 3).
ii) u
1−
α
2
=
0, 04.

100

0, 2.0, 8
= 1
T`ım ¯d

u

o
.
c
1 −
α
2
= 0, 84 =⇒ 1 − α = 0, 68
Vˆa
.
y ¯dˆo
.
tin cˆa
.
y l`a 68%.
iii)1−α = 0, 99 =⇒ α = 0, 01 =⇒ 1−
α
2
= 0, 995. T`ım ¯d

u

o
.
c u
0,995
= 2, 576.
Do ¯d´o
n ≈
(2, 576)
2
.0, 2.0, 8
(0, 04)
2
= 6, 635.100 = 663, 5
Vˆa
.
y n = 664
2.4

U
´

oc l

u

o
.
ng ph

u

ong sai
Gi

a s


u ¯da
.
i l

u

o
.
ng ng
˜
ˆau nhiˆen X c´o phˆan ph
´
ˆoi chu

ˆan v
´

oi ph

u

ong sai V ar(X) = σ
2
ch

ua bi
´
ˆet. Cho 0 < α < 0.05.

U
´

oc l

u

o
.
ng ph

u

ong sai V ar(X) l`a ch

i ra kho

ang (σ
2
1
, σ
2
2
)
ch
´

ua σ
2
sao cho P (σ
2
1
< σ
2
< σ
2
2
) = 1 − α.
T
`

u X lˆa
.
p m
˜
ˆau ng
˜
ˆau nhiˆen W
X
= (X
1
, X
2
, . . . , X
n
) v`a x´et c´ac tr

u
`

ong h

o
.
p
a) Bi
´
ˆet E(X) = µ.
Cho
.
n th
´
ˆong kˆe χ
2
=
n

i=1
(X
i
− µ)
2
σ
2
Ta th
´
ˆay χ
2
c´o phˆan ph
´
ˆoi ”khi-b`ınh ph

u

ong” v
´

oi n bˆa
.
c t

u
.
do.
Cho
.
n α
1
v`a α
2
kh´a b´e sao cho α
1
+ α
2
= α. Ta t`ım ¯d

u

o
.
c c´ac phˆan vi
.
χ
2
α
1
v`a χ
2
1−α
2
th

oa m˜an
P (χ
2
α
1
< χ
2
< χ
2
1−α
2
) = 1 − α (4.7)
Thay bi

ˆeu th
´

uc c

ua χ
2
v`ao (4.7) v`a gi

ai ra ta ¯d

u

o
.
c

(X
i
− µ)
2
χ
2
1−α
2
< σ
2
<

(X
i
− µ)
2
χ
2
α
1
Cho
.
n α
1
= α
2
=
α
2
th`ı

(X
i
− µ)
2
χ
2
1−
α
2
< σ
2
<

(X
i
− µ)
2
χ
2
α
2
(4.8)
V
´

oi m
˜
ˆau cu
.
th

ˆe w
x
= (x
1
, x
2
, . . . , x
n
), t´ınh c´ac t

ˆong

(x
i
− µ)
2
v`a d

u
.
a v`ao (4.8) ta
t`ım ¯d

u

o
.
c kho

ang tin cˆa
.
y (σ
2
1
, σ
2
2
), trong ¯d´o
80 Ch ’u ’ong 4.

U
´

oc l

u

ong tham s
´
ˆo c

ua ¯da
.
i l

u

ong ng
˜
ˆau nhiˆen
σ
2
1
=

(x
i
− µ)
2
n
i
χ
2
n,1−
α
2
σ
2
2
=

(x
i
− µ)
2
n
i
χ
2
n,
α
2
v
´

oi
χ
2
n,1−
α
2
l`a phˆan vi
.
”khi−b`ınh ph

u

ong” m
´

uc 1 −
α
2
v
´

oi n bˆa
.
c t

u
.
do.
χ
2
n,
α
2
l`a phˆan vi
.
”khi−b`ınh ph

u

ong” m
´

uc
α
2
v
´

oi n bˆa
.
c t

u
.
do.
b) Ch

ua bi
´
ˆet E(X).
Cho
.
n th
´
ˆong kˆe χ
2
=
(n − 1)S
2
σ
2
Th
´
ˆong kˆe n`ay c´o phˆan ph
´
ˆoi ”khi−b`ınh ph

u

ong v
´

oi n − 1 bˆa
.
c t

u
.
do. T

u

ong t

u
.
nh

u
trˆen ta t`ım ¯d

u

o
.
c kho

ang tin cˆa
.
y (σ
2
1
, σ
2
2
) v
´

oi
σ
2
1
=
(n − 1)s
2
χ
2
n−1,1−
α
2
; σ
2
2
=
(n − 1)s
2
χ
2
n−1,
α
2
• V´ı du
.
6 M
´

