BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: . You are given:
(i) \(\quad S_{0}(t)=\left(1-\frac{t}{\omega}\right)^{\frac{1}{4}}\), for \(0 \leq t \leq \omega\)
(ii) \(\quad \mu_{65}=\frac{1}{180}\)

Calculate \(e_{106}\), the curtate expectation of life at age 106.

Options:
A
2.2
B
2.5
C
2.7
D
3.0
E
3.2

Correct answer: B

Solution: Since \(S_{0}(t)=1-F_{0}(t)=\left(1-\frac{t}{\omega}\right)^{\frac{1}{4}}\), we have \(\ln \left[S_{0}(t)\right]=\frac{1}{4} \ln \left[\frac{\omega-t}{\omega}\right]\).
Then \(\mu_{t}=-\frac{d}{d t} \log S_{0}(t)=\frac{1}{4} \frac{1}{\omega-t}\), and \(\mu_{65}=\frac{1}{180}=\frac{1}{4} \frac{1}{\omega-65} \Rightarrow \omega=110\).
\(e_{106}=\sum_{t=1}^{3}{ }_{t} p_{106}\), since \({ }_{4} p_{106}=0\)
\({ }_{t} p_{106}=\frac{S_{0}(106+t)}{S_{0}(106)}=\frac{\left(1-\frac{106+t}{110}\right)^{1 / 4}}{\left(1-\frac{106}{110}\right)^{1 / 4}}=\left(\frac{4-t}{4}\right)^{1 / 4}\)
\(e_{106}=\sum_{i=1}^{i=4}{ }_{t} p_{106}=\frac{1}{4^{0.25}}\left(1^{0.25}+2^{0.25}+3^{0.25}\right)=2.4786\)
Question: Scientists are searching for a vaccine for a disease. You are given:
(i) 100,000 lives age \(x\) are exposed to the disease
(ii) Future lifetimes are independent, except that the vaccine, if available, will be given to all at the end of year 1
(iii) The probability that the vaccine will be available is 0.2
(iv) For each life during year \(1, q_{x}=0.02\)
(v) For each life during year 2, \(q_{x+1}=0.01\) if the vaccine has been given, and \(q_{x+1}=0.02\) if it has not been given

Calculate the standard deviation of the number of survivors at the end of year 2.

Options:
A
100
B
200
C
300
D
400
E
500

Correct answer: D

Solution: This is a mixed distribution for the population, since the vaccine will apply to all once available.

\begin{tabular}{|l|l|l|l|l|l|}
\hline \multicolumn{2}{|l|}{\multirow[b]{2}{*}{Available?}} & \multirow[b]{3}{*}{\({ }_{2} p \mid A\)} & \multicolumn{3}{|c|}{S = \# of survivors} \\
\hline & & & & & \\
\hline (A) & \(\operatorname{Pr}(A)\) & & \(E(S \mid A)\) & \(\operatorname{Var}(S \mid A)\) & \(E\left(S^{2} \mid A\right)\) \\
\hline Yes & 0.2 & 0.9702 & 97,020 & 2,891 & 9,412,883,291 \\
\hline No & 0.8 & 0.9604 & 96,040 & 3,803 & 9,223,685,403 \\
\hline & & & \(E(S)\) & & \(E\left(S^{2}\right)\) \\
\hline & & & 96,236 & & 9,261,524,981 \\
\hline & & & \(\operatorname{Var}(S)\) & 157,285 & \\
\hline & & & SD(S) & 397 & \\
\hline
\end{tabular}

As an example, the formulas for the "No" row are
\(\operatorname{Pr}(\mathrm{No})=1-0.2=0.8\)
\({ }_{2} p\) given No \(=(0.98\) during year 1\()(0.98\) during year 2\()=0.9604\).
\(E(S \mid\) No \(), \operatorname{Var}(S \mid\) No \()\) and \(E\left(S^{2} \mid\right.\) No \()\) are just binomial, \(n=100,000 ; \mathrm{p}\) (success) \(=0.9604\)
\(E(S), E\left(S^{2}\right)\) are weighted averages,
\(\operatorname{Var}(S)=E\left(S^{2}\right)-E(S)^{2}\)
Or, by the conditional variance formula:
\[
\begin{aligned}
\operatorname{Var}(S) & =\operatorname{Var}[E(S \mid A)]+E[\operatorname{Var}(S \mid A)] \\
& =0.2(0.8)(97,020-96,040)^{2}+0.2(2,891)+0.8(3,803) \\
& =153,664+3,621=157,285
\end{aligned}
\]
\(\operatorname{StdDev}(S)=397\)
Question: . You are given that mortality follows Gompertz Law with \(B=0.00027\) and \(c=1.1\). Calculate \(f_{50}(10)\).

