Egzamin dla Aktuariuszy z 8 stycznia 2007 r.
Prawdopodobieństwo i Statystyka
Zadanie 1
(
)
( )
2
2
1
;
2
1
2
2
1
;
2
1
2
2
2
t
F
t
Y
X
X
P
t
T
P
Beta
Beta
≅
=
<
+
=
<
4
8
47
6
( )
( )
2
2
1
2
1
2
1
2
1
2
1
2
2
1
2
1
1
2
1
)
5
,
0
(
)
1
(
2
)
(
)
(
t
t
t
t
t
t
t
t
f
t
T
P
dt
d
−
Π
=
−
Π
=
−
Γ
Γ
=
=
<
−
−
−
Zadanie 2
Patrz zadanie 3 z 15.06.2002 (identyczne) – odpowiedź (E) prawidłowa
Zadanie 3
(
)
∑
∑
=
=
−
=
−
20
1
20
1
2
2
2
20
i
i
i
i
X
X
X
X
(
) (
)
∑
=
+
+
+
=
20
1
2
2
2
2
1
2
2
2
10
10
i
i
µ
σ
µ
σ
X
E
(
)
2
1
2
1
5
,
0
5
,
0
10
10
20
1
µ
µ
µ
µ
X
E
+
=
+
=
(
)
2
2
2
40
3
20
10
400
1
var
σ
σ
σ
X
=
+
=
2
2
1
1
µ
X
E
µ
X
E
=
=
( )
2
2
2
2
2
1
2
,
0
20
100
1
var
1
,
0
10
100
1
var
σ
σ
X
σ
σ
X
=
=
=
=
(
)
2
2
2
1
2
1
20
1
2
2
2
2
20
ˆ
X
X
X
X
b
X
X
a
σ
i
i
+
−
+
−
=
∑
=
[
]
=
−
+
+
+
+
+
+
+
−
+
+
=
2
1
2
2
2
2
1
2
2
1
2
2
2
1
2
2
2
2
1
2
2
2
2
,
0
1
,
0
2
1
4
1
4
1
40
3
20
10
10
30
ˆ
µ
µ
µ
σ
µ
σ
b
µ
µ
µ
µ
σ
µ
µ
σ
a
σ
E
(
)
(
)
(
)
=
−
−
+
+
−
+
+
−
+
+
+
−
=
b
a
µ
µ
b
a
a
µ
b
a
a
µ
b
b
a
a
σ
2
10
5
10
5
10
2
,
0
1
,
0
2
3
30
2
1
2
2
2
1
2
(
)
2
1
2
2
2
1
2
2
)
5
(
10
3
2
57
µ
µ
µ
µ
b
a
b
a
σ
−
+
+
+
+
=
Z tego:
27
5
,
27
1
1
10
3
2
27
0
5
−
=
=
→
=
+
=
+
b
a
b
a
b
a
Zadanie 4
≅
n
σ
µ
N
X
2
;
(
)
X
kw
n
σ
µ
σ
n
µ
X
P
n
σ
µ
X
P
zm.
1
,
0
.
2816
,
1
1
,
0
2816
,
1
2816
,
1
−
−
→
=
−
<
−
=
−
<
analogicznie:
n
σ
µ
2816
,
1
+
- kwantyl 0,9
i
x
X
m
dla
5
:
3
−
(
) (
) (
)
=
<
−
<
=
<
<
=
<
<
1
,
0
5
:
3
9
,
0
5
:
3
9
,
0
5
:
3
1
,
0
)
9
,
0
1
,
0
(
z
X
P
z
X
P
z
X
z
P
kw
m
kw
P
∑
∑
=
=
−
−
=
−
=
−
=
5
3
5
3
5
5
9
,
0
1
,
0
5
1
,
0
9
,
0
5
)
1
,
0
;
5
;
3
(
)
9
,
0
;
5
;
3
(
i
i
i
i
i
i
i
i
Q
Q
98288
,
0
1
,
0
9
,
0
1
,
0
5
9
,
0
1
,
0
10
9
,
0
1
,
0
9
,
0
5
1
,
0
9
,
0
10
5
4
2
3
5
4
2
3
=
−
⋅
−
⋅
−
+
⋅
+
⋅
=
Zadanie 5
1
2
σ
σ
>
(
)
∑
∑
=
−
−
−
=
→
−
=
=
Π
Π
Π
Π
∑
∑
∑
10
1
2
2
2
2
1
2
1
2
2
2
2
2
10
2
1
2
10
1
2
10
2
2
2
1
2
1
2
2
2
2
1
2
2
1
2
2
2
2
i
i
i
σ
i
X
σ
i
X
σ
i
X
σ
i
X
iX
STAT
σ
σ
σ
σ
i
X
e
σ
σ
e
i
σ
e
i
σ
i
i
i
i
)
1
(
)
1
,
0
(
2
2
2
χ
σ
i
X
N
σ
i
X
i
i
≅
→
≅
(
)
228
,
73
307
,
18
4
05
,
0
4
)
10
(
2
2
4