uc hao ph´ı nhiˆen liˆe
.
u cho mˆo
.
t ¯d

on vi
.
s

an ph

ˆam l`a ¯da
.
i l

u

o
.
ng ng
˜
ˆau nhiˆen
c´o phˆan ph
´
ˆoi chu

ˆan. X´et trˆen 25 s

an ph

ˆam ta thu ¯d

u

o
.
c k
´
ˆet qu

a sau:
X 19,5 20 20,5
n
i
5 18 2
H˜ay

u
´

oc l

u

o
.
ng ph

u

ong sai v
´

oi ¯dˆo
.
tin cˆa
.
y 90 % trong c´ac tr

u
`

ong h

o
.
p sau:
i) Bi
´
ˆet k`y vo
.
ng µ = 20g.
ii) Ch

ua bi
´
ˆet k`y vo
.
ng.
Gi

ai
i) Bi
´
ˆet µ = 20g.
x
i
n
i
x
i
− 20 (x
i
− 20)
2
(x
i
− 20)
2
n
i
19,5 5 -0,5 0,25 1,25
20 18 0 0 0
20,5 2 0,5 0,25 0,5

n=25 1,75
D
¯
ˆo
.
tin cˆa
.
y 1 − α = 0, 9 =⇒ α = 0, 1 =⇒
α
2
= 0, 05 =⇒ 1 −
α
2
= 0.95
Tra b

ang phˆan vi
.
χ
2
v
´

oi n = 25 bˆa
.
c t

u
.
do ta ¯d

u

o
.
c
χ
2
25;0,05
= 14, 6; χ
2
25;0,95
= 37, 7
3. B`ai t
.
ˆap 81
Do ¯d´o
σ
2
1
=

(x
i
− 20)
2
n
i
χ
2
25;0,95
=
1, 75
37, 7
= 0, 046
σ
2
2
=

(x
i
− 20)
2
n
i
χ
2
25;0,05
=
1, 75
14, 6
= 0, 12
Vˆa
.
y kho

ang tin cˆa
.
y l`a (0, 046; 0, 12).
ii) Khi ch

ua bi
´
ˆet k`y vo
.
ng ta t`ım s
2
= 0, 0692.
Tra b

ang phˆan vi
.
khi b`ınh ph

u

ong v
´

oi bˆa
.
c t

u
.
do n − 1 = 24.
χ
2
0,05
= 13, 85; χ
2
0,95
= 36, 4
v`a t´ınh
σ
2
1
=
24s
2
χ
2
0,95
=
24 × 0, 0692
36, 4
= 0, 046
σ
2
2
=
24s
2
χ
2
0,05
=
24 × 0, 0692
13, 85
= 0, 12
Vˆa
.
y kho

ang tin cˆa
.
y l`a (0, 046; 0, 12).
3. B
`
AI T
ˆ
A
.
P
1. Mˆo
.
t m
˜
ˆau c´ac tro
.
ng l

u

o
.
ng t

u

ong
´

ung l`a 8,3; 10,6; 9,7; 8,8; 10,2 v`a 9,4 kg. X´ac ¯di
.
nh

u
´

oc l

u

o
.
ng khˆong chˆe
.
ch c

ua
a) trung b`ınh c

ua t

ˆong th

ˆe,
b) ph

u

ong sai c

ua t

ˆong th

ˆe.
2. Mˆo
.
t m
˜
ˆau ¯dˆo
.
¯do 5 ¯d

u
`

ong k´ınh c

ua qu

a c
`
ˆau l`a 6,33; 6,37; 6,36; 6,32 v`a 6,37cm. X´ac
¯di
.
nh

u
´

oc l

u

o
.
ng khˆong chˆe
.
ch c

ua trung b`ınh v`a ph

u

ong sai c

ua ¯d

u
`

ong k´ınh qu

a
c
`
ˆau.
3. D
¯

ˆe x´ac ¯di
.
nh ¯dˆo
.
ch´ınh x´ac c

ua mˆo
.
t chi
´
ˆec cˆan ta
.
khˆong c´o sai s
´
ˆo hˆe
.
th
´
ˆong, ng

u
`

oi ta
ti
´
ˆen h`anh 5 l
`
ˆan cˆan ¯dˆo
.
c lˆa
.
p (c`ung mˆo
.
t vˆa
.
t), k
´
ˆet qu

a nh

u sau:
94, 1 94, 8 96, 0 95, 2 kg
X´ac ¯di
.
nh

u
´

oc l

u

o
.
ng khˆong chˆe
.
ch c

ua ph

u

ong sai s
´
ˆo ¯do trong hai tr

u
`

ong h

o
.
p:
a) bi
´
ˆet kh
´
ˆoi l

u

o
.
ng vˆa
.
t cˆan l`a 95kg;
b) khˆong bi
´
ˆet kh
´
ˆoi l

u

o
.
ng vˆa
.
t cˆan.
4. D
¯

u
`

ong k´ınh c

ua mˆo
.
t m
˜
ˆau ng
˜
ˆau nhiˆen c

ua 200 viˆen bi ¯d

u

o
.
c s

an xu
´
ˆat b


oi mˆo
.
t m´ay
trong mˆo
.
t tu
`
ˆan c´o trung b`ınh 20,9mm v`a ¯dˆo
.
lˆe
.
ch tiˆeu chu

ˆan 1,07mm.

