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NOTES
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Survival Models  1
PAST CAS AND SOA EXAMINATION QUESTIONS ON SURVIVAL
A.
Time of Death for a Person Aged x
A1.
The lx column of a mortality table is calculated by the formula: lx = 1000000  100x2 for 0 ≤ x ≤ 100.
It is desired to know the probability that a life now age 10 will survive 30 years and then die within the
ensuing ten years. In which of the following ranges does this probability lie?
B. ≥ .075 but < .085
A. < .075
(80S–4–39)
A2.
C. ≥ .085 but < .095
D. ≥ .095 but < .105
E. ≥ .105
The following mortality table is applied to a certain population.
Age
lx
0
1
2
3
4
5
1,000
960
918
874
828
779
dx
Age x
40
42
44
46
49
51
6
7
8
9
10
lx
dx
728
674
618
559
497
54
56
59
62
64
After several years, the table is revised to reflect the fact that mortality has increased at ages 5 and above
such that only 716 lives reach age 6. However, the probability that a life age 2 will die between ages 6
and 7 is unchanged by the mortality increase. In which of the following ranges is the number of lives
reaching age 7 under the revised table?
A. < 600
A3.
B. ≥ 600 but < 620
C. ≥ 620 but < 640
D. ≥ 640 but < 660
E. ≥ 660
(81S–4–14)
A survival function is defined by:
s(x) = 1  x/250 for 0 ≤ x ≤ 25
s(x) = (1.2)(1  x/100) for 25 < x ≤ 100
In which of the following ranges is the probability that a survivor age 10 will die between the ages 40
and 60?
A. < .20
A4.
B. ≥ .20 but < .24
C. ≥ .24 but < .28
D. ≥ .28 but < .32
E. ≥ .32
(81F–4–9)
For a certain population, the number of lives at age x is determined by applying the following formula:
lx = 10,000 121  x for 0 ≤ x ≤ 121
K is the probability that a life now age 21 will die after attaining age 40, but before attaining age 57. In
which of the following ranges is K?
A. < .7
A5.
B. ≥ .07 but < .09
C. ≥ .09 but < .11
D. ≥ .11 but < .13
E. ≥ .13
(81F–4–15)
Given the following, calculate (qx+1 + qx+2) to the nearest .01.
1|qx+1
A. .15
= .095
B. .20
2|qx+1
C. .25
© 2014 ACTEX Publications, Inc.
= .171
D. .27
qx+3 = .200
E. .30
(84S–4–B7)
Exam MLC Review Questions
2 Survival Models
A1.
l10 = 1000000  (100)(10)2 = 990,000
l40 = 1000000  (100)(40)2 = 840,000
l50 = 1000000  (100)(50)2 = 750,000
30|10p10
=
l40  l50
840,000  750,000
=
= .09091
990,000
l10
Answer: C
A2.
l'7 = l'6(l7/l6) = (716)(674/728) = 663
Answer: E
A3.
s(10) = 1  10/250 = .96
s(40) = (1.2)(1  40/100) = .72
s(60) = (1.2)(1  60/100) = .48
30|20q10
=
s(40)  s(60)
.72  .48
=
= .25
s(10)
.96
Answer: C
A4.
l21 = 10,000 121  21 = 100,000
l40 = 10,000 121  40 = 90,000
l57 = 10,000 121  57 = 80,000
19|17q21
=
l40  l57
90,000  80,000
=
= .10
100,000
l21
Answer: C
A5.
2px+1
= 2|qx+1/qx+3 = .171/.200 = .855
px+1 = 2px+1 + 1|qx+1 = .855 + .095 = .95
px+2 = 2px+1/px+1 = .855/.95 = .9
qx+1 + qx+2 = (1  px+1) + (1  px+2) = (1  .95) + (1  .90) = .15
Answer: A
© 2014 ACTEX Publications, Inc.
Exam MLC Review Questions
Survival Models  3
A6.