2
=
→
=
→
=
>
=
>
∑
=
t
t
t
χ
P
t
iX
P
i
σ
(
)
584
,
18
94
,
3
228
,
73
228
,
73
94
,
3
95
,
0
228
,
73
)
10
(
228
,
73
2
2
2
2
2
≈
>
→
>
→
≥
>
=
>
=
∑
σ
σ
σ
χ
P
iX
P
moc
i
Zadanie 6
( )
−
∈
⋅
=
wpp
0
)
4
;
4
(
4
2
1
8
i
X
LICZ
−
∈
=
wpp
0
)
2
;
2
(
4
1
8
i
X
MIAN
−
∈
=
wpp
0
)
2
;
2
(
256
1
i
X
MIAN
LICZ
1
,
0
0
256
1
0
1
0
=
→
=
>
γ
f
f
P
{
}
)
2
;
2
(
X
i
lub
odrzucamy
0,1
z
wtedy
)
2
;
2
(
:
i
−
∉
∃
=
−
∈
∀
γ
X
i
K
i
{
}
0,9
z
i
)
2
;
2
(
:
=
−
∈
′
γ
x
K
( )
0035
,
0
9
,
0
2
1
8
1
≈
⋅
=
′
K
P
H
Zadanie 7
)
1
,
0
(
∈
t
(
) (
) (
)
(
)
2
2
2
2
)
,
min(
=
+
<
=
+
<
+
=
+
<
=
=
+
<
Y
X
t
X
P
Y
X
t
Y
P
Y
X
t
X
P
Y
X
t
Y
X
P
Γ
≅
+
λ
Y
X
1
,
2
(
)
(
)
t
e
λ
dx
e
e
λ
Y
X
t
X
P
λ
t
λ
x
λ
x
=
=
=
+
<
−
−
−
−
∫
2
2
0
2
2
2
1
2
f(t)=1 dla
12
1
var
)
1
,
0
(
)
1
,
0
(
=
→
≅
∈
J
t
Zadanie 8
(
)
(
)
[
]
5
,
0
var
var
var
,
cov
N
N
N
N
N
N
S
T
S
T
S
T
−
−
+
=
−
+
−
=
2
2
1
2
1
var
µ
q
q
µ
q
q
r
T
N
( )
−
+
−
=
−
+
−
=
2
2
2
2
2
1
2
1
)
(
1
1
var
µ
p
q
q
µ
p
q
q
r
IX
E
q
q
IX
E
q
q
r
S
N
µ
p
EIEX
IX
E
=
=
)
(
(
)
2
2
2
2
2
2 µ
p
EX
EI
X
I
E
=
=
(
)
+
−
+
+
−
=
+
)
(
1
)
(
1
var
2
2
IX
I
E
q
q
IX
X
E
q
q
r
S
T
N
N
(
)
)
2
1
(
2
2
1
)
(
2
2
2
2
p
p
µ
I
I
E
EX
IX
X
E
+
+
=
+
+
=
+
)
1
(
)
(
p
µ
IX
X
E
+
=
+
(
)
−
+
−
−
+
+
−
+
+
+
−
=
2
2
2
2
2
2
2
1
2
1
2
1
1
2
4
2
1
{
2
1
µ
q
q
µ
q
rq
p
p
µ
q
q
µ
p
µ
p
µ
q
q
r
ODP
=
−
+
−
−
}
1
2
1
2
2
2
µ
p
q
q
µ
p
q
rq
(
)
(
)
=
−
−
+
+
−
+
−
−
+
−
=
2
2
2
2
2
2
2
2
2
2
2
2
2
2
)
1
(
2
2
6
2
1
2
1
µ
p
µ
p
µ
µ
p
µ
q
q
r
µ
p
µ
µ
p
µ
q
q
r
=
−
+
−
=
−
+
−
=
2
2
2
2
2
2
1
2
1
2
)
1
(
4
1
2
1
µ
p
q
q
µ
p
q
rq
µ
p
q
q
r
µ
p
q
q
r
2
2
2
2
2
2
2
2
2
2
2
2
)
1
(
)
2
(
)
1
(
2
2
)
1
(
)
1
(
2
q
q
rq
µ
p
q
µ
p
rq
r
q
µ
p
rq
µ
p
q
µ
p
rq
q
rq
µ
p
−
−
=
−
+
−
=
−
+
−
=
Zadanie 9
( )
(
)
1
10
3
3
1
+
+
Π
=
θ
i
θ
X
θ
L
( )
(
)
∑
=
+
+
−
=
10
1
3
ln
)
1
(
3
ln
10
1
ln
i
i
θ
X
θ
θ
L
(
)
(
)
(
)
∑
∑
=
+
−
+
=
+
−
+
=
∂
∂
0
3
ln
10
3
ln
10
3
ln
3
3
ln
3
3
10
i
i
θ
θ
θ
X
θ
X
θ
θ
θ
Z tego:
(
)
2
3
ln
3
ln
10
10
ˆ
=
+
−
−
=
∑
i
X
θ
(
)
∑
−
=
+
−
10
3
ln
2
3
ln
20
i
X
(
)
}
(
)
(
)
∑
∑
∑
=
=
+
=
+
→
+
=
+
+
→
+
=