Options:
A
0.048
B
0.050
C
0.052
D
0.054
E
0.056

Correct answer: A

Solution: \[
\begin{aligned}
& f_{x}(t)=-\frac{d}{d t} S_{x}(t)=-\frac{d}{d t}\left(e^{-\frac{B}{\ln c}\left(c^{x}\right)\left(c^{t}-1\right)}\right) \\
& =-e^{-\frac{B}{\ln c}\left(c^{x}\right)\left(c^{t}-1\right)} \cdot\left(-\frac{B}{\ln c} \cdot c^{x}\right) \cdot c^{t} \cdot \ln c \\
& =e^{-\frac{B}{\ln c}\left(c^{x}\right)\left(c^{t}-1\right)} \cdot B c^{x+t} \\
& =0.00027 \times 1.1^{x+t} \cdot e^{-\frac{0.00027}{\ln (1.1)}\left(1.1^{x}\right)\left(1.1^{t}-1\right)} \\
& f_{50}(10)=0.00027 \times 1.1^{50+10} \cdot e^{-\frac{0.00027}{\ln (1.1)}\left(1.1^{50}\right)\left(1.1^{10}-1\right)}=0.04839
\end{aligned}
\]

Alternative Solution:
\[
f_{x}(t)={ }_{t} p_{x} \cdot \mu_{x+t}
\]

Then we can use the formulas given for Makeham with \(A=0, B=0.00027\) and \(c=1.1\)
\[
f_{x}(t)=\left(e^{-\frac{0.00027}{\ln (1.1)}\left(1.1^{50}\right)\left(1.1^{10}-1\right)}\right)\left(0.00027 \times 1.1^{50+10}\right)=0.04839
\]
Question: . You are given \({ }_{t} q_{0}=\frac{t^{2}}{10,000}\) for \(0<t<100\).

Calculate \(\stackrel{\circ}{e}_{75 \text { iol }}\).

Options:
A
6.6
B
7.0
C
7.4
D
7.8
E
8.2

Correct answer: E

Solution: \(\stackrel{\circ}{e}_{75: \overline{10}}=\int_{t=0}^{t=10}{ }_{t} p_{75} d t\) where \({ }_{t} p_{x}=\frac{{ }_{t+x} p_{0}}{{ }_{x} p_{0}}=\frac{1-\frac{(t+x)^{2}}{10000}}{1-\frac{x^{2}}{10000}}=\frac{10000-(t+x)^{2}}{10000-x^{2}}\) for \(0<t<100-x\)
\(=\int_{0}^{10} \frac{10000-75^{2}-150 t-t^{2}}{10000-75^{2}} d t\)
\(=\frac{1}{4375} \cdot\left[4375 t-75 t^{2}-\frac{t^{3}}{3}\right]_{t=0}^{t=10}=8.21\)
Question: . You are given the following:
(i) \(e_{40: \overline{20}}=18\)
(ii) \(e_{60}=25\)
(iii) \(\quad{ }_{20} q_{40}=0.2\)
(iv) \(\quad q_{40}=0.003\)

Calculate \(e_{41}\).

Options:
A
\(\quad 36.1\)
B
37.1
C
38.1
D
39.1
E
40.1

Correct answer: B

Solution: \(e_{40}=e_{40: 20}+{ }_{20} p_{40} \cdot e_{60}\)
\(=18+(1-0.2)(25)\)
\(=38\)
\(e_{40}=e_{40: 11}+p_{40} \cdot e_{41}\)
\(\Rightarrow e_{41}=\frac{e_{40}-e_{40: 1}}{p_{40}}=\frac{e_{40}-p_{40}}{p_{40}}=\frac{38-0.997}{0.997}=37.11434\)
Question: . You are given the survival function:
\(S_{0}(x)=\left(1-\frac{x}{60}\right)^{\frac{1}{3}}, \quad 0 \leq x \leq 60\).
Calculate \(1000 \mu_{35}\).