You are given: t|qx = .10, for t = 0, 1, . . . 9. Calculate 2px+5.
A. .40
B. .60
C. .72
D. .80
E. .81
(86S–4–13)
A7.
t+uqx
≥ uqx+t for t ≥ 0 and u ≥ 0. (87S–150–7–MC)
A8.
uqx+t
≥ t|uqx for t ≥ 0 and u ≥ 0. (87S–150–7–MC)
A9.
A survival function is defined by:
f(t) = (kt/2)e-t/, t > 0,  > 0
Determine k.
A. 1/4
A10.
B. 1/2
C. 1
D. 2
E. 4
(88S–160–4)
Which of the following are equivalent to tpx?
A. t|uqx t+upx B. t+uqx  tqx + t+upx C. tqx  t+uqx + tpx+u D. tqx  t+uqx  tpx+u
E. None of these expressions are equivalent to tpx. (88–4–16–1)
A11.
Define in words the following:
a. T(x)
A12.
b. t|uqx
c. F(x).
(89S–150–A1-ai-aiii–1.2)
Which of the following are true?
1. t|uqx = tpx uqx+t
2. t|uqx =
A. 1
D. 1,2
B. 2
C. 3
lx+t+u  lx+t
lx
E. 1,3
3. t|uqx = tpx t+upx
(89–4–10–1)
A13.
t+rpx
≥ rpx+t for t ≥ 0 and r ≥ 0. (90S–150–7–MC)
A14.
rqx+t
≥ t|rqx for t ≥ 0 and r ≥ 0. (90S–150–7–MC)
A15.
You are given:
qx = .04
(x + t) = .04 + .001644t, 0 ≤ t ≤ 1
(y + t) = .08 + .003288t, 0 ≤ t ≤ 1
Calculate qy.
A. .0784
A16.
B. .0792
C. .0800
D. .0808
E. .0816
(90F–150–5)
Given s(x) = [1  (x/100)]1/2, for 0 ≤ x ≤ 100, calculate the probability that a life age 36 will die between
ages 51 and 64.
A. < .15
B. ≥ .15 but < .20
© 2014 ACTEX Publications, Inc.
C. ≥ .20 but < .25
D. ≥ .25 but < .30
E. ≥ .30
(04F–3C–8–2)
Exam MLC Review Questions
4 Survival Models
A6.
.1 = t|qx = tpx  t+1px
Since 0px = 1, this equation gives us: 1px = .9, 5px = .5, and 7px = .3.
p
=
p
/
p
=
.3/.5
=
.60
2 x+5
7 x5 x
Answer: B
A7.
T – t+uqx = tqx + (1  tqx) uqx+t = uqx+t + tqx(1  uqx+t). Since the second term must be ≥ 0, the sum
of the two terms is ≥ uqx+t.
A8.
T – t|uqx = (tpx)(uqx+t) ≤ uqx+t.
A9.
Since 
 f(t) = 1 and f(t) = (t/2)e-t/ is the probability density function for an inverse exponential, k
∞
0
must equal 1.
Answer: C
A10.
tq x
= 1  tpx
A.
B.
C.
D.
t|uqx
= tpx  t+upx
t+uqx
= 1  t+upx
Thus we get:
(tpx  t+upx)  t+upx = tpx  2(t+upx)
(1  t+upx)  (1  tpx ) + t+upx = tpx
(1  tpx )  (1  t+upx) + t+upx = 2(t+upx)  tpx
(1  tpx)  (1  t+upx)  t+upx = tpx
Answer: B
A11.
a.
b.
c.
T(x) is the time-until-death random variable.
t|uqx is the probability (x) will die between ages (x + t) and (x + t + u).
F(x) is the continuous distribution function of the newborn's age at death random variable.
A12.
1.
2.
3.
T – tpx uqx+t = tpx(1  upx+t) = tpx  t+upx
F – The right-hand side equals t+upx  tpx, which is negative.
