mc course3
DESCRIPTION
mc courceTRANSCRIPT
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**BEGINNING OF EXAMINATION**
COURSE/EXAM 3: MAY 2000 - 1 - GO ON TO NEXT PAGE
1. Given:
(i) eo0 25=
(ii) l x xx = - w w, 0
(iii) T xb g is the future lifetime random variable.
Calculate Var T 10b g .
(A) 65
(B) 93
(C) 133
(D) 178
(E) 333
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COURSE/EXAM 3: MAY 2000 - 2 - GO ON TO NEXT PAGE
2. Lucky Tom finds coins on his way to work at a Poisson rate of 0.5 coins/minute. Thedenominations are randomly distributed:
(i) 60% of the coins are worth 1;
(ii) 20% of the coins are worth 5; and
(iii) 20% of the coins are worth 10.
Calculate the conditional expected value of the coins Tom found during his one-hour walk today,given that among the coins he found exactly ten were worth 5 each.
(A) 108
(B) 115
(C) 128
(D) 165
(E) 180
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COURSE/EXAM 3: MAY 2000 - 3 - GO ON TO NEXT PAGE
3. For a fully discrete two-year term insurance of 400 on ( x):
(i) i = 01.
(ii) 400 74 3321Px: .=
(iii) 400 16581 21V x: .=
(iv) The contract premium equals the benefit premium.
Calculate the variance of the loss at issue.
(A) 21,615
(B) 23,125
(C) 27,450
(D) 31,175
(E) 34,150
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COURSE/EXAM 3: MAY 2000 - 4 - GO ON TO NEXT PAGE
4. You are given:
(i) The claim count N has a Poisson distribution with mean L .
(ii) L has a gamma distribution with mean 1 and variance 2.
Calculate the probability that N = 1.
(A) 0.19
(B) 0.24
(C) 0.31
(D) 0.34
(E) 0.37
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COURSE/EXAM 3: MAY 2000 - 5 - GO ON TO NEXT PAGE
5. An insurance company has agreed to make payments to a worker age x who was injured at work.
(i) The payments are 150,000 per year, paid annually, starting immediately and continuingfor the remainder of the workers life.
(ii) After the first 500,000 is paid by the insurance company, the remainder will be paid by areinsurance company.
(iii) t x
t
pt
t=
1
600 000600 000
03
Calculate the expected value of his bonus.
(A) 16,700
(B) 31,500
(C) 48,300
(D) 50,000
(E) 56,600
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COURSE/EXAM 3: MAY 2000 - 26 - GO ON TO NEXT PAGE
26. For a fully discrete 3-year endowment insurance of 1000 on xb g:(i) q qx x= =+1 020.
(ii) i = 0 06.
(iii) 1000 373633Px: .=
Calculate 1000 2 3 1 3V Vx x: : .-e j
(A) 320
(B) 325
(C) 330
(D) 335
(E) 340
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COURSE/EXAM 3: MAY 2000 - 27 - GO ON TO NEXT PAGE
27. For an insurer with initial surplus of 2:
(i) The annual aggregate claim amount distribution is:
Amount Probability038
0.60.30.1
(ii) Claims are paid at the end of the year.
(iii) A total premium of 2 is collected at the beginning of each year.
(iv) i = 0 08.
Calculate the probability that the insurer is surviving at the end of year 3.
(A) 0.74
(B) 0.77
(C) 0.80
(D) 0.85
(E) 0.86
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COURSE/EXAM 3: MAY 2000 - 28 - GO ON TO NEXT PAGE
28. For a mortality study on college students:
(i) Students entered the study on their birthdays in 1963.
(ii) You have no information about mortality before birthdays in 1963.
(iii) Dick, who turned 20 in 1963, died between his 32 nd and 33rd birthdays.
(iv) Jane, who turned 21 in 1963, was alive on her birthday in 1998, at which time she left thestudy.
(v) All lifetimes are independent.
(vi) Likelihoods are based upon the Illustrative Life Table.
Calculate the likelihood for these two students.