Options:
A
5.6
B
6.7
C
13.3
D
16.7
E
20.1

Correct answer: C

Solution: \[
\begin{aligned}
\mu_{x} & =-\frac{d}{d_{x}} \ln S_{0}(x)=-\frac{1}{3} \frac{d}{d_{x}} \ln \left(1-\frac{x}{60}\right) \\
& =\frac{1}{180}\left(1-\frac{x}{60}\right)^{-1}=\frac{1}{3(60-x)}
\end{aligned}
\]

Therefore, \(1000 \mu_{35}=(1000) \frac{1}{3(25)}=\frac{1000}{75}=13.3\).
Question: You are given the following survival function of a newborn:
\[
S_{0}(x)= \begin{cases}1-\frac{x}{250}, & 0 \leq x<40 \\ 1-\left(\frac{x}{100}\right)^{2}, & 40 \leq x \leq 100\end{cases}
\]

Calculate the probability that (30) dies within the next 20 years.

Options:
A
0.13
B
0.15
C
0.17
D
0.19
E
0.21

Correct answer: B

Solution: \[
\begin{aligned}
& { }_{20} q_{30}=\frac{S_{0}(30)-S_{0}(50)}{S_{0}(30)}=\frac{\left(1-\frac{30}{250}\right)-\left(1-\left[\frac{50}{100}\right]^{2}\right)}{1-\frac{30}{250}}=\frac{\frac{220}{250}-\frac{3}{4}}{\frac{220}{250}} \\
& =\frac{440-375}{440}=\frac{65}{440}=\frac{13}{88}=0.1477
\end{aligned}
\]
Question: . In a population initially consisting of \(75 \%\) females and \(25 \%\) males, you are given:
(i) For a female, the force of mortality is constant and equals \(\mu\)
(ii) For a male, the force of mortality is constant and equals \(1.5 \mu\)
(iii) At the end of 20 years, the population is expected to consist of \(85 \%\) females and 15\% males

Calculate the probability that a female survives one year.

Options:
A
0.89
B
0.92
C
0.94
D
0.96
E
0.99

Correct answer: C

Solution: The 20-year female survival probability \(=e^{-20 \mu}\)
The 20-year male survival probability \(=e^{-30 \mu}\)
We want 1-year female survival \(=e^{-\mu}\)
Suppose that there were \(M\) males and \(3 M\) females initially. After 20 years, there are expected to be \(M e^{-30 \mu}\) and \(3 M e^{-20 \mu}\) survivors, respectively. At that time we have:
\(\frac{3 M e^{-20 \mu}}{M e^{-30 \mu}}=\frac{85}{15} \Rightarrow e^{10 \mu}=\frac{85}{45}=\frac{17}{9} \Rightarrow e^{-\mu}=\left(\frac{9}{17}\right)^{1 / 10}=0.938\)
Question: . You are given that mortality follows Makeham's Law with the following parameters:
i) \(\quad A=0.004\)
ii) \(B=0.00003\)
iii) \(\quad c=1.1\)

Let \(L_{15}\) be the random variable representing the number of lives alive at the end of 15 years if there are 10,000 lives age 50 at time 0 .

Calculate \(\operatorname{Var}\left[L_{15}\right]\).

Options:
A
1,317
B
1,328
C
1,339
D
1,350
E
1,361

Correct answer: E

Solution: The distribution is binomial with 10,000 trials.
\(\operatorname{Var}\left[L_{15}\right]=n p q=10,000\left({ }_{15} p_{50}\right)\left({ }_{15} q_{50}\right)\)
\({ }_{15} p_{50}=e^{\left[-A(15)-\frac{B}{\ln C} * c^{50}\left(c^{15}-1\right)\right]}=0.837445\)
\({ }_{15} q_{50}=1-{ }_{15} p_{50}=0.162555\)
\(\operatorname{Var}\left[L_{15}\right]=10,000(0.837445)(0.162555)=1361.3\)
Question: . You are given:
i) \(\mu_{x+t}=\beta t^{2}, t \geq 0\)
ii) \(l_{x}=1000\)
iii) \(l_{x+10}=400\)

Calculate \(1000 \beta\).