T
Answer: E
A13.
F – t+rpx = tpx rpx+t ≤ rpx+t
A14.
T – rqx+t ≥ t|rqx = tpx rqx+t
A15.
qy = 1  py = 1  exp  y+t dt

[
1
] = 1  exp[2 (x + t) dt] = 1  (p )
1
x
0
qy = 1  (1  qy
)2
= 1  (1 
2
0
.04)2
= 0.0784
Answer: A
A16.
1  36/100 = 0.8
s(51) = 1  51/100 = 0.7
s(51)  s(64) .7  .6
= .8 = 0.125
15|13q36 =
s(36)
s(36) =
s(64) =
1  64/100 = 0.6
Answer: A
© 2014 ACTEX Publications, Inc.
Exam MLC Review Questions
Survival Models  27
D.
Life Table Characteristics: Expectation of Life
D1.
The curtate expectation of life (ex) for a life age x is 10.9 years. The probability that a life age x dies
within the next year is .05. In which of the following ranges is the curtate expectation of life in years for
a life aged (x + 1)?
A. < 10.0
(80F–4–39)
D2.
B. ≥ .93 but < .95
C. ≥ .95 but < .97
B. ≥ .80 but < .85
C. ≥ .85 but < .90
D. ≥ .97 but < .99
E. ≥ .99
(81S–4–4)
D. ≥ .90 But < .95
E. ≥ .95
(81S–4–39)
This inequality cannot be true.
The inequality is true if and only if ex+1 > px/qx+1.
The inequality is true if and only if ex+1 > px/(px+1)(qx+1).
The inequality is true if and only if ex+1 > px+1/qx.
The inequality is true if and only if ex+1 > px/qx. (82–4–33–2)
Given the values shown below, in which of the following ranges is the probability a life aged 86 will
survive one year? Assume all values are exact.
x
e° x
Tx
85
86
87
4.35
4.07
3.82
56,670
44,030
34,630
A. < .830
B. ≥ .830 but < .835
(83–4–30–2)
D6.
E. ≥ 11.5
Which of the following statements is true concerning the inequality ex+1 > ex?
A.
B.
C.
D.
E.
D5.
D. ≥ 11.0 but < 11.5
You are given the following curtate expectations of life, ex:
Age x
75
76
77
10.5
10.0
9.5
ex
In which of the following ranges is the probability that a life age 75 will survive to age 77?
A. ≤ .80
D4.
C. ≥ 10.5 but < 11.0
The curtate expectation of life (ex) for a life age x is 14.7 years. The curtate expectation of life age
(x + 1) is 14.0 years. In which of the following ranges is the probability that a life age x will survive to
age (x + 1)?
A. < .93
D3.
B. ≥ 10.0 but < 10.5
C. ≥ .835 but < .840
D. ≥ .840 but < .845
E. ≥ .845
You are given the following:
lx = 75  .5x, 0 ≤ x ≤ 50
lx = 100  x, 50 ≤ x ≤ 100
In which of the following ranges is the curtate expectation of life at age x?
A. < 57.0
(83F–4–30)
B. ≥ 57.0 but < 57.5
© 2014 ACTEX Publications, Inc.
C. ≥ 57.5 but ≤ 58.0
D. ≥ 58.0 but ≤ 58.5
E. ≥ 58.5
Exam MLC Review Questions
28 Survival Models
D1.
px = 1  qx = 1  .05 = .95
ex+1 =
ex  px
10.9  .95
=
= 10.47 .
.95
px
Answer: B
D2.
ex
14.7
= 14 + 1 = .98.
px = e
+
1
x+1
Answer: D
D3.
2p75
e75
e76
10.5
10.0
= e + 1 e + 1 = 10.0 + 1 9.5 + 1 = .909.
76
77
Answer: D
D4.
Since ex = px(1 + ex+1), we get: ex+1  ex = (ex+1)qx  px
For ex+1 to be greater than ex, ex+1 must be greater than px/qx.