(A) 0.00138
(B) 0.00146
(C) 0.00149
(D) 0.00156
(E) 0.00169
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COURSE/EXAM 3: MAY 2000 - 29 - GO ON TO NEXT PAGE
29. For a whole life annuity-due of 1 on (x), payable annually:
(i) qx = 0 01.
(ii) qx+ =1 0 05.
(iii) i = 0 05.
(iv) && .ax+ =1 6 951
Calculate the change in the actuarial present value of this annuity-due if px+1 is increasedby 0.03.
(A) 0.16
(B) 0.17
(C) 0.18
(D) 0.19
(E) 0.20
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COURSE/EXAM 3: MAY 2000 - 30 - GO ON TO NEXT PAGE
30. X is a random variable for a loss.
Losses in the year 2000 have a distribution such that:
E X d d d d = - + - =0025 1475 2 25 10 11 12 262. . . , , , ,...,
Losses are uniformly 10% higher in 2001.
An insurance policy reimburses 100% of losses subject to a deductible of 11 up to a maximumreimbursement of 11.
Calculate the ratio of expected reimbursements in 2001 over expected reimbursements in theyear 2000.
(A) 110.0%
(B) 110.5%
(C) 111.0%
(D) 111.5%
(E) 112.0%
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COURSE/EXAM 3: MAY 2000 - 31 - GO ON TO NEXT PAGE
31. Company ABC issued a fully discrete three-year term insurance of 1000 on Pat whose stated ageat issue was 30. You are given:
(i)x qx30313233
0.010.020.030.04
(ii) i = 0 04.
(iii) Premiums are determined using the equivalence principle.
During year 3, Company ABC discovers that Pat was really age 31 when the insurance wasissued. Using the equivalence principle, Company ABC adjusts the death benefit to the leveldeath benefit it should have been at issue, given the premium charged.
Calculate the adjusted death benefit.
(A) 646
(B) 664
(C) 712
(D) 750
(E) 963
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COURSE/EXAM 3: MAY 2000 - 32 - GO ON TO NEXT PAGE
32. Insurance for a citys snow removal costs covers four winter months.
(i) There is a deductible of 10,000 per month.
(ii) The insurer assumes that the citys monthly costs are independent and normallydistributed with mean 15,000 and standard deviation 2,000.
(iii) To simulate four months of claim costs, the insurer uses the Inverse Transform Method(where small random numbers correspond to low costs).
(iv) The four numbers drawn from the uniform distribution on [0,1] are:
0.5398 0.1151 0.0013 0.7881
Calculate the insurers simulated claim cost.
(A) 13,400
(B) 14,400
(C) 17,800
(D) 20,000
(E) 26,600
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COURSE/EXAM 3: MAY 2000 - 33 - GO ON TO NEXT PAGE
33. In the state of Elbonia all adults are drivers. It is illegal to drive drunk. If you are caught, yourdrivers license is suspended for the following year. Drivers licenses are suspended only fordrunk driving. If you are caught driving with a suspended license, your license is revoked andyou are imprisoned for one year. Licenses are reinstated upon release from prison.
Every year, 5% of adults with an active license have their license suspended for drunk driving.Every year, 40% of drivers with suspended licenses are caught driving.
Assume that all changes in driving status take place on January 1, all drivers act independently,and the adult population does not change.
Calculate the limiting probability of an Elbonian adult having a suspended license.
(A) 0.019
(B) 0.020
(C) 0.028
(D) 0.036
(E) 0.047
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COURSE/EXAM 3: MAY 2000 - 34 - GO ON TO NEXT PAGE
34. For a last-survivor whole life insurance of 1 on (x) and (y):
(i) The death benefit is payable at the moment of the second death.
(ii) The independent random variables T x T y Z* , * ,b g b g and are the components of acommon shock model.
(iii) T x*b g has an exponential distribution with mxT x t t* . , .b gb g = 003 0
(iv) T y*b g has an exponential distribution with myT y t t* . , .b g b g = 0 05 0(v) Z, the common shock random variable, has an exponential distribution with
mZ t tb g = 002 0. , .(vi) d = 006.
Calculate the actuarial present value of this insurance.