Options:
A
2.75
B
2.80
C
2.85
D
2.90
E
2.95

Correct answer: A

Solution: \({ }_{10} p_{x}=\frac{l_{x+10}}{l_{x}}=e^{-\int_{0}^{10} \mu_{x+t} \cdot d t} \Rightarrow \frac{400}{1000}=e^{-\int_{0}^{10} \beta t^{2} \cdot d t} \Rightarrow 0.4=e^{-\beta t^{3} / 3 t_{0}^{10}}\)
\(==>0.4=e^{-\beta \cdot 100^{3} / 3}=>\ln (0.4)=-\beta\left(\frac{1000}{3}\right)==>\beta=-\ln (0.4)(.003)=0.0027489\)
Question: . You are given:
i) \(q_{80}=0.04\)
ii) \(\quad q_{81}=0.06\)
iii) \(\quad q_{82}=0.08\)
iv) Deaths between ages 80 and 81 are uniformly distributed
v) Deaths between ages 81 and 82 are subject to a constant force of mortality

Calculate the probability that a person aged 80.6 will die between ages 81.1 and 81.6.

Options:
A
0.0294
B
0.0296
C
0.0298
D
0.0300
E
0.0302

Correct answer: C

Solution: Let \(l_{80}=1000 \Rightarrow l_{81}=960 \Rightarrow l_{82}=902.4\)
Answer \(=\frac{l_{81.1}-l_{81.6}}{l_{80.6}}=\frac{(960)^{0.9}(902.4)^{0.1}-(960)^{0.4}(902.4)^{0.6}}{(0.4)(1000)+(0.6)(960)}=0.02978\)
The value for \(l_{80}\) is arbitrary. Any other starting value gives the same result.
Another form for the survivors between 81 and 82 would be
\(\mu=-\ln (902 / 960)=0.0095833 ;\)
\(l_{81.1}=960 \times e^{-0.1 \times 0.0095833}=959.08\);
\(l_{81.6}=960 \times e^{-0.6 \times 0.0095833}=954.496\).
Question: . For a new light bulb, you are given:
i) \({ }_{t} q_{0}=\frac{t^{2}+t}{72}\) for \(0 \leq t \leq 8\)
ii) \(\quad T_{0}\) is the random variable representing the future lifetime

Calculate \(\operatorname{Var}\left[T_{0}\right]\).

Options:
A
3.9
B
4.1
C
4.3
D
4.5
E
4.7

Correct answer: A

Solution: \(E\left[T_{0}\right]=\int_{0}^{8}{ }_{t} p_{0} d t=\int_{0}^{8}\left(1-\frac{t^{2}+t}{72}\right) d t \rightarrow \frac{1}{72}\left[72 t-\frac{t^{3}}{3}-\frac{t^{2}}{2}\right]_{0}^{8}=5.1852\)
\(E\left[T_{0}{ }^{2}\right]=2 \int_{0}^{8}\left({ }_{t} p_{0} \times t\right) d t=\frac{2}{72} \int_{0}^{8}\left(72 t-t^{3}-t^{2}\right) d t=\frac{2}{72}\left[36 t^{2}-\frac{t^{4}}{4}-\frac{t^{3}}{3}\right]_{0}^{8}=30.815\)
\(\operatorname{Var}\left[T_{0}\right]=E\left[T_{0}{ }^{2}\right]-\left(E\left[T_{0}\right]\right)^{2}=3.9287\)
Question: . You are given:
(i) An excerpt from a select and ultimate life table with a select period of 3 years:

\begin{tabular}{|c|c|c|c|c|c|}
\hline\(x\) & \(l_{[x]}\) & \(l_{[x]+1}\) & \(l_{[x]+2}\) & \(l_{x+3}\) & \(x+3\) \\
\hline 60 & 80,000 & 79,000 & 77,000 & 74,000 & 63 \\
\hline 61 & 78,000 & 76,000 & 73,000 & 70,000 & 64 \\
\hline 62 & 75,000 & 72,000 & 69,000 & 67,000 & 65 \\
\hline 63 & 71,000 & 68,000 & 66,000 & 65,000 & 66 \\
\hline
\end{tabular}
(ii) Deaths follow a constant force of mortality over each year of age Calculate \(1000_{2 \mid 3} q_{[60]+0.75} \cdot\)

