| SOA真题November2005ExamM |
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Exam M Fall 2005
FINAL ANSWER KEY
Question # Answer Question # Answer
1 C 21 E
2 C 22 B
3 C 23 E
4 D 24 E
5 C 25 C
6 B 26 E
7 A 27 E
8 D 28 D
9 B 29 A
10 A 30 D
11 A 31 A
12 A 32 A
13 D 33 B
14 C 34 C
15 A 35 A
16 D 36 A
17 D 37 C
18 D 38 C
19 B 39 E
20 B 40 B
Exam M: Fall 2005 -1- GO ON TO NEXT PAGE
**BEGINNING OF EXAMINATION**
1. For a special whole life insurance on (x), you are given:
(i) Z is the present value random variable for this insurance.
(ii) Death benefits are paid at the moment of death.
(iii) ( ) 0.02, 0 xt t %26micro; = ≥
(iv) 0.08 δ=
(v) 0.03, 0 t
tb e t = ≥
Calculate ( ) Var Z .
(A) 0.075
(B) 0.080
(C) 0.085
(D) 0.090
(E) 0.095
Exam M: Fall 2005 -2- GO ON TO NEXT PAGE
2. For a whole life insurance of 1 on (x), you are given:
(i) Benefits are payable at the moment of death.
(ii) Level premiums are payable at the beginning of each year.
(iii) Deaths are uniformly distributed over each year of age.
(iv) 0.10 i =
(v) 8 x a = %26#1048581;%26#1048581;
(vi) 10 6 x a + = %26#1048581;%26#1048581;
Calculate the 10th year terminal benefit reserve for this insurance.
(A) 0.18
(B) 0.25
(C) 0.26
(D) 0.27
(E) 0.30
Exam M: Fall 2005 -3- GO ON TO NEXT PAGE
3. A special whole life insurance of 100,000 payable at the moment of death of (x) includes a
double indemnity provision. This provision pays during the first ten years an additional
benefit of 100,000 at the moment of death for death by accidental means.
You are given:
(i) %26micro; τ
x t t b gb g= ≥ 0 001 0 . ,
(ii) %26micro;x t t 1 0 0002 0 b gb g= ≥ . , , where %26micro;x
1 b g is the force of decrement due to death by
accidental means.
(iii) δ= 006 .
Calculate the single benefit premium for this insurance.
(A) 1640
(B) 1710
(C) 1790
(D) 1870
(E) 1970
Exam M: Fall 2005 -4- GO ON TO NEXT PAGE
4. Kevin and Kira are modeling the future lifetime of (60).
(i) Kevin uses a double decrement model:
x ( )
x l τ ( ) 1
x d ( ) 2
x d
60 1000 120 80
61 800 160 80
62 560 %26#8722; %26#8722;
(ii) Kira uses a non-homogeneous Markov model:
(a) The states are 0 (alive), 1 (death due to cause 1), 2 (death due to cause 2).
(b) 60 Q is the transition matrix from age 60 to 61; 61 Q is the transition matrix
from age 61 to 62.
(iii) The two models produce equal probabilities of decrement.
Calculate 61 Q .
(A)
1.00 0.12 0.08
0 1.00 0
0 0 1.00
%26#9115; %26#9118;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9117; %26#9120;
(B)
0.80 0.12 0.08
0.56 0.16 0.08
0 0 1.00
%26#9115; %26#9118;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9117; %26#9120;
(C)
0.76 0.16 0.08
0 1.00 0
0 0 1.00
%26#9115; %26#9118;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9117; %26#9120;
(D)
0.70 0.20 0.10
0 1.00 0
0 0 1.00
%26#9115; %26#9118;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9117; %26#9120;
(E)
0.60 0.28 0.12
0 1.00 0
0 0 1.00
%26#9115; %26#9118;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9117; %26#9120;
Exam M: Fall 2005 -5- GO ON TO NEXT PAGE
5. A certain species of flower has three states: sustainable, endangered and extinct. Transitions
between states are modeled as a non-homogeneous Markov chain with transition matrices i Q
as follows:
1
Endangered Sustainable Extinct
Sustainable 0.85 0.15 0
0 0.7 0.3 Endangered
Extinct 0 0 1
Q
%26#9115; %26#9118;
%26#9116; %26#9119;
%26#9116; %26#9119; = %26#9116; %26#9119;
%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9117; %26#9120;
2
0.9 0.1 0
0.1 0.7 0.2
0 0 1
Q
%26#9115; %26#9118;
%26#9116; %26#9119; =%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9117; %26#9120;
3
0.95 0.05 0
0.2 0.7 0.1
0 0 1
Q
%26#9115; %26#9118;
%26#9116; %26#9119; =%26#9116; %26#9119;
%26#9116; %26#9119;
%26#9117; %26#9120;
0.95 0.05 0
0.5 0.5 0 , 4,5,...
0 0 1
iQ i
%26#9115; %26#9118;
%26#9116; %26#9119; = = %26#9116; %26#9119;
%26#9116; %26#9119;
%26#9117; %26#9120;
Calculate the probability that a species endangered at the start of year 1 will ever become
extinct.
(A) 0.45
(B) 0.47
(C) 0.49
(D) 0.51
(E) 0.53
Exam M: Fall 2005 -6- GO ON TO NEXT PAGE
6. For a special 3-year term insurance:
(i) Insureds may be in one of three states at the beginning of each year: active, disabled,
or dead. All insureds are initially active. The annual transition probabilities are as
follows:
Active Disabled Dead
Active 0.8 0.1 0.1
Disabled 0.1 0.7 0.2
Dead 0.0 0.0 1.0
(ii) A 100,000 benefit is payable at the end of the year of death whether the insured was
active or disabled.