(A) 0.216
(B) 0.271
(C) 0.326
(D) 0.368
(E) 0.423
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COURSE/EXAM 3: MAY 2000 - 35 - GO ON TO NEXT PAGE
35. The distribution of Jacks future lifetime is a two-point mixture:
(i) With probability 0.60, Jacks future lifetime follows the Illustrative Life Table, withdeaths uniformly distributed over each year of age.
(ii) With probability 0.40, Jacks future lifetime follows a constant force of mortalitym= 002. .
A fully continuous whole life insurance of 1000 is issued on Jack at age 62.
Calculate the benefit premium for this insurance at i = 0 06. .
(A) 31
(B) 32
(C) 33
(D) 34
(E) 35
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COURSE/EXAM 3: MAY 2000 - 36 - GO ON TO NEXT PAGE
36. A new insurance salesperson has 10 friends, each of whom is considering buying a policy.
(i) Each policy is a whole life insurance of 1000, payable at the end of the year of death.
(ii) The friends are all age 22 and make their purchase decisions independently.
(iii) Each friend has a probability of 0.10 of buying a policy.
(iv) The 10 future lifetimes are independent.
(v) S is the random variable for the present value at issue of the total payments to those whopurchase the insurance.
(vi) Mortality follows the Illustrative Life Table.
(vii) i = 0 06.
Calculate the variance of S.
(A) 9,200
(B) 10,800
(C) 12,300
(D) 13,800
(E) 15,400
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COURSE/EXAM 3: MAY 2000 - 37 - GO ON TO NEXT PAGE
37. Given:
(i) pk denotes the probability that the number of claims equals k for k = 0,1,2,
(ii)pp
mn
n
m= !
! , m n 0 0,
Using the corresponding zero-modified claim count distribution with p pM M0 101= . , .calculate
(A) 0.1
(B) 0.3
(C) 0.5
(D) 0.7
(E) 0.9
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COURSE/EXAM 3: MAY 2000 - 38 - GO ON TO NEXT PAGE
38. For Shoestring Swim Club, with three possible financial states at the end of each year:
(i) State 0 means cash of 1500. If in state 0, aggregate member charges for the next year areset equal to operating expenses.
(ii) State 1 means cash of 500. If in state 1, aggregate member charges for the next year areset equal to operating expenses plus 1000, hoping to return the club to state 0.
(iii) State 2 means cash less than 0. If in state 2, the club is bankrupt and remains in state 2.
(iv) The club is subject to four risks each year. These risks are independent. Each of the fourrisks occurs at most once per year, but may recur in a subsequent year.
(v) Three of the four risks each have a cost of 1000 and a probability of occurrence 0.25 peryear.
(vi) The fourth risk has a cost of 2000 and a probability of occurrence 0.10 per year.
(vii) Aggregate member charges are received at the beginning of the year.
(viii) i = 0
Calculate the probability that the club is in state 2 at the end of three years, given that it is instate 0 at time 0.
(A) 0.24
(B) 0.27
(C) 0.30
(D) 0.37
(E) 0.56
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COURSE/EXAM 3: MAY 2000 - 39 - GO ON TO NEXT PAGE
39. For a continuous whole life annuity of 1 on (x):
(i) T xb g, the future lifetime of ( x), follows a constant force of mortality 0.06.(ii) The force of interest is 0.04.
Calculate Pr .a aT x xb ge j>
(A) 0.40
(B) 0.44
(C) 0.46
(D) 0.48
(E) 0.50
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COURSE/EXAM 3: MAY 2000 - 40 - STOP
40. Rain is modeled as a Markov process with two states:
(i) If it rains today, the probability that it rains tomorrow is 0.50.
(ii) If it does not rain today, the probability that it rains tomorrow is 0.30.
Calculate the limiting probability that it rains on two consecutive days.
(A) 0.12
(B) 0.14
(C) 0.16
(D) 0.19
(E) 0.22
**END OF EXAMINATION**
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Course 3 May 2000
Solutions to Course 3 Exam May 2000
Question # 1Key: C
et
dto
0
2
01
2 225 50= -FHG
IKJ = - = = =z w w ww w w
w
et
dto
10
2
0
401
4040
402 40
20= -FHGIKJ = - =z b gb g
Var T x tt
dt
t t
b g b g
b g
= -FHGIKJ -
= -
LNM
OQP
-
=
z2 1 40 202
2 3 4020
133
2
0
40
2 3
0
402
Question # 2Key: C
A priori, expect 30 coins: 18 worth one, 6 worth 5, 6 worth 10.