Options:
A
104
B
117
C
122
D
135
E
142

Correct answer: B

Solution: Under constant force over each year of age, \(l_{x+k}=\left(l_{x}\right)^{1-k}\left(l_{x+1}\right)^{k}\) for \(x\) an integer and \(0 \leq k \leq 1\).
\[
{ }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}}
\]
\(l_{[60]+0.75}=(80,000)^{0.25}(79,000)^{0.75}=79,249\)
\(l_{[60]+2.75}=(77,000)^{0.25}(74,000)^{0.75}=74,739\)
\(l_{[60]+5.75}=(67,000)^{0.25}(65,000)^{0.75}=65,494\)
\[
{ }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}}=\frac{74,739-65,494}{79,249}=0.11679
\]
\[
1000_{2 \mid 3} q_{[60]+0.75}=116.8
\]
Question: You are given:
(i) The following extract from a mortality table with a one-year select period:

\begin{tabular}{|c|c|c|c|c|}
\hline\(x\) & \(l_{[x]}\) & \(d_{[x]}\) & \(l_{x+1}\) & \(x+1\) \\
\hline 65 & 1000 & 40 & - & 66 \\
\hline 66 & 955 & 45 & - & 67 \\
\hline
\end{tabular}
(ii) Deaths are uniformly distributed over each year of age
\(\stackrel{\circ}{e}_{\text {[65] }}=15.0\)
Calculate \(\stackrel{\circ}{e}_{[66]}\).

Options:
A
14.1
B
14.3
C
14.5
D
14.7
E
14.9

Correct answer: D

Solution: \[
\begin{aligned}
& l_{65+1}=1000-40=960 \\
& l_{66+1}=955-45=910 \\
& \stackrel{\circ}{e}_{[65]}=\int_{0}^{1} t p_{[65]} d t+p_{[65]} \int_{0}^{1} p_{66} d t+p_{[65]} p_{66} \stackrel{\circ}{e}_{67} \\
& 15.0=\left[1-\left(\frac{1}{2}\right)\left(\frac{40}{1000}\right)\right]+\frac{960}{1000}\left[1-\left(\frac{1}{2}\right)\left(\frac{50}{960}\right)\right]+\left(\frac{960}{1000}\right)\left(\frac{910}{960}\right) \stackrel{\circ}{e}_{67} \\
& \stackrel{\circ}{e}_{67}=\frac{15(1000)-(980+935)}{910}=14.37912 \\
& \stackrel{\circ}{e}_{[66]}=\int_{0}^{1} t_{[66]} d t+p_{[66]} \stackrel{\circ}{e}_{67}=\left[1-\left(\frac{1}{2}\right)\left(\frac{45}{955}\right)\right]+\left(\frac{910}{955}\right) \stackrel{\circ}{e}_{67} \\
& \stackrel{\circ}{e}_{[66]}=\left[1-\left(\frac{1}{2}\right)\left(\frac{45}{955}\right)\right]+\left(\frac{910}{955}\right)(14.37912)=14.678
\end{aligned}
\]

Note that because deaths are uniformly distributed over each year of age, \(\int_{0}^{1}{ }_{t} p_{x} d t=1-0.5 q_{x}\).
Question: . The SULT Club has 4000 members all age 25 with independent future lifetimes. The mortality for each member follows the Standard Ultimate Life Table.
Calculate the largest integer \(N\), using the normal approximation, such that the probability that there are at least \(N\) survivors at age 95 is at least \(90 \%\).

Options:
A
800
B
815
C
830
D
845
E
860

Correct answer: B

Solution: Let \(S\) denote the number of survivors.
This is a binomial random variable with \(n=4000\) and success probability \(\frac{21,178.3}{99,871.1}=0.21206\)
\[
E(S)=4,000(0.21206)=848.24
\]

The variance is \(\operatorname{Var}(S)=(0.21206)(1-0.21206)(4,000)=668.36\)
\[
\operatorname{StdDev}(S)=\sqrt{668.36}=25.853
\]

The \(90 \%\) percentile of the standard normal is 1.282
Let \(S^{*}\) denote the normal distribution with mean 848.24 and standard deviation 25.853. Since \(S\) is discrete and integer-valued, for any integer \(s\),
\[
\begin{aligned}
\operatorname{Pr}(S \geq s) & =\operatorname{Pr}(S>s-0.5) \approx \operatorname{Pr}\left(S^{*}>s-0.5\right) \\
& =\operatorname{Pr}\left(\frac{S^{*}-848.24}{25.853}>\frac{s-0.5-848.24}{25.853}\right) \\
& =\operatorname{Pr}\left(Z>\frac{s-0.5-848.24}{25.853}\right)
\end{aligned}
\]

