BjornvanBraak/ActuarialMathBench huggingface dataset:
Question: . (50) just had surgery to remove a life-threatening tumor and is purchasing a 3 -year term life insurance policy with a face amount of 100,000 . You are given:
i) The probability of (50) surviving the first year after surgery is 55\% of the Standard Ultimate Life Table survival probability
ii) If (50) survives the first year, subsequent mortality follows the Standard Ultimate
Life Table
iii) Benefits are payable at the end of the year of death
iv) \(i=0.05\)

Calculate the expected present value of the death benefit.

Options:
A
43,000
B
44,000
C
45,000
D
46,000
E
47,000

Correct answer: A

Solution: Probability (50) survives one year under Standard Ultimate Life Table \(=1-0.001209=\) 0.998791

Probability (50) survives one year following surgery \(=0.55 \times 0.998791=0.5493=p_{50}\)
\(q_{50}=1-p_{50}=0.4507\)
\[
\begin{aligned}
A_{50: 31} & =q_{50}\left(\frac{1}{1.05}\right)+p_{50} q_{51}\left(\frac{1}{1.05^{2}}\right)+p_{50} p_{51} q_{52}\left(\frac{1}{1.05^{3}}\right) \\
& =(0.4507)\left(\frac{1}{1.05}\right)+(0.5493)(0.001331)\left(\frac{1}{1.05^{2}}\right)+(0.5493)(0.99867)(0.001469)\left(\frac{1}{1.05^{3}}\right) \\
& =0.4306
\end{aligned}
\]

Therefore, answer \(=100,000 \times 0.4306 \approx 43,000 \quad(\mathrm{~A})\)
Question: You are given:
(i) \(\quad \delta_{t}=0.06, \quad t \geq 0\)
(ii) \(\quad \mu_{x}(t)=0.01, \quad t \geq 0\)
(iii) \(Y\) is the present value random variable for a continuous annuity of 1 per year, payable for the lifetime of ( \(x\) ) with 10 years certain
Calculate \(\operatorname{Pr}(Y>\mathrm{E}[Y])\).

Options:
A
0.705
B
0.710
C
0.715
D
0.720
E
0.725

Correct answer: A

Solution: \(E(Y)=\bar{a}_{\overline{10} \mid}+e^{-\delta(10)} e^{-\mu(10)} \bar{a}_{x+10}\)
\(=\frac{\left(1-e^{-0.6}\right)}{0.06}+e^{-0.7} \frac{1}{0.07}\)
=14.6139
\(Y>E(Y) \Rightarrow\left(\frac{1-e^{-0.06 T}}{0.06}\right)>14.6139\)
\(\Rightarrow T>34.90\)
\(\operatorname{Pr}[Y>E(Y)]=\operatorname{Pr}(T>34.90)=e^{-34.90(0.01)}=0.705\)
Question: You are given:
(i) \(A_{x}=0.30\)
(ii) \(\quad A_{x+n}=0.40\)
(iii) \(\quad A_{x: \frac{1}{n}}=0.35\)
(iv) \(\quad i=0.05\)

Calculate \(a_{x: n} \cdot\)

Options:
A
9.3
B
9.6
C
9.8
D
10.0
E
10.3

Correct answer: B

Solution: \(A_{x: \frac{1}{n}}={ }_{n} E_{x}\)
\(A_{x}=A_{x: n \mid}^{1}+{ }_{n} E_{x} A_{x+n}\)
\(0.3=A_{x: n \mid}^{1}+(0.35)(0.4) \Rightarrow A_{x: n \mid}^{1}=0.16\)
\(A_{x: n \mid}=A_{x: n \mid}^{1}+{ }_{n} E_{x}=0.16+0.35=0.51\)
\(\ddot{a}_{x: \bar{n}}=\frac{1-A_{x: \bar{n}}}{d}=\frac{1-0.51}{(0.05 / 1.05)}=10.29\)
\(a_{x: \bar{n} \mid}=\ddot{a}_{x: \bar{n} \mid}-1+{ }_{n} E_{x}=10.29-0.65=9.64\)
Question: . You are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) Deaths are uniformly distributed over each year of age
(iii) \(\quad i=0.05\)

Calculate \(\frac{d}{d t}(\overline{\bar{I}})_{40: t}\) at \(t=10.5\).