Given 10 worth 5, expect: 18 @ 1; 10 @ 5; 6 @ 10.
Total = 18 + 50 + 60 = 128
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Course 3 May 2000
Question # 3Key: E
Need to determine q qx xand +1.
Formula 7.23: 1658400
1174 33 0251
11. .
. .= = - =+ +Vq
qx x
Formula 8.39: 0 0 400 1V vq v Vpx x= = - +p
qV
Vx= -
-=p 11
4000171
1
..
Loss (Example 8.51):
0 4001
1174 33 289 30
.. .FHG
IKJ - =
1 4001
117433 1
1
1118868
2
..
..FHG
IKJ - +
FHG
IKJ =
2 - +FHGIKJ = -7433 1
1
1114190.
..
Since E L = 0,
Var = + - + - - =289 30 017 18868 0 25 1 017 14190 1 017 1 0 25 341502 2 2. . . . . . . .b g b g b g b gb g b g b gb g
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Course 3 May 2000
Question # 4Key: A
The distribution is negative binomial
E N = 1
Var Var Var
Var
N E N E N
E
= +
= += + =
L L
L L
L L
L L
c h c h
1 2 3
From the KPW appendix, we have:
rb = 1
rb b1 3+ =b g1
31
2+ = =b bb gr rb = =1 1
2
Pr .Nr
r= = += =+1
1
1
3019245
1 32
bbb g
Question # 5Key: B
If the worker survives for three years, the reinsurance will pay 100,000 at t=3, and everythingafter that. So the actuarial present value of the reinsurers portion of the claim
= FHGIKJ +
FHG
IKJ +
FHG
IKJ
FHG
IKJ100 000
7105
150 0007
1057
105
3 4 5
,..
,..
..
= 79 012,
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Course 3 May 2000
Question # 6Key: D
Surplus = U t t S tb g b g= + -60 20 .
P P tS t
Ruinb g b g= < -FHGIKJ
60
20
Hence claims of 100, 200 causing ruin can only occur on the intervals
0100 60
202 0
200 60
207, , ,
- =LNMOQP
- =LNMOQP respectively.
P t P tt
Claim by no claimb g b g= - =+
11
Therefore,
P Ps
Ruin p s claim of size s causes ruinb g b g b g= =
++
+=60% 2
1 240%
71 7
75%
Question # 7Key: D
Ll l
qd
md
L
dd
md
L
dd
d d d d d
q
xx x
xi x
i
xd x
d
x
xd
xd
xw x
w
x
xw
xw
x xi
xd
xw
xi
xi
tt t
t
t
t
b gb g b g
b g b g
b g b gb g
b g b g
b g b gb g
b g b g
b g b g b g b g b g
b g
= + =
=
= = = =
= = = =
= = + + =
= =
+1
224 000
25 000
24 0000 02 480
24 0000 05 1200
2000 320
32025 000
00128
,
,
,.
,.
,.
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Course 3 May 2000
Question # 8Key: C
Zv k
kE Z E Z S E Z N
vq v q vq v q
k
xs
x
s
x xN
x
N
===
RST= +
= + + +
+10
0
0 1
2 31 3 2 3
10 1 3 2 3
5 1
5 21
21
,
, ,...
e j e j{ }
q px x= -1 p ex =-FHG
IKJ
1
q
t
1 2q p px x x= - 2
2
p ex =-FHG
IKJq
t
Smokers Nonsmokersp x 0.64118 0.77880
2 p x 0.16901 0.36788
Now plug in
= + FHGIKJ
LNMM
OQPP + +
FHG
IKJ
LNMM
OQPP
RS|T|
UV|W|
= + =
101
105035882
1
1050 47217 1 3
1
10502212
1
105041092 2 3
58 338 2 3 77 000 1 3 64 559
52 2
..
..
..
..