For this probability to be at least \(90 \%\), we must have \(\frac{s-0.5-848.24}{25.853}<-1.282\)
\[
\Rightarrow s<815.6
\]

So \(s=815\) is the largest integer that works.
Question: . You are given:

\begin{tabular}{|c|c|}
\hline\(x\) & \(l_{x}\) \\
\hline 60 & 99,999 \\
\hline 61 & 88,888 \\
\hline 62 & 77,777 \\
\hline 63 & 66,666 \\
\hline 64 & 55,555 \\
\hline 65 & 44,444 \\
\hline 66 & 33,333 \\
\hline 67 & 22,222 \\
\hline
\end{tabular}
\(a={ }_{3.4 \mid 2.5} q_{60}\) assuming a uniform distribution of deaths over each year of age \(b={ }_{3.4 \mid 2.5} q_{60}\) assuming a constant force of mortality over each year of age Calculate 100,000( \(a-b\) ).

Options:
A
-24
B
9
C
42
D
73
E
106

Correct answer: E

Solution: Using UDD
\(l_{63.4}=(0.6) 66,666+(0.4)(55,555)=62,221.6\)
\(l_{65.9}=(0.1)(44,444)+(0.9)(33,333)=34,444.1\)
\({ }_{3.4 \mid 2.5} q_{60}=\frac{l_{63.4}-l_{65.9}}{l_{60}}=\frac{62,221.6-34,444.1}{99,999}=0.277778\)
(a)

Using constant force
\[
\begin{aligned}
l_{63.4} & =l_{63}\left(\frac{l_{64}}{l_{63}}\right)^{0.4}=l_{63}^{0.6} l_{64}^{0.4} \\
& =\left(66,666^{0.6}\right)\left(55,555^{0.4}\right) \\
& =61,977.2 \\
l_{65.9} & =l_{65}^{0.1} l_{66}^{0.9}=\left(44,444^{0.1}\right)\left(33,333^{0.9}\right) \\
& =34,305.9
\end{aligned}
\]
\[
\begin{aligned}
3.4 \mid 2.5 q_{60} & =\frac{61,977.2-34,305.9}{99.999} \\
& =0.276716
\end{aligned}
\]
(b)
\(100,000(a-b)=100,000(0.277778-0.276716)=106\)
Question: . You are given the following extract from a table with a 3 -year select period:

\begin{tabular}{|c|c|c|c|c|c|}
\hline\(x\) & \(q_{[x]}\) & \(q_{[x]+1}\) & \(q_{[x]+2}\) & \(q_{x+3}\) & \(x+3\) \\
\hline 60 & 0.09 & 0.11 & 0.13 & 0.15 & 63 \\
\hline 61 & 0.10 & 0.12 & 0.14 & 0.16 & 64 \\
\hline 62 & 0.11 & 0.13 & 0.15 & 0.17 & 65 \\
\hline 63 & 0.12 & 0.14 & 0.16 & 0.18 & 66 \\
\hline 64 & 0.13 & 0.15 & 0.17 & 0.19 & 67 \\
\hline
\end{tabular}
\(e_{64}=5.10\)
Calculate \(e_{[61]}\).

Options:
A
5.30
B
5.39
C
5.68
D
5.85
E
6.00

Correct answer: D

Solution: \(e_{[61]}=e_{[61] ; 3]}+{ }_{3} p_{[61]}\left(e_{64}\right)\)
\(p_{[61]}=0.90\),
\({ }_{2} p_{[61]}=0.9(0.88)=0.792\),
\({ }_{3} p_{[61]}=0.792(0.86)=0.68112\)
\(e_{[61]: 3]}=\sum_{k=1}^{3}{ }_{k} p_{[61]}=0.9+0.792+0.68112=2.37312\)
\(e_{[61]}=2.37312+0.68112 e_{64}=2.37312+0.68112(5.10)=5.847\)
Question: . For a mortality table with a select period of two years, you are given:

\begin{tabular}{|c|c|c|c|c|}
\hline\(x\) & \(q_{[x]}\) & \(q_{[x]+1}\) & \(q_{x+2}\) & \(x+2\) \\
\hline 50 & 0.0050 & 0.0063 & 0.0080 & 52 \\
\hline 51 & 0.0060 & 0.0073 & 0.0090 & 53 \\
\hline 52 & 0.0070 & 0.0083 & 0.0100 & 54 \\
\hline 53 & 0.0080 & 0.0093 & 0.0110 & 55 \\
\hline
\end{tabular}

The force of mortality is constant between integral ages.
Calculate \(1000_{2.5} q_{[50]+0.4}\).