Answer: E
D5.
lx = Tx/e° x
l86 = 44,030/4.07 = 10,818
l87 = 34,630/3.82 = 9,065
p86 = l87/l86 = 9,065/10,818 = .838.
Answer: C
D6.
ex = (74/74.5 + 73.5/74.5 + . . . + 50.5/74.5) + (50/74.5 + 49/74.5 + . . . + 1/74.5)
ex = 40.11 + 17.11 = 57.22.
Answer: B
© 2014 ACTEX Publications, Inc.
Exam MLC Review Questions
Survival Models  29
D7.
Given the following, calculate the instantaneous absolute rate of increase of H(x) at 60.
e° x = 20
A. .10
D8.
60 = .02
B. .20
C. .30
H(x) = x + e° x
D. .35
E. .40
(84F–5–28)
You are given the following survival distribution;
b  x/a
S(x) =
The median age is 75. Determine e° 75.
A. 8.3
D9.
B. 12.5
C. 16.7
D. 20
E. 33.3
(87S–150–1)
Life expectancies and death rates from one life table are denoted by superscript A. Corresponding values
B
from another life table are denoted by superscript B. Given the following, Calculate q0 .
B
/A
A
ex ex = 1.1 for x = 0, 1, 2, . . .
A. .049
D10.
B. .051
C. .053
D. .055
A. 16
(87S–160–4)
B. 18
C. 20
D. 22
0 ≤ x < 100
E. 24
(87F–160–8)
Given the following information about a group of lives, what is the complete expectation of life e° 39?
i)
ii)
iii)
iv)
The total number of years lived beyond age 38 (T38) equals 95,000.
The total expected number of years lived between ages 38 and 39 (L38) equals 2,475.
The central death rate at age 38 (m38) equals .021.
The number of survivors to age 38 (l38) 2,500.
A. < 37.0
B. ≥ 37.0 but < 37.2
(88–4–17–2)
D12.
E. .057
A
e0 = 57
Given the following force of mortality for a survival distribution, determine e° 64.
x = (.5)(100  x)-1,
D11.
q0 = .05
C. ≥ 37.2 but < 37.4
D. ≥ 37.4 but < 37.6
E. ≥ 37.6
You are given:
lx = (100  x).5, where 0 ≤ x ≤ 100
__ = 24.67
e° 36:28|
28
Calculate 
 t tp36 (36 + t) dt.
0
A. 3.67
D13.
B. 5.00
C. 11.33
D. 19.67
E. 24.67
(88S–150–11)
Define in words e° x. (89S–150–A1a-v)
© 2014 ACTEX Publications, Inc.
Exam MLC Review Questions
30 Survival Models
D7.
d/dx (x + e° x) = 1 + [e° x (x)  1] = e° x (x)
d/dx (x + e° x) = (20)(.02) = .40.
Answer: E
D8.
b  0/a
1 = S(0) =
100


e° 75 =
b  75/a
.50 = S(75) =
1  x/100
75
=
S(75)
b=1
a = 100
|
(200/3)(1  x/100)1.5 100
75 = 16.7.
.5
Answer: C
D9.
B
e1
=
A
1.1e1
B
q0 = 1 
=
(1.1)[eA0  (1  qA0)]
A
1.1e0
B
1 + e1
1
A
q0
=
[1.1][57  (1  .05)]
= 64.9
1  .05
(1.1)(57)
= 1  1 + 64.9 = .049.
Answer: A
D10.
tp64
[
t
= exp 
 (.5)(100  64  t)-1 dt)
0
] = exp[.5 log (36  t)]|
36
e° 64
= 

t
0
=
(36  t)/36
|
36
(36  t)/36 dt = (1/9)(36 t)3/2 0 = 24.
0
Answer: E
D11.
l39 = l38  L38 m38 = 2,500  (2,475)(.021) = 2,448
T39 = T38  L38 = 95,000  2,475 = 92,525
e° 39 = T39/l39 = 92,525/2,448 = 37.80.