, , ,
b g b g b g b g
b g b g
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Course 3 May 2000
Question # 9Key: A
510
10 35 10 35 35 5V a P a se j e j= :
=P aa
s10 3545
35 10e j
:
a v p dt p dt i
et
tt t45 45 45
0
40
0
40
45
2
0
40
0
140
402
20
= = =
= -FHG
IKJ
zz= =
b go
s a E a v p
t dtll
35 10 35 10 10 35 35 1010
10 35
35
450
10150
50
45040
: : :/ /= =
= -
=
z b g
= =P a10 35
20450
40 1778e j .
510
10 35 35 535
400
5
2
0
5
1778
5050
1
501778
50
4550
2938
V P a s t dtll
tt
= = -
= FHGIKJ -FHG
IKJ =
ze j b g b gb g
:
.
. .
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Course 3 May 2000
Question # 10Key: D
Let X(t) be the number leaving by cab in a t hour interval and let Yi denote the numberof people in the ith group. Then:
E Y
E Y
i
i
= + + =
= + + =
1 0 6 2 0 3 3 01 15
1 06 2 0 3 3 01 2 72 2 2 2. . . .
. . . .
b g b g b gc h c h c h
E X
Var X
72 10 72 15 1 080
72 10 72 2 7 1944
b gb g
= =
= =
. ,
. ,
P X 72 1050b g , = P X 72 10495b g .=
-
-LNM
OQPP
X 72 1080
1944
10495 1080
1944
b g .
= - -1 0 691754F .b g= F 0 691754.b g= 0 7553.
The answer is D with or without using the 0.5 continuity correction.
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Course 3 May 2000
Question # 11Key: E
Calculate convolutions of f xx b g:x f f *2
1020
0.30.3
0.000.09
n 0 1 2 3 4Pr N n=b g 0.37 0.37 0.18 0.06 0.02
f
f
f
s
s
s
0 037
10 037 03 011
20 018 009 03 037 013
b gb gb g
=
= =
= + =
.
. . .
. . . . .
F
F
F
s
s
s
0 037
10 037 011 0 48
20 0 48 013 0 61
b gb gb g
=
= + =
= + =
.
. . .
. . .
E S = + + =1 03 10 03 20 0 4 50 29. . .b gb g b gb g b gb g
E S E S F F FS S S- = - - - - - -
= - - - - = - =+30 10 1 0 10 1 10 10 1 20
29 10 3 037 048 061 29 154 136
b g b gc h b gc h b gc hb g. . . . .
Question # 12Key: A
0 0408 8151 1 2
0040081
8181. . .= = -
=mb g b gq
qq
Similarly, q and q80 820 0200 0 0600= =. .
2 80 5 1 2 80 5 1 2 80 5 81 81 1 2 82
0010 99
098099
004 096 0 03 0 0782
q q p q p q. / . / . /..
.
.. . . .
= + +
= + + =b g
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Course 3 May 2000
Question # 13Key: A
E Z e
e
b g b g=+
= FHGIKJ
=
- +
-
1000
1000512
1255
10
1 2
mm d
m d
.
.
Var Z e eb g b g=+
- FHGIKJ
FHG
IKJ
=
- + -10002
5
12
23 61016
2 10 22
2 4mm d
m d .
, .
E S E Zb g b g= =400 50 200,
Var VarS Zb g b g= =400 9 444 064, ,
09550 200
9 444 0641645 9 444 064 50 200 55 255. Pr
,
, ,. , , , ,=
-
-FHG
IKJ
= + =S E S
S
kk
b gb gVar
-
Course 3 May 2000
Question # 14Key: B
According to the theorem on p. 67, the expected number of interations is c 178. , as follows:
f xg x
x x x
ddx
f xg x
x x
x x
x
b gb g d ib gb g d i
b gb g
= - +
FHG
IKJ = - + =
- - =
=
12 2
12 1 4 3 0
12 3 1 1 0
1
3
2 3
2
ddx
f xg x
2
21
3
b gb g
FHG
IKJ = - + = - + = -
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Course 3 May 2000
Question # 15Key: B
p p p30 301
302 09 07 063tb g b g b g b gb g= = =. . .
l p l31 30 30 630t t tb g b g b g= =
d q l311
1 301
30 75b g b g b g= =t
d l l31 31 32 630 472 158t t tb g b g b g= - = - =
d d d312
31 311 158 75 83b g b g b g= - = - =t
qd
l312 31
2
31
83630
01317 013b gb gb g= = = t . .
