Egzamin dla Aktuariuszy z 18 stycznia 1997 r.
Prawdopodobieństwo i Statystyka
Zadanie 1
(
)
( )
3
1
1
1
=
∩
A
P
C
A
P
(
)
( )
2
1
2
2
=
∩
A
P
C
A
P
(
)
6
1
1
=
∩
C
A
P
(
)
4
1
2
=
∩
C
A
P
(
)
4
1
2
1
2
1
2
1
=
=
∩
A
A
P
(
)
0
2
1
=
∩
∩
C
A
A
P
Z tego wynika:
Z obrazka:
(
)
(
) (
)
( ) ( ) (
)
2
1
2
1
2
1
2
1
A
A
P
A
P
A
P
C
A
P
C
A
P
A
A
C
P
∩
−
+
∩
+
∩
=
∪
)
(
9
5
18
10
3
4
24
6
4
4
1
2
1
2
1
4
1
6
1
C
→
=
=
+
=
−
+
+
=
Zadanie 2
λ
λ
λ
λ
e
e
λ
e
λ
e
−
−
−
−
=
+
:
2
9
8
2
2
9
4
1
λ
λ
=
+
15
225
0
9
9
4
2
=
∆
=
∆
=
−
−
λ
λ
3
8
24
8
15
9
8
15
9
1
=
=
+
=
∉
−
=
λ
D
λ
Zadanie 3
7
4
;
2
1
;
7
4
2
2
=
=
=
=
EX
EY
EY
EX
[
]
20
2
1
880
7
4
735
−
=
⋅
−
⋅
=
−
∑
∑
i
i
Y
X
E
[
]
400
4
1
2
1
880
49
16
7
4
735
var
=
−
+
−
=
−
∑
∑
i
i
Y
X
84
,
0
)
1
(
20
20
)
0
(
≈
Φ
=
<
=
<
Y
P
X
P
Zadanie 4
(
) (
) (
) (
)
(
) (
)
(
)
1
1
1
3
1
3
1
3
2
3
1
2
1
0
3
2
2
1
−
=
→
+
=
+
+
+
=
=
=
X
X
E
X
X
E
X
X
E
X
X
E
X
X
E
X
X
E
X
X
E
4
8
47
6
(
)
0
a
ale
1
,
1
3
3
1
<
+
=
→
−
=
b
aX
X
X
X
ρ
0
0
0
3
=
→
=
+
⋅
=
b
b
a
EX
1
...
var
2
3
−
=
→
=
a
a
X
z tego:
(
)
1
3
1
=
−
=
X
X
P
Zadanie 5
}
[ ]
∫
∫
∫
∞
−
−
=
+
=
+
=
+
=
+
=
+
≤
=
5
,
0
0
0
5
,
0
0
2
5
,
0
0
3
5
,
0
0
2
3
4
1
4
12
1
1
4
1
8
1
3
2
1
3
2
2
1
2
2
1
x
x
ye
x
x
EY
X
X
E
ODP
y
nzl
Zadanie 6
10
)
1
(
1
=
−
≅
−
−
=
n
n
t
n
S
µ
X
t
(
)
)
975
,
0
.
(
2622
,
2
95
,
0
.
1
st
kwantyl
u
t
P
=
=
<
2622
,
2
3
2622
,
2
<
−
<
−
S
µ
X
.......
L
S
X
µ
S
X
u
=
⋅
−
>
>
⋅
+
=
3
2622
,
2
3
2622
,
2
3
2622
,
2
262
,
4
,
2622
,
2
2
3
524
,
4
262
,
4
3
2622
,
2
262
,
0
3
2622
,
2
S
X
S
S
X
S
X
⋅
−
=
⋅
⋅
=
→
=
⋅
+
−
=
⋅
−
(
)
-2,8215
0,99
kwantyl
01
,
0
.
2
2
=
=
<
u
t
P
8215
,
2
3
−
<
−
S
µ
X
3
8215
,
2
S
µ
X
⋅
−
<
−
821
,
4
3
8215
,
2
3
2622
,
2
262
,
4
3
8215
,
2
≈
⋅
+
⋅
−
=
=
⋅
+
>
S
S
W
S
X
µ
Zadanie 7
≅
−
=
>
−
=
≤
−
µ
n
wykl
e
t
P
t
P
µ
tn
1
)
(min
1
)
(min
µ
n
µ
n
nE
µ
E
µ
µ
n
n
µ
E
=
=
=
=
=
min
ˆ
1
ˆ
2
1
oba nieobciążone
2
2
2
2
2
2
2
2
1
ˆ
var
1
ˆ
var
µ
n
µ
n
µ
n
µ
µ
n
n
µ
=
=
=
=
ponieważ n>1
)
(
2
2
E
µ
n
µ
→
<
→
Zadanie 8
(
)
1
1
)
(
+
−
−
+
=
α
x
x
α
e
e
α
x
f
(
)
∏
=
+
−
−
+
=
n
i
α
x
x
n
i
i
e
e
α
L
1
1
1
(
)
(
)
(
)
∑
∑
−
+
−
−
+
+
−
−
+
=
+
+
=
i
i
i
x
i
α
x
x
e
α
x
α
n
e
e
α
n
L
1
ln
)
1
(
ln
1
ln
ln
ln
1
(
)
∑
=
+
−
=
∂
∂
−
0
1
ln
i
x
e
α
n
α
(
)
∑
=
+
−
−
0
1
ln
i
x
e
α
n
(
)
∑
=
−
+
=
n
i
x
i
e
n
α
1
1
ln
ˆ
Zadanie 9
>
=
Κ
t
x
1
5
4
(
)
α
t
x
P
H
=
>
4
5
0
α
t
t
X
P
t
H
=
−
=
=
>
∫
4
1
1
5
4
5
1
5
25
,
0
0
α
t
−
=
1
5
4
1
4
)
1
(
5
α
t
−
=
4
)
1
(
5
α
t
−
=
[ ]
∫
−
−
=
−
=
=
=
>
=
1
5
4
5
4
4
5
1
5
5
4
4
25
,
0
25
,
0
1
5
)
1
(
5
1
5
1
5
5
t
t
H
α
t
x
x
t
X
P
moc
Zadanie 10
6
1
,
12
5
,
12
5
36
2
6
3
2
1
=
=
=
=
p
p
p
75
,
1
6
1
120
6
1
120
25
12
5
120
12
5
120
50
12
5
120
12
5
120
45
2
2
2
2
=
⋅
⋅
−
+
⋅
⋅
−
+
⋅
⋅
−
=
p
χ
10,597
0,005
4,605
0,1
5,991
0,05
9,21
kw
0,01
dla
=
=
=
=
α
α
α
α
>
2
p
χ
wszystkich
→
nie prowadzi do odrzucenia w żadnym
wypadku czyli prawidłowa jest odpowiedź (D)