Answer: E
28
D12.
__  28 p = 24.67 
 t tp36 (36 + t) dt = e° 36:28|
28 36

0
28
28 s(64)
(28)(100  64).5
=
24.67


s(36)
(100  36).5

 t tp36 (36 + t) dt = 3.67.
0
Answer: A
D13.
e° x is the complete expectation of life.
© 2014 ACTEX Publications, Inc.
Exam MLC Review Questions
Survival Models  51
F7.
A mortality table for a subset of the population with better than average health is constructed by dividing
the force of mortality in the standard table by 2. The probability of an 80-year-old dying within the next
year is defined in the standard table as q80 and in the revised table it is defined as q'80. In the standard
table q80 = .30. Determine the value of q'80 in the revised table.
A. < .150
B. ≥ .150 but < .155
(93S–4A–12–2)
F8.
C. ≥ .155 but < .160
D. ≥ .160 but < .165
E. ≥ .165
You are given:
i)
[
1
R = 1  exp 
 (x + t) dt
[
]
0
1
ii)
S = 1  exp 
 [(x + t)  k] dt
iii)
k is a positive constant.
]
0
Determine an expression for k such that S = 2R/3.
A. log([1 – px] /[1 – 2qx/3])
D. log([1 – qx] /[1 – 2qx/3])
F9.
B. log([1 – 2qx/3]/[1 – px])
E. log([1 – 2qx/3]/[1 – qx])
C. log([1 – 2px/3]/[1 – px])
(93F–150–15)
You are given:
i)
ii)
iii)
iv)
v)
(35 + t) =  for 0 ≤ t ≤ 1
p35 = .985
'(35 + t) is the force of mortality for (35) subject to an additional hazard, for 0 ≤ t ≤ 1.
'(35 + t) = + c
The additional force of mortality decreases uniformly from c to 0 between age 35.5 and 36.
Determine the probability that (35) subject to the additional hazard will not survive to age 36.
A. .015 e-.25c
F10.
C. 1  .985e-c
D. 1  .985e-.5c
E. 1  .985e-.75c
(95S–150–15)
A life table for severely disabled lives is created by modifying an existing life table by doubling the
force of mortality at all ages. In the original table, q75 = .12. Calculate q75 in the modified table.
A. < .21
F11.
B. .015e.25c
B. ≥ .21 but < .23
C. ≥ .23 but < .25
D. ≥ .25 but < .27
E. ≥ .27
(96S–4A–16–2)
You are given the following life table information about the normal population:
t
lx+t
0
1,000
1
700
2
480
3
300
4
175
Assume that you insure a group of insureds who are subject to mortality 50% greater than the normal
population. Determine the probability that an insured from this group will live 3 years from time t = 0
(i.e., 3p'x).
A. < .05
B. ≥ .05 but < .10
© 2014 ACTEX Publications, Inc.
C. ≥ .10 but < .15
D. ≥ .15 but < .20
E. ≥ .20
(97F–4A–18–2)
Exam MLC Review Questions
52 Survival Models
F7.
See F2.
q'80 = 1 
1  q80 = 1 
1  .30 = 1  .83666 = .1633
Answer: D
F8.
S = 1  pxek = 2R/3 = 2qx/3
ek =
1  2qx/3
px
k = log
(1 1 2qq /3).
x
x
Answer: E
F9.
(35 + t) + 2(1  t)c for .5 ≤ t ≤ 1
[
.5
] [
1
]
q'35 = 1  .5p'.5 .5p'35.5 = 1  exp 
 ((35) + c) dt exp 
 [(35 + t) + 2(1  t)c] dt
0
[
.5
| ] = 1  (.985)exp[.5c + 2c  c  c + .25c]
1
q'35 = 1  p35 exp .5c + (2tc  t2c) .5
q'35 = 1  (.985)e-.75c.