U
´

oc l

u

o
.
ng
trung b`ınh ¯d

u
`

ong k´ınh c

ua viˆen bi v
´

oi ¯dˆo
.
tin cˆa
.
y (a) 95%, (b) 99%.
82 Ch ’u ’ong 4.

U
´

oc l

u

ong tham s
´
ˆo c

ua ¯da
.
i l

u

ong ng
˜
ˆau nhiˆen
5. D
¯

ˆe kh

ao s´at s
´

uc b
`
ˆen chi
.
u l

u
.
c c

ua mˆo
.
t loa
.
i
´
ˆong cˆong nghiˆe
.
p ng

u
`

oi ta ti
´
ˆen h`anh ¯do
9
´
ˆong v`a thu ¯d

u

o
.
c c´ac s
´
ˆo liˆe
.
u sau
4500 6500 5000 5200 4800 4900 5125 6200 5375
T
`

u kinh nghiˆe
.
m ngh
`
ˆe nghiˆe
.
p ng

u
`

oi ta bi
´
ˆet r
`
˘
ang s
´

uc b
`
ˆen ¯d´o c´o phˆan ph
´
ˆoi chu

ˆan
v
´

oi ¯dˆo
.
lˆe
.
ch chu

ˆan σ = 300. X´ac ¯di
.
nh kho

ang tin cˆa
.
y 95% cho s
´

uc b
`
ˆen trung b`ınh
c

ua loa
.
i
´
ˆong trˆen.
6. Ta
.
i mˆo
.
t v`ung r
`

ung nguyˆen sinh, ng

u
`

oi ta ¯deo v`ong cho 1000 con chim. Sau mˆo
.
t
th
`

oi gian, b
´
˘
at la
.
i 200 con th`ı th
´
ˆay c´o 40 con c´o ¯deo v`ong. Th


u

u
´

oc l

u

o
.
ng s
´
ˆo chim
trong v`ung r
`

ung ¯d´o v
´

oi ¯dˆo
.
tin cˆa
.
y 99%.
7. Bi
´
ˆet t

y lˆe
.
n

ay m
`
ˆam c

ua mˆo
.
t loa
.
i ha
.
t gi
´
ˆong l`a 0,9. V
´

oi ¯dˆo
.
tin cˆa
.
y 0,95, n
´
ˆeu ta
mu
´
ˆon ¯dˆo
.
d`ai kho

ang tin cˆa
.
y c

ua t

y lˆe
.
n

ay m
`
ˆam khˆong v

u

o
.
t qu´a 0,02 th`ı c
`
ˆan ph

ai
gieo bao nhiˆeu ha
.
t?
8. K
´
ˆet qu

a quan s´at v
`
ˆe h`am l

u

o
.
ng vitamine C c

ua mˆo
.
t loa
.
i tr´ai cˆay cho


o b

ang sau:
H`am l

u

o
.
ng vitamine C (%) S
´
ˆo tr´ai
6 − 7 5
7 − 8 10
8 − 9 20
9 − 10 35
10 − 11 25
11 − 12 5
a) H˜ay

u
´

oc l

u

o
.
ng h`am l

u

o
.
ng vitamine C trung b`ınh trong mˆo
.
t tr´ai v
´

oi ¯dˆo
.
tin cˆa
.
y
95%.
b) Qui

u
´

oc nh
˜

ung tr´ai c´o h`am l

u

o
.
ng vitamine C trˆen 10% l`a tr´ai loa
.
i A.

U
´

oc l

u

o
.
ng
t

y lˆe
.
tr´ai loa
.
i A v
´

oi ¯dˆo
.
tin cˆa
.
y 90%.
c) Mu
´
ˆon ¯dˆo
.
ch´ınh x´ac khi

u
´

oc l

u

o
.
ng h`am l

u

o
.
ng vitamine C trung b`ınh l`a 0,1 v`a
¯dˆo
.
ch´ınh x´ac khi

u
´

oc l

u

o
.
ng t

y lˆe
.
tr´ai loa
.
i A l`a 5% v
´

oi c`ung ¯dˆo
.
tin cˆa
.
y 95% th`ı c
`
ˆan
quan s´at thˆem bao nhiˆeu tr´ai n
˜

ua? A
9. D
¯
o ¯d

u
`

ong k´ınh c

ua 100 chi ti
´
ˆet m´ay do mˆo
.
t phˆan x

u


ong s

an xu
´
ˆat, ta ¯d

u

o
.
c k
´
ˆet qu

a
cho


o b

ang sau:
D
¯

u
`

ong k´ınh (mm) S
´
ˆo chi ti
´
ˆet m´ay
9,85 8
9,90 12
9,95 20
10,00 30
10,05 14
10,10 10
10,15 6

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