Options:
A
5.8
B
6.0
C
6.2
D
6.4
E
\(\quad 6.6\)

Correct answer: C

Solution: \((\bar{I} \bar{a})_{40: \bar{t}}=\int_{0}^{t} s_{s} p_{40} v^{s} d s \Rightarrow \frac{d(\bar{I} \bar{a})_{40: \bar{t}}}{d t}=t_{t} p_{40} v^{t}\)
At \(t=10.5\),
\(10.5_{10.5} E_{40}=10.5_{10} p_{40.5} p_{50} v^{10.5}\)
\(=10.5_{10} E_{400.5} p_{50} v^{0.5}\)
\(=10.5 \times 0.60920 \times(1-0.5 \times 0.001209)(0.975900073)\)
= 6.239
Question: (40) wins the SOA lottery and will receive both:
- A deferred life annuity of \(K\) per year, payable continuously, starting at age \(40+\stackrel{\circ}{e}_{40}\) and
- An annuity certain of \(K\) per year, payable continuously, for \(\stackrel{\circ}{e}_{40}\) years

You are given:
(i) \(\quad \mu=0.02\)
(ii) \(\delta=0.01\)
(iii) The actuarial present value of the payments is 10,000 Calculate \(K\).

Options:
A
214
B
216
C
218
D
220
E
222

Correct answer: A

Solution: \(\stackrel{\circ}{e}_{40}=\frac{1}{\mu}=50\) So, receive \(K\) for 50 years guaranteed and for life thereafter.
\(10,000=K\left[\bar{a}_{\overline{50 \mid}}+{ }_{50 \mid} \bar{a}_{40}\right]\)
\(\bar{a}_{\overline{50 \mid}}=\int_{0}^{50} e^{-\delta t}=\frac{1-e^{-50 \delta}}{\delta}=\frac{1-e^{-50(0.01)}}{0.01}=39.35\)
\({ }_{50 \mid} \bar{a}_{40}={ }_{50} E_{40} \bar{a}_{40+50}=e^{-(\delta+\mu) 50} \frac{1}{\mu+\delta}=e^{-1.5} \frac{1}{0.03}=7.44\)
\(K=\frac{10,000}{39.35+7.44}=213.7\)
Question: . For an annuity-due that pays 100 at the beginning of each year that (45) is alive, you are given:
(i) Mortality for standard lives follows the Standard Ultimate Life Table
(ii) The force of mortality for standard lives age \(45+t\) is represented as \(\mu_{45+t}^{S U L T}\)
(iii) The force of mortality for substandard lives age \(45+t, \mu_{45+t}^{S}\), is defined as:
\[
\mu_{45+t}^{S}= \begin{cases}\mu_{45+t}^{S U L T}+0.05, & \text { for } 0 \leq t<1 \\ \mu_{45+t}^{S U T}, & \text { for } t \geq 1\end{cases}
\]
(iv) \(\quad i=0.05\)

Calculate the actuarial present value of this annuity for a substandard life age 45.

Options:
A
1700
B
1710
C
1720
D
1730
E
1740

Correct answer: A

Solution: \(\ddot{a}_{45}^{S}=1+v p_{45}^{S} \ddot{a}_{46}^{S U L T}\)
\(p_{45}^{S}=e^{-\int_{0}^{1} \mu_{45+t}^{S} d t}=e^{-\int_{0}^{1}\left(\mu_{45+t}^{S U T T}+0.05\right) d t}=e^{-\int_{0}^{1}\left(\mu_{45+t}^{S U T}\right) d t} e^{-\int_{0}^{1}(0.05) d t}=p_{45}^{S U L T} \cdot e^{-0.05}=\left(\frac{98,957.6}{99,033.9}\right) e^{-0.05}=0.9504966\)
\(\ddot{a}_{45}^{S}=1+v p_{45}^{S} \ddot{a}_{46}^{S U L T}=1+(1.05)^{-1}(0.9504966)(17.6706)=17.00\)
\(100 \ddot{a}_{45}^{S}=1700\)
Question: For a group of 100 lives age \(x\) with independent future lifetimes, you are given:
(i) Each life is to be paid 1 at the beginning of each year, if alive
(ii) \(A_{x}=0.45\)
(iii) \({ }^{2} A_{x}=0.22\)
(iv) \(\quad i=0.05\)
(v) \(\quad Y\) is the present value random variable of the aggregate payments.