Question # 16Key: C
Let S = the aggregate loss.
E Sb g = 8 10 000,
Var Sb g b g b g= + =8 3937 3 10 000 32 0002 2 2 2, ,
P S E S PS
P Z
> = - >FHGIKJ
= >
= -
= -
1580 000
32 000
40 000
32 000
125
125
1 125
.,
,
,
,
.
.
.
b gc hb gb g
b gF
F
-
Course 3 May 2000
Question # 17Key: E
s x X x E X x e d
e e
x
x
x x
b g b g c hc h
= > = > =
=-
-
- -
zPr Pr ..
q qq01
01
1
11
11
So: s 05 01205. .b g =
Question # 18Key: A
pm
md m
d m
md m
d m= = +
+
= =- +
- +zz
1000 1000
11000 1500
0
e dt
e dt
t
t
b g
b g
Expected loss = 1000 80 80A a- p .
1000 1000 1029708672 66575 6855380 80Ai
A= = =d
. . .b gb g
a801 068553
53969= - =. .d
Expected loss = 68553 150 53969 124. .- = -b g
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Course 3 May 2000
Question # 19Key: D
E S E N X E X Nb g b g b g b g b g= +Var Var2
N E N nq N nq qis binomial Varb g b g b g= = -, 1 , where q is the probability of a claim.
X E X Xis uniform. Max Var Maxb g b g= =2 12
2
,
Hence,
Total variance
= + - FHGIKJ
RS|T|UV|W|
= + FHGIKJ
LNMM
OQPP
+ + FHGIKJ
LNMM
OQPP
=
#policies Max12Max
policies
q q q2 2
2 2 2 2
12
100 005400
12005 0 95
400
2200 0 06
300
12006 094
300
2
600 46667
b g
b gb g b gb g. . . . . ., .
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Course 3 May 2000
Question # 20Key: A
APV v p p= -5 4 70 80 5 70 80: :c h
4 7074
70
4 8084
80
4 70 80 4 4 80 4 70 4 80
5 7075
70
5 8085
80
5 70 80 5 70 5 80 5 70 5 80
56 64051
66 161550856094
26 607343914365
0679736
0953912
53 960816616155
0815592
23 58246
39143650 602459
0926690
70
pll
pll
p p p p p
pll
pll
p p p p p
= = =
= = =
= + - =
= = =
= = =
= + - =
, .
, ..
, ., .
.
.
, ., .
.
, .
, ..
.
:
:
APV v= - = =0953912 0926690 0027222103
0 023485 5. ..
..b g
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Course 3 May 2000
Question # 21Key: E
The expected values at the end of 20 years are 11 11620 20 20. . .b g b g= -e m
Actuarial present values at time zero are 1120 20.b g v and 116 20 20 20.b g e v- m
So 11 116. .= -e m
So m = FHGIKJ =log
.
..
116
110 0531
S x e xx0 5 0 50 5 0505 06931
005311305. .. .
log . .
..b g = = = - = =-m
m
Question # 22Key: B
Note: These values of l are 1/100 of the ones in the Exam booklet, which does not affect theanswer.
l
l
l l
K
30
30
81 82
95 014
1 0 63 35155
35155
30 51
=- =< = - >FHG
IKJ
= > -+
FHG
IKJ
LNM
OQP
= = -+
FHG
IKJ
=
=+
FHG
IKJ
= FHGIKJ =
-
e j 1
1
1
6
1004648
0
0
32
0
d
dm
d m
dm
d m
mm d
m
md
where
-
Course 3 May 2000
Question # 40Key: D
p = limiting probability of rain.
p p p
p
= + -
= =
05 0 3 1
0 3
080 375
. .
.
..
b g
Prob (rain on two consecutive days) = Prob (rain on first) x Prob (rain on second, given rain on first)
= (0.375)(0.5)= 0.1875