+
8
1
8
1
6
po
5
3
ln
6
3
ln
5
3
ln
10
3
ln
9
ln
2
2
10
3
ln
20
3
ln
i
i
i
i
i
X
X
X
(
) (
)
θ
θ
i
i
Y
P
X
P
3
1
6
3
3
6
6
=
+
=
>
=
=
( )
(
)
∏
=
+
+
=
8
1
1
8
2
3
1
3
3
1
i
θ
i
θ
θ
X
θ
L
( )
(
)
∑
+
+
−
+
−
=
i
θ
θ
X
θ
θ
L
3
ln
)
1
(
3
ln
8
3
ln
2
ln
(
)
(
)
∑
→
+
−
+
+
−
=
∂
∂
i
θ
θ
θ
X
θ
θ
θ
3
ln
3
ln
3
3
3
8
3
ln
2
(
)
6
,
1
5
8
3
ln
4
5
3
ln
4
8
9
ln
2
5
3
ln
10
3
ln
6
8
3
ln
3
ln
6
8
ˆ
=
=
−
+
=
+
−
−
−
=
+
−
−
=
→
∑
i
X
θ
Zadanie 10
W – dwie kule tego samego koloru
(
)
)
(
)
(
IIIb
P
W
P
W
IIIb
P
ODP
=
(
) (
) (
) (
) (
)
W
c
P
c
IIIb
P
W
b
P
b
IIIb
P
W
IIIb
P
2
2
2
2
+
=
(
) (
) ( ) (
) ( )
( )
( )
b
B
P
b
A
P
b
B
P
B
b
IIIb
P
b
A
P
A
b
IIIb
P
b
IIIb
P
2
7
4
2
7
5
2
,
2
2
,
2
2
+
=
+
=
(
) (
) ( ) (
) ( )
( )
( )
c
B
P
c
A
P
c
B
P
B
c
IIIb
P
c
A
P
A
c
IIIb
P
c
IIIb
P
2
7
2
2
7
3
2
,
2
2
,
2
2
+
=
+
=
( ) ( )
25
,
0
5
,
0
3
,
0
5
,
0
1
,
0
5
,
0
1
,
0
)
2
(
)
(
2
2
=
⋅
+
⋅
⋅
=
=
b
P
A
P
A
b
P
b
A
P
( )
75
,
0
2
=
b
B
P
( ) ( )
75
,
0
5
,
0
1
,
0
5
,
0
3
,
0
5
,
0
3
,
0
)
2
(
)
(
2
2
=
⋅
+
⋅
⋅
=
=
c
P
A
P
A
c
P
c
A
P
( )
25
,
0
2
=
c
B
P
(
)
28
17
28
12
28
5
4
3
7
4
4
1
7
5
2
=
+
=
+
=
b
IIIb
P
(
)
28
11
4
1
7
2
4
3
7
3
2
=
+
=
c
IIIb
P
(
)
2
1
5
,
0
1
,
0
5
,
0
3
,
0
5
,
0
3
,
0
5
,
0
1
,
0
5
,
0
3
,
0
5
,
0
1
,
0
)
(
)
2
(
2
=
⋅
+
⋅
+
⋅
+
⋅
⋅
+
⋅
=
=
W
P
b
P
W
b
P
(
)
2
1
)
(
)
2
(
2
=
=
W
P
c
P
W
c
P
(
)
2
1
2
1
28
11
2
1
28
17
=
+
=
W
IIIb
P
(
)
4
,
0
5
,
0
3
,
0
5
,
0
1
,
0
2
)
2
(
)
2
(
)
(
=
⋅
+
⋅
=
+
=
c
P
b
P
W
P
(
)
(
)
)
,
(
,
)
(
)
(
c
b
P
c
b
IIIb
P
W
P
W
IIIb
P
IIIb
P
+
=
(
) (
) (
) (
) (
)
(
)
(
)
c
b
B
P
c
b
A
P
c
b
B
P
B
c
b
IIIb
P
c
b
A
P
A
c
b
IIIb
P
c
b
IIIb
P
,
7
3
,
7
4
,
,
,
,
,
,
,
+
=
+
=
(
) (
)
2
1
5
,
0
6
,
0
2
5
,
0
6
,
0
)
,
(
)
(
,
,
=
⋅
⋅
⋅
=
=
c
b
P
A
P
A
c
b
P
c
b
A
P
(
) (
)
2
1
5
,
0
6
,
0
2
5
,
0
6
,
0
)
,
(
)
(
,
,
=
⋅
⋅
⋅
=
=
c
b
P
B
P
B
c
b
P
c
b
B
P
(
)
2
1
2
1
7
3
2
1
7
4
,
=
+
=
c
b
IIIb
P
2
1
6
,
0
2
1
4
,
0
2
1
)
(
=
+
=
IIIb
P
(
)
(
)
6
,
0
5
,
0
6
,
0
5
,
0
6
,
0
)
(
,
)
(
,
)
,
(
=
⋅
+
⋅
=
+
=
B
P
B
c
b
P
A
P
A
c
b
P
c
b
P
4
,
0
5
,
0
4
,
0
5
,
0
=
⋅
=
ODP