Answer: E
F10.
(
1
)
q75 = 1  exp 
 (75 + s) ds
1
= .12
( (75 + s) ds) = log .88 = .12783
0
0
1
( 2(75 + s) ds) = 25567
q'75 = 1  exp (.25567) = .226.
0
Answer: B
F11.
q'0 = (1.5)(1  p0) = (1.5)(1  700/1,000) = .45
q'1 = (1.5)(1  p1) = (1.5)(1  480/700) = .47143
q'2 = (1.5)(1  p2) = (1.5)(1  300/480) = .5625
3p'x
= (1  q'0)(1  q'1)(1  q'2 ) = (1  .45)(1  .47143)(1  .5625) = .1271.
Answer: C
© 2014 ACTEX Publications, Inc.
Exam MLC Review Questions
Survival Models  53
F12.
You are given:
i)
ii)
iii)
t
R = 1 – exp –
 X(t) dt
0
t
S = 1 – exp –
 (X(t) + k) dt
0
k is a constant such that S = .75R
[
]
[
]
Determine an expression for k.
A. log((1  qx)/(1  .75qx))
D. log((1  px)/(1  .75qx))
F13.
C. log((1  .75px)/(1  px))
(02F–3–35) (Sample–M–59)
Individuals with flapping gum disease are known to have a constant force of mortality . Historically,
10% will die within twenty years. A new, more serious strain of the disease has surfaced with a constant
force of mortality equal to 2. Calculate the probability of death in the next twenty years for an
individual with this new strain.
A. 17%
F14.
B. log((1  .75qx)/(1  px))
E. log((1  .75qx)/(1  qx))
B. 18%
C. 19%
D. 20%
E. 21%
(05F–3–11–2)
For a group of lives aged 30, containing an equal number of smokers and nonsmokers, you are given:
i)
ii)
For nonsmokers, n(x) = .08,
For smokers, s(x) = .16,
x ≥ 30
x ≥ 30
Calculate q80 for a life randomly selected from those surviving to age 80.
A. .078
F15.
B. .086
C. .095
D. .104
E. .112
(05F–M–32)
Use the Illustrative Life Table with the following values and adjustments:
*
qx = 4qx for 67 ≤ x ≤ 68
*
q69 = q66
Values from the Illustrated Life Table are the following:
1,000q66 = 23.2871
1,000q67 = 25.4391
1,000q68 = 27.7932
1,000q69 = 30.3680
Calculate 2|2q66.
A. < .1150 B. ≥ .1150 but < .1175
(08S–3L–14–2)
F16.
C. ≥ .1175 but < .1200
D. ≥ .1200 but < .1225
E. ≥ .1225
For a certain life aged 48, mortality follows the Illustrative Life Table. However, from ages 50 to 52, it
is found that mortality is four times greater than the Illustrative Life Table, i.e. q50* = 4q50. Given the
following values, calculate 1,0004p48.
l48 = 90,456.78
A. ≤ 925
l49 = 90,000.55
B. ≥ 925 but < 935
© 2014 ACTEX Publications, Inc.
l50 = 89,509.00
C. ≥ 935 but < 945
l51 = 88,979.11
D. ≥ 945 but < 950
E. ≥ 950
l52 = 88,407.68
(10F–3L–2–2)
Exam MLC Review Questions
54 Survival Models
F12.
S = 1  pxe-k = 3R/4 = 3qx/4
k = log
e-k =
1  3qx/4
px
ek =
px
1  3qx/4
(1 1 3qq /4)
x
x
Answer: A
F13.
ln .90 = .20
S(20) = .90 = e-20
 =
ln .90
(.10536)
=
= .00527
20
20
S'(20) = e-(20)(2) = e-(20)(2)(.00527) = .80994
= 1  S'(20) = 1  .80994 = .19006.