Using the normal approximation to \(Y\), calculate the initial size of the fund needed in order to be \(95 \%\) certain of being able to make the payments for these life annuities.

Options:
A
1170
B
1180
C
1190
D
1200
E
1210

Correct answer: D

Solution: Let \(Y_{i}\) be the present value random variable of the payment to life \(i\).
\[
E\left[Y_{i}\right]=\ddot{a}_{x}=\frac{1-A_{x}}{d}=11.55 \quad \operatorname{Var}\left[Y_{i}\right]=\frac{{ }^{2} A_{x}-\left(A_{x}\right)^{2}}{d^{2}}=\frac{0.22-0.45^{2}}{(0.05 / 1.05)^{2}}=7.7175
\]

Then \(Y=\sum_{i=1}^{100} Y_{i}\) is the present value of the aggregate payments.
\(E[Y]=100 E\left[Y_{i}\right]=1155\) and \(\operatorname{Var}[Y]=100 \operatorname{Var}\left[Y_{i}\right]=771.75\)
\(\operatorname{Pr}[Y \leq F]=\operatorname{Pr}\left[Z \leq \frac{F-1155}{\sqrt{771.75}}\right]=0.95 \Rightarrow \frac{F-1155}{\sqrt{771.75}}=1.645\)
\(\Rightarrow F=1155+1.645 \sqrt{771.75}=1200.699\)
Question: You are given:
(i) \(\quad A_{35}=0.188\)
(ii) \(A_{65}=0.498\)
(iii) \({ }_{30} p_{35}=0.883\)
(iv) \(\quad i=0.04\)

Calculate \(1000 \ddot{a}_{35: 30}^{(2)}\) using the two-term Woolhouse approximation.

Options:
A
\(\quad 17,060\)
B
17,310
C
17,380
D
17,490
E
17,530

Correct answer: C

Solution: \[
\begin{aligned}
\ddot{a}_{35: \overline{30}}^{(2)} & \approx \ddot{a}_{35: \overline{30}}-\frac{(m-1)}{2 m}\left(1-v^{30}{ }_{30} p_{35}\right) \\
\ddot{a}_{35: 301} & =\frac{1-A_{35: 301}}{d}=\frac{1-A_{35: 301}^{1}-{ }_{30} E_{35}}{d} \\
& =\frac{1-\left(A_{35}-{ }_{30} E_{35} \times A_{65}\right)-{ }_{30} E_{35}}{d}
\end{aligned}
\]

Since \({ }_{30} E_{35}=v^{30}{ }_{30} p_{35}=0.2722\), then
\[
\begin{aligned}
\ddot{a}_{35: 301} & =\frac{1-\left(A_{35}-v^{30}{ }_{30} p_{35} \times A_{65}\right)-v^{30}{ }_{30} p_{35}}{d} \\
& =\frac{1-(0.188-(0.2722)(0.498))-0.2722}{(0.04 / 1.04)} \\
& =17.5592
\end{aligned}
\]
\(\ddot{a}_{35: 301}^{(2)} \approx 17.5592-\frac{1}{4}(1-0.2722)=17.38\)
\(1000 \ddot{a}_{35: 301}^{(2)} \approx 1000 \times 17.38=17,380\)
Question: . For an annual whole life annuity-due of 1 with a 5 -year certain period on (55), you are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) \(\quad i=0.05\)

Calculate the probability that the sum of the undiscounted payments actually made under this annuity will exceed the expected present value, at issue, of the annuity.