(iii) Premiums are paid at the beginning of each year when active. Insureds do not pay
any annual premiums when they are disabled.
(iv) d = 0.10
Calculate the level annual benefit premium for this insurance.
(A) 9,000
(B) 10,700
(C) 11,800
(D) 13,200
(E) 20,800
Exam M: Fall 2005 -7- GO ON TO NEXT PAGE
7. Customers arrive at a bank according to a Poisson process at the rate of 100 per hour. 20%
of them make only a deposit, 30% make only a withdrawal and the remaining 50% are there
only to complain. Deposit amounts are distributed with mean 8000 and standard deviation
1000. Withdrawal amounts have mean 5000 and standard deviation 2000.
The number of customers and their activities are mutually independent.
Using the normal approximation, calculate the probability that for an 8-hour day the total
withdrawals of the bank will exceed the total deposits.
(A) 0.27
(B) 0.30
(C) 0.33
(D) 0.36
(E) 0.39
Exam M: Fall 2005 -8- GO ON TO NEXT PAGE
8. A Mars probe has two batteries. Once a battery is activated, its future lifetime is exponential
with mean 1 year.
The first battery is activated when the probe lands on Mars. The second battery is activated
when the first fails.
Battery lifetimes after activation are independent.
The probe transmits data until both batteries have failed.
Calculate the probability that the probe is transmitting data three years after landing.
(A) 0.05
(B) 0.10
(C) 0.15
(D) 0.20
(E) 0.25
Exam M: Fall 2005 -9- GO ON TO NEXT PAGE
9. For a special fully discrete 30-payment whole life insurance on (45), you are given:
(i) The death benefit of 1000 is payable at the end of the year of death.
(ii) The benefit premium for this insurance is equal to 45 1000P for the first 15 years
followed by an increased level annual premium of π for the remaining 15 years.
(iii) Mortality follows the Illustrative Life Table.
(iv) 0.06 i =
Calculate π.
(A) 16.8
(B) 17.3
(C) 17.8
(D) 18.3
(E) 18.8
Exam M: Fall 2005 -10- GO ON TO NEXT PAGE
10. For a special fully discrete 2-year endowment insurance on (x):
(i) The pure endowment is 2000.
(ii) The death benefit for year k is ( ) 1000k plus the benefit reserve at the end of year k,
1, 2 k = .
(iii) π is the level annual benefit premium.
(iv) i = 0.08
(v) 1 0.9, 1, 2 x k p k + %26#8722; = =
Calculate π.
(A) 1027
(B) 1047
(C) 1067
(D) 1087
(E) 1107
Exam M: Fall 2005 -11- GO ON TO NEXT PAGE
11. For a group of 250 individuals age x, you are given:
(i) The future lifetimes are independent.
(ii) Each individual is paid 500 at the beginning of each year, if living.
(iii) 0.369131 x A =
(iv) 2 0.1774113 x A =
(v) 0.06 i =
Using the normal approximation, calculate the size of the fund needed at inception in order to
be 90% certain of having enough money to pay the life annuities.
(A) 1.43 million
(B) 1.53 million
(C) 1.63 million
(D) 1.73 million
(E) 1.83 million
Exam M: Fall 2005 -12- GO ON TO NEXT PAGE
12. For a double decrement table, you are given:
Age ( )
x l τ ( ) 1
x d ( ) 2
x d
40 1000 60 55
41 %26#8722; %26#8722; 70
42 750 %26#8722; %26#8722;
Each decrement is uniformly distributed over each year of age in the double decrement table.
Calculate ( ) 1
41 q′ .
(A) 0.077
(B) 0.078
(C) 0.079
(D) 0.080
(E) 0.081
Exam M: Fall 2005 -13- GO ON TO NEXT PAGE
13. The actuarial department for the SharpPoint Corporation models the lifetime of pencil
sharpeners from purchase using a generalized DeMoivre model with ( ) ( ) 1 / s x x α ω = %26#8722; , for
0 α%26gt; and 0 x ω ≤ ≤ .
A senior actuary examining mortality tables for pencil sharpeners has determined that the
original value of α must change. You are given:
(i) The new complete expectation of life at purchase is half what it was previously.
(ii) The new force of mortality for pencil sharpeners is 2.25 times the previous force of
mortality for all durations.
(iii) ω remains the same.
Calculate the original value of α.
(A) 1
(B) 2
(C) 3
(D) 4
(E) 5
Exam M: Fall 2005 -14- GO ON TO NEXT PAGE
14. You are given:
(i) T is the future lifetime random variable.
(ii) ( ) t %26micro; %26micro; = , 0 t ≥
(iii) [ ] Var 100 T = .
Calculate [ ] E 10 T ∧ .
(A) 2.6
(B) 5.4
(C) 6.3
(D) 9.5
(E) 10.0
Exam M: Fall 2005 -15- GO ON TO NEXT PAGE
15. For a fully discrete 15-payment whole life insurance of 100,000 on (x), you are given:
(i) The expense-loaded level annual premium using the equivalence principle is 4669.95.
(ii) 100,000 51,481.97 x A =
(iii) :15 11.35 x a = %26#1048581;%26#1048581;
(iv) 0.02913 d =
(v) Expenses are incurred at the beginning of the year.
(vi) Percent of premium expenses are 10% in the first ye[1] [2] [3] 下一页 |
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