Answer: C
20qx
F14.
n
q80 = 1  e-.08 = 1  .92312 = .07688
s
q80 = 1  e-.16 = .14786
n
35p80
= e-(.08)(50) = e-4 = .01832
= e-(.16)(50) = e-8 = .00034
n
35p80
+ 35p80 = .01832 + .00034 = .01866
s
s
35p80
wn = .01832/.01866 = .98178
ws = .00034/.01866 = .01822
n
s
q80 = wn q80 + ws q80 = (.98178)(.07688) + (.01822)(.14786) = .07817.
Answer: A
F15.
*
q69 = q66 = .02329
*
q67 = (4)(.025491) = .10196
*
q68 = (4)(.0277932) = .11117
[1  q66][1  q*67 ][1  (1  q*68 )(1  q*69 )]
2|2q66
=
2|2q66
= [1  .02329][1  .10196][1  (1  .11117)(1  .02329)] = .11567.
Answer: B
F16.
d50 = l50 – l51 = 89,509.00 – 88,979.11 = 529.89
d51 = l51 – l52 = 88,979.11 – 88,407.68 = 571.43
d50 = 3d50 = (3)(529.89) = 1,589.67
d51 = 3d51(l51 – d50)/l51 = (3)(571.43)(88,979.11 – 1,589.67)/88,979.11 = 1,683.66
l’52 = l52 – d50 – d51 = 88,407.68 – 1,589.67 – 1,683.66 = 85,134.35
1,0004p48 = (1,000)(l’52/ l48) = (1,000)(85,134.35/90,456.78) = 941.16.
Answer: C
© 2014 ACTEX Publications, Inc.
Exam MLC Review Questions
Survival Models  67
H28.
You are given:
(i)
(ii) a 
3.4 2.5
x
lx
60
61
62
63
64
65
66
67
99,999
88,888
77,777
66,666
55,555
44,444
33,333
22,222
q60 assuming a uniform distribution of deaths over each year of age.
(iii) b  3.4 2.5 q60 assuming a constant force of mortality over each year of age.
Calculate 100,000(a − b).
A. −24
B. 9
© 2014 ACTEX Publications, Inc.
C. 42
D. 73
E. 106
(13F-MLC-25)
Exam MLC Review Questions
68 Survival Models
H28.
First we note that
3.4 2.5
q60 
3.4
p60  2.5 q63.4 
l63.4 2.5 d63.4


l60 l63.4
d63.4 l63.4  l65.9

.
l60
l60
2.5
Using the uniform distribution of deaths over each year of age assumption, a.k.a., UDD, we have
l63.4  0.6 l63  0.4  l64  0.6 66,666  0.4 55,555  62,221.60,
UDD
l65.9  0.1 l65  0.9  l66  0.1 44,444  0.9 33,333  34,444.10.
UDD
Therefore,
a
3.4 2.5
62,221,6  34, 444.1
 0.27777778.
UDD
99,999
q60 
On the other hand, under Constant Force (CF)
Constant
Force

 

0.4
l 
l63.4
0.4
 ln p
0.4
 0.4 p63  e  63 
 p63   64  ,
CF
l63
 l63 
so that
0.4 0.6
l63.4  l64
 l63  55,5550.4 66,6660.6  61,977.1925.
 
CF
Similarly,
0.1 0.9
l65.9  l65
 l66  44,4440.1 33,3330.9  34,305.8572.
CF
Therefore,
b
3.4 2.5
61,977.1925  34, 305.8572
 0.27671612.
CF
99,999
q60 
Finally,
100,000  a  b   100,000  0.27777778  0.27671612   106.165758.
Answer E.
© 2014 ACTEX Publications, Inc.
Exam MLC Review Questions
Survival Models  69
I.
De Moivre's Law of Mortality
I1.
For every 125 lives born at the same time, a mortality table shows one death each year until there are no
survivors. Which of the following is the youngest age for which the probability of living to age 65 is at
least 2/3?
A. 25
I2.
