background image

Egzamin dla Aktuariuszy z 13 kwietnia 2002 r. 
 
Prawdopodobieństwo i Statystyka 
 
Zadanie 1 
 

=

=





































=

!

13

!

13

!

26

!

26

!

13

!

39

!

39

!

13

!

52

!

9

!

9

!

18

!

18

!

9

!

27

!

27

!

9

!

36

!

4

!

4

!

8

!

8

!

4

!

12

!

12

!

4

!

16

13

26

13

39

13

52

9

18

9

27

9

36

4

8

4

12

4

16

p

 









=





=

=

16

52

4

13

!

52

!

36

!

16

4

13

!

52

)

!

13

(

)

!

9

(

)

!

4

(

!

36

!

16

4

4

4

4

4

 

 
Zadanie 2 
 

x

p

x

p

x

p

C

B

A

+

=

+

=

+

=

60

30

,

60

20

,

60

10

 

x

EN

x

EN

x

EN

C

B

A

+

=

+

=

+

=

60

210

,

60

140

,

60

70

 

2

2

2

)

60

(

)

30

(

210

var

,

)

60

(

)

40

(

140

var

,

)

60

(

)

50

(

70

var

x

x

N

x

x

N

x

x

N

C

B

A

+

+

=

+

+

=

+

+

=

 

(

)

(

)

(

)

(

)

B

A

B

A

B

A

B

A

B

A

B

A

EN

EN

N

N

N

N

EN

EN

N

N

N

N

E

+

+

=

+

=

2

1

var

var

var

,

cov

 

(

)

(

)

(

)

C

A

C

A

C

A

C

A

EN

EN

N

N

N

N

N

N

E

+

+

=

5

,

0

var

var

var

 

(

)

(

)

(

)

C

B

C

B

C

B

C

B

EN

EN

N

N

N

N

N

N

E

+

+

=

5

,

0

var

var

var

 

(

)

(

)

(

)

2

2

2

)

60

(

)

10

(

50

7

var

,

)

60

(

)

20

(

40

7

var

,

)

60

(

)

30

(

30

7

var

x

x

N

N

x

x

N

N

x

x

N

N

C

B

C

A

B

A

+

+

=

+

+

+

=

+

+

+

=

+

 

war: 

(

)

(

)(

)

C

B

B

A

C

B

B

C

A

B

A

EN

EN

EN

EN

N

N

N

N

N

N

N

E

+

+

=

+

+

+

2

 

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

2

2

2

2

2

2

2

)

60

(

)

30

(

210

)

60

(

)

50

(

70

)

60

(

)

20

(

280

5

,

0

)

60

(

140

70

)

60

(

)

40

(

140

)

60

(

)

50

(

70

)

60

(

)

30

(

210

5

,

0

x

x

x

x

x

x

x

x

x

x

x

x

x

2

2

2

2

2

2

2

2

2

)

60

(

350

210

)

60

(

210

140

)

60

(

)

30

(

210

)

60

(

)

40

(

140

)

60

(

)

10

(

350

5

,

0

)

60

(

140

)

60

(

)

40

(

140

)

60

(

210

70

x

x

x

x

x

x

x

x

x

x

x

x

+

=

+

+

+

+

+

+

+

+

+

+

+

+

+

+

+

0,5(-2800)+9800+0,5(-4200)+14700+5600+140x+19600+0,5(-8400)+29400=73500 
140x=2100 
x=15 
 
Zadanie 3 
 

(

)

(

)

=

=

=

1

1

)

(

,...,

max

n

n

n

N

P

X

X

E

EM

 

 

background image

1

max

+

=

n

n

E

 

=

=

=

=

=

=

=

=

+

=

+

=

1

2

2

2

1

1

1

!

)!

1

(

)!

1

(

1

1

!

1

n

t

t

t

λ

t

λ

t

λ

t

λ

n

e

t

λ

e

t

λ

e

t

λ

t

t

t

n

e

n

λ

n

n

ODP

 

(

)

λ

e

e

λ

e

λ

e

e

λ

e

λ

e

λ

λ

λ

λ

λ

λ

λ

=

+

+

=

=

1

1

1

1

1

1

1

 

 
Zadanie 4 
 
Z CTG 
 

)

1

,

0

(

N

n

σ

mn

X

i

 

(

)

n

σ

N

mn

X

i

2

,

0

 

(

)

n

σ

mn

N

X

i

2

,

 

(

)

2

σ

n

m

N

n

X

i

 

=

=

µ

K

S

Y

I

µ

X

I

Y

n

n

i

i

i

i

i

,

 

(

)

µ

p

X

I

E

i

i

=

 

( )

µ

p

I

µ

E

i

=

 

0

=

i

EY

 