Options:
A
15.2
B
16.4
C
17.7
D
19.0
E
20.2

Correct answer: B

Solution: \[
\begin{aligned}
{ }_{2.5} q_{[50]+0.4} & =1-{ }_{2.5} p_{[50]+0.4}=1-{ }_{2.9} p_{[50]} /\left(p_{[50]}\right)^{0.4} \\
& =1-\left\{p_{[50]} p_{[50]+1}\left(p_{52}\right)^{0.9}\right\} /\left(1-q_{[50]}\right)^{0.4} \\
& =1-\left\{\left(1-q_{[50]}\right)\left(1-q_{[50]+1}\right)\left(1-q_{52}\right)^{0.9}\right\} /\left(1-q_{[50]}\right)^{0.4} \\
& =1-\left\{(1-0.0050)(1-0.0063)(1-0.0080)^{0.9}\right\} /(1-0.0050)^{0.4} \\
& =0.01642
\end{aligned}
\]
\(1000_{2.5} q_{[50]+0.4}=16.42\)
Question: . A club is established with 2000 members, 1000 of exact age 35 and 1000 of exact age 45. You are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) Future lifetimes are independent
(iii) \(\quad N\) is the random variable for the number of members still alive 40 years after the club is established

Using the normal approximation, without the continuity correction, calculate the smallest \(n\) such that \(\operatorname{Pr}(N \geq n) \leq 0.05\).

Options:
A
1500
B
1505
C
1510
D
1515
E
1520

Correct answer: B

Solution: \(E(N)=1000\left({ }_{40} p_{35}+{ }_{40} p_{45}\right)=1000\left(\frac{85,203.5}{99,556.7}+\frac{61,184.9}{99,033.9}\right)=1473.65\)
\(\operatorname{Var}(N)=1000_{40} p_{35}\left(1-{ }_{40} p_{35}\right)+1000_{40} p_{45}\left(1-{ }_{40} p_{45}\right)=359.50\)
Since \(1473.65+1.645 \sqrt{359.50}=1504.84\)
\(N=1505\)
Question: . For the country of Bienna, you are given:
(i) Bienna publishes mortality rates in biennial form, that is, mortality rates are of the form:
\({ }_{2} q_{2 x}\), for \(x=0,1,2, \ldots\)
(ii) Deaths are assumed to be uniformly distributed between ages \(2 x\) and \(2 x+2\), for \(x=0,1,2, \ldots\)
(iii) \(\quad{ }_{2} q_{50}=0.02\)
(iv) \(\quad{ }_{2} q_{52}=0.04\)

Calculate the probability that (50) dies during the next 2.5 years.

Options:
A
0.02
B
0.03
C
0.04
D
0.05
E
0.06

Correct answer: B

Solution: \({ }_{2.5} q_{50}={ }_{2} q_{50}+{ }_{2} p_{50 \quad 0.5} q_{52}=0.02+(0.98)\left(\frac{0.5}{2}\right)(0.04)=0.0298\)
Question: . X and Y are both age \(61 . \mathrm{X}\) has just purchased a whole life insurance policy. Y purchased a whole life insurance policy one year ago.

Both X and Y are subject to the following 3-year select and ultimate table:

\begin{tabular}{|c|c|c|c|c|c|}
\hline\(x\) & \(\ell_{[x]}\) & \(\ell_{[x]+1}\) & \(\ell_{[x]+2}\) & \(\ell_{x+3}\) & \(x+3\) \\
\hline 60 & 10,000 & 9,600 & 8,640 & 7,771 & 63 \\
\hline 61 & 8,654 & 8,135 & 6,996 & 5,737 & 64 \\
\hline 62 & 7,119 & 6,549 & 5,501 & 4,016 & 65 \\
\hline 63 & 5,760 & 4,954 & 3,765 & 2,410 & 66 \\
\hline
\end{tabular}

The force of mortality is constant over each year of age.
Calculate the difference in the probability of survival to age 64.5 between X and Y .