B. 30
C. 35
D. 40
E. 45
(79F–4–17)
The mortality of a population is as follows:
i)
ii)
For 100 males born at the same time, a mortality table shows one death each year until there are
no survivors.
For 110 females born at the same time, a mortality table shows one death each year until there
are no survivors.
On January 1 of a given calendar year, 150 males and 121 females are born. Which of the following is
closest to the age of the group when there are exactly 5 more surviving males than females.
A. 45
I3.
B. 50
C. 55
D. 60
E. 65
(79F–4–19)
The terminal age  of a mortality table is 120, and a constant number of deaths occur each year. It is
desired to know the probability that a person now age 24 will not die between the ages of 36 and 72. In
which of the following ranges does this probability lie?
A. < .300
(80S–4–21)
B. ≥ .300 but < .450
C. ≥ .450 but < .600
D. ≥ .600 but < .750
I4.
(x) = dx-2/lx if lx is of the form k(  x). (84–4–16–2)
I5.
Mortality follows de Moivre's law and e° 16 = 42. Calculate Var(T(16)) to the nearest integer.
A. 595
I6.
B. 588
C. 505
D. 472
E. 300
E. ≥ .750
(84F–4–21)
Equal numbers of males and females are born in a population. You are given the following survival
distribution functions:
i)
i)
For males, s(x) = 1  x/90 for 0 ≤ x ≤ 90.
For females, s(x) = 1  x/100 for 0 ≤ x ≤ 100.
Calculate the force of mortality at age 60 for a cohort observed from birth.
A. .0288
I7.
B. .0280
C. .0290
D. .0291
E. .0292
(86S–5–A10)
You are given that mortality follows de Moivre's law and that e° 30 = 30. Calculate q30.
A. 1/30
B. 1/60
© 2014 ACTEX Publications, Inc.
C. 1/61
D. 1/62
E. 1/70
(86S–4–22)
Exam MLC Review Questions
70 Survival Models
I1.
Mortality follows de Moivre's law with  = 125.
  65
125  65
2/3 = 65px =
=
x = 35
x
125  x
Answer: C
I2.
Mortality follows de Moivre's law with  = 100 for males and  = 110 for females.
Number of Males at Age x = 150  1.5x
Number of Females at Age x = 121  1.1x
Difference in Number of Males and Females = (150  1.5x)  (121  1.1x) = 5
x = 60
Answer: D
I3.
Mortality follows de Moivre's law with  = 120.
36
36
1  12|36q24 = 1 
= 1 
= .625
  24
120  24
Answer: D
I4.
T – dx-2 = k(lx-2  lx-1) = k[  (x  2)]  k[  (x  1)] = k
-x
-x
I5.
(x) =
k
1
=
k(  x)
x
|
t2
x
xt
-x
e° x = 
 tpx dt = 
   x dt = t  2  x 0 = 2

0
0
 = 2e° x + x = (2)(42) + 16 = 100
x
-x
Var[T(x)] = 2 
 t tpx dt 
(e° x)2
= 2 
 t
0
0
|
xt
dt  (e° x)2
x
-x
Var[T(x)] = [t2  2t3/3(  x)] 0 – (  x/2)2 = (  x)2/12
Var(T(16)) = (  x)2/12 = (100  16)2/12 = 588.
Answer: B
I6.
x
90  60
=
= 1/3
90

1
1
m
60 =
=
= 1/30
x
90  60
m
60p0
60 =
=
(60p0m)(60m ) + (60pf0)(f60)
m
60p0
+
f
60p0
100  60
= 2/5
100
1
f
60 =
= 1/40
100  60
f
60p0
=
=
(1/3)(1/30) + (2/5)(1/40)
= .0288
1/3 + 2/5
Answer: A
I7.
= 2e° x + x = (2)(30) + 30 = 90
x1
90  30  1
= 1 
= 1/60.
q30 = 1  p30 = 1 
x
90  30
See I5.
Answer: B
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Exam MLC Review Questions