(

) (

)

p

σ

p

µ

µ

p

µ

µ

σ

p

I

µ

X

I

µ

X

I

E

EY

i

i

i

i

i

i

2

2

2

2

2

2

2

2

2

2

2

2

=

+

+

=

+

=

 

(

)

p

σ

N

n

µ

K

S

n

n

2

,

0

 

 
Zadanie 5 
 

(

)

(

)

(

)

(

)

=

=

<

=

+

=

<

=

=

X

W

P

W

W

X

W

W

E

X

W

P

W

W

X

W

W

E

ODP

2

1

2

2

1

1

2

1

1

1

,

,

 

(

)

=

>

+

=

4

4

4

4

8

4

4

4

4

7

6

A

X

W

W

W

W

E

X

2

2

1

1

,

2

1

 

(

)

λ

x

e

e

λ

xe

X

W

P

e

λ

t

A

x

λ

x

λ

x

λ

x

t

λ

1

1

1

+

=

+

=

>

=

 

λ

X

λ

X

X

ODP

2

1

1

2

1

+

=





+

+

=

 

 
 
 
 
 
 

background image

Zadanie 6 
 

4

4

4

4

8

4

4

4

4

7

6

A

n

m

m

n

n

m

n

X

X

X

n

m

X

)

(

...

1

+

+

+

=

+

 

LICZNIK nzl od 

m

X

n

m

 

LICZNIK nzl od A (oczywiste) 
Z tego: 
LICZNIK nzl od 

n

X

 nzl od MIANOWNIKA, z tego LICZNIK nzl od MIANOWNIKA 

Z tego: 
Ma rozkład F(m-1,n-1) 

Czyli 

1

1

=

n

m

Er

 

 
Zadanie 7 
 

2

1

var

λ

X

=

 

Γ

)

,

(

λ

n

X

i

 

( )

(

)

2

2

2

2

2

2

1

)

1

(

1

λ

n

n

λ

n

n

n

n

X

E

X

E

i

+

=

+

=

=

 

1

1

)

1

(

2

2

+

=

=

+

n

n

c

λ

λ

n

n

c

 

 
Zadanie 8 
 

(

)

2

10

1

,

cov

σ

X

X

j

i

=

 

(

)

=

+

=







+

+

+

2

2

2

100

1

10

1

9900

100

,

100

,

cov

2

100

2

100

;

100

...

σ

σ

µ

N

X

X

σ

µ

N

X

X

j

i

 

(

)

2

1090

,

100

σ

µ

N

=

 

2

2

100

1090

,

σ

µ

N

X

 

(

)

45

,

0

1090

6

,

19

1090

6

,

19

...

96

,

1

1090

100

...

10

96

,

1





=





=

Y

P

σ

µ

X

P

σ

P

 

 
Zadanie 9 
 

01

,

0

)

1

(

1

)

101

(

101

2

2

0

=

>

+

+

t

x

x

P

 

 

background image

(

)

(

)

=

+

>

+

=

+

>

+

=



>

+

+

t

x

x

P

t

x

x

P

t

x

x

P

)

101

(

101

)

1

(

)

101

(

101

)

1

(

)

101

(

101

)

1

(

2

2

2

2

0

 

(

)

(

)

01

,

0

101

101

101

1

1

01

,

0

101

101

101

101

101

101

0

=

+

=



>

=

>

=

t

t

t

t

x

P

t

t

x

P

01

,

0

100

101

=

t

t

 

t

t

=

101

 

101

2

=

t

 

2

101

=

t

 

4

101

=

t

 

moc: 

505

,

0

200

101

99

101

101

101

101

2

101

101

101

2

101

101

1

=

=

+

=

+

=





>

A

x

P

A

4

4 3

4

4 2

1

 

 
Zadanie 10 
 

(

)

(

)

(

)

43

42

1

4

4

8

4

4

7

6

5

1

1

5

1

1

2

1

1

2

lim

=

=

=

=

n

n

n

X

P

X

P

X

X

P

 

rozkład stacjonarny: 

Π

=

Π

+

Π

Π

=

Π

Π

=

Π

+

Π

+

Π

3

3

2

2

1

1

3

2

1

3

2

3

2

2

1

3

1

3

1

2

1

 

Rozwiązujemy układ z warunkiem: 

1

3

2

1

=

Π

+

Π

+

Π

 i mamy: 

5

2

,

5

1

,

5

2

3

2

1

=

Π

=

Π

=

Π

 

(

)

(

)

(

)

(

)

(

)

(

)

(

)

=

=

=

=

+

=

=

=

+

=

=

=

=

=

3

3

1

2

2

1

1

1

1

1

0

0

1

0

0

1

0

0

1

1

X

P

X

X

P

X

P

X

X

P

X

P

X

X

P

X

P

36

13

36

6

4

3

6

1

9

1

12

1

2

1

3

1

3

1

3

1

6

1

2

1

=

+

+

=

+

+

=

+

+

=