Options:
A
0.035
B
0.045
C
0.055
D
0.065
E
0.075

Correct answer: C

Solution: \({ }_{3.5} p_{[61]}-{ }_{3.5} p_{[60]+1}={ }_{0.5} p_{64}\left({ }_{3} p_{[61]}-{ }_{3} p_{[60]+1}\right)\)
\(=\left(\frac{l_{65}}{l_{64}}\right)^{0.5}\left(\frac{l_{64}}{l_{[61]}}-\frac{l_{64}}{l_{[60]+1}}\right)\)
\(=\left(\frac{4016}{5737}\right)^{0.5}\left(\frac{5737}{8654}-\frac{5737}{9600}\right)\)
\(=0.05466\)
Question: . A life is subject to the following 3 -year select and ultimate table:

\begin{tabular}{|c|c|c|c|c|c|}
\hline\([x]\) & \(\ell_{[x]}\) & \(\ell_{[x]+1}\) & \(\ell_{[x]+2}\) & \(\ell_{x+3}\) & \(x+3\) \\
\hline 55 & 10,000 & 9,493 & 8,533 & 7,664 & 58 \\
\hline 56 & 8,547 & 8,028 & 6,889 & 5,630 & 59 \\
\hline 57 & 7,011 & 6,443 & 5,395 & 3,904 & 60 \\
\hline 58 & 5,853 & 4,846 & 3,548 & 2,210 & 61 \\
\hline
\end{tabular}

You are also given:
(i) \(\quad e_{60}=1\)
(ii) Deaths are uniformly distributed over each year of age Calculate \(e_{[58]+2}\).

Options:
A
1.5
B
1.6
C
1.7
D
1.8
E
1.9

Correct answer: B

Solution: \(\stackrel{\circ}{e}_{[58]+2}=e_{[58]+2}+0.5\)
\(e_{[58]+2}=p_{[58]+2}\left(1+e_{61}\right)=p_{[58]+2}\left[1+\frac{e_{60}}{p_{60}}-1\right]\)
\(=\frac{l_{61}}{l_{[58]+2}} \times \frac{e_{60}}{p_{60}}=\frac{2210}{3548} \times \frac{1}{(2210 / 3904)}=\frac{3904}{3549}=1.100338\)
\(\stackrel{e}{e}_{[58]+2}=1.100338+0.5=1.6\)
Question: . You are given the following information from a life table:

\begin{tabular}{ccccc}
\hline\(x\) & \(l_{x}\) & \(d_{x}\) & \(p_{x}\) & \(q_{x}\) \\
\hline 95 & - & - & - & 0.40 \\
96 & - & - & 0.20 & - \\
97 & - & 72 & - & 1.00 \\
\hline
\end{tabular}

You are also given:
(i) \(\quad l_{90}=1000\) and \(l_{93}=825\)
(ii) Deaths are uniformly distributed over each year of age.

Calculate the probability that (90) dies between ages 93 and 95.5.

Options:
A
0.195
B
0.220
C
0.345
D
0.465
E
0.668

Correct answer: C

Solution: We need to determine \({ }_{3 \mid 2.5} q_{90} \cdot\)
\({ }_{3 \mid 2.5} q_{90}=\frac{l_{90+3}-l_{90+3+2.5}}{l_{90}}=\frac{l_{93}-l_{95.5}}{l_{90}}=\frac{l_{93}-\left(l_{95}-0.5 d_{95}\right)}{l_{90}}=\frac{825-[600-0.5(240)]}{1,000}=0.3450\)
where \(l_{90}=1,000, l_{93}=825, l_{97}=\frac{d_{97}}{q_{97}}=\frac{72}{1}=72, l_{96}=\frac{l_{97}}{p_{96}}=\frac{72}{0.2}=360\),
\(l_{95}=\frac{l_{96}}{p_{95}}=\frac{360}{1-0.4}=600\), and \(d_{95}=l_{95}-l_{96}=600-360=240\).
Question: . You are given the following survival function:
\[
S_{0}(x)=\left(1-\frac{x}{100}\right)^{0.5}, 0 \leq x \leq 100
\]

Calculate \(1000 \mu_{25}\).

Options:
A
6.7
B
10.0
C
13.3
D
16.7
E
20.0

Correct answer: A

Solution: \(\mu_{25}=\frac{-s^{\prime}(25)}{s(25)}=\frac{0.5(0.01)(1-0.01 \times 25)^{-0.5}}{(1-0.01 \times 25)^{0.5}}=0.00667\)