Options:
A
0.88
B
0.90
C
0.92
D
0.94
E
\(\quad 0.96\)

Correct answer: C

Solution: The expected present value is:
\[
\ddot{a}_{51}+{ }_{5} E_{55} \ddot{a}_{60}=4.54595+0.77382 \times 14.9041=16.07904
\]

The probability that the sum of the undiscounted payments will exceed the expected present value is the probability that at least 17 payments will be made. This will occur if (55) survives to age 71 . The probability is therefore:
\[
{ }_{16} p_{55}=\frac{l_{71}}{l_{55}}=\frac{90,134.0}{97,846.2}=0.92118
\]
Question: . For a select and ultimate mortality model with a one-year select period, you are given:
(i) \(\quad p_{[x]}=(1+k) p_{x}\), for some constant \(k\)
(ii) \(\quad \ddot{a}_{x: \bar{n}}=21.854\)
(iii) \(\quad \ddot{a}_{[x]: n}=22.167\)

Calculate \(k\).

Options:
A
0.005
B
0.010
C
0.015
D
0.020
E
0.025

Correct answer: C

Solution: \[
\ddot{a}_{[x]: \bar{n}]}=1+v p_{[x]} \ddot{a}_{x+1: \overline{n-1}]}=1+(1+k)\left(v p_{x} \ddot{a}_{x+1: \overline{n-1}}\right)=1+(1+k)\left(\ddot{a}_{x: \bar{n}]}-1\right)
\]

Therefore, we have
\[
k=\frac{\ddot{a}_{[x]: n]}-1}{\ddot{a}_{x: n]}-1}-1=\frac{21.167}{20.854}-1=0.015
\]
Question: . For a 10 -year certain and life annuity-due on (65) with annual payments you are given:
i) Mortality follows the Standard Ultimate Life Table
ii) \(\quad i=0.05\)

Calculate the probability that the sum of the payments on a non-discounted basis made under the annuity will exceed the expected present value of the annuity at issue.

Options:
A
0.826
B
0.836
C
\(\mathbf{0 . 8 4 6}\)
D
0.856
E
\(\quad 0.866\)

Correct answer: C

Solution: \[
\ddot{a}_{\overline{65: \overline{10}}}=\ddot{a}_{10 \mid}+{ }_{10} E_{65} \times \ddot{a}_{75}=8.10782+0.55305 \times 10.3178=13.8141
\]

Assuming payments of 1 (any other payment amount would just cancel, giving the same number of years), 14 payments are needed for the sum of payments to exceed 13.8141. That requires surviving to the start of year 14, thus surviving 13 years.
\[
{ }_{13} p_{65}=\frac{l_{78}}{l_{65}}=\frac{80,006.2}{94,579.7}=0.846
\]
Question: . For a 3 -year temporary life annuity due, you are given:
i) The life annuity pays 10 at the beginning of each year
ii) \(\quad v=0.93\)
iii) \(\quad p_{x}=0.95, p_{x+1}=0.9, p_{x+2}=0.8\)

Calculate the standard deviation of the present value random variable for this annuity.

Options:
A
\(\quad 4.4\)
B
\(\quad 4.5\)
C
\(\quad 4.6\)
D
4.7
E
\(\quad 4.8\)

Correct answer: B

Solution: \begin{tabular}{|c|c|}
\hline Discounted Payment & Probability \\
\hline 10 & \((1-0.95)=0.05\) \\
\hline \(10+10 v=19.3\) & \((0.95)(1-0.9)=0.095\) \\
\hline \(10+10 v+10 v^{2}=27.949\) & \((0.95)(0.9)=0.855\) \\
\hline
\end{tabular}
\(\operatorname{Var}(X)=E\left(X^{2}\right)-[E(X)]^{2}\)
\(E(X)=10(1-0.95)+[10+10 v](0.95)(1-0.9)+\left[10+10 v+10 v^{2}\right](0.95)(0.9)=26.23\)
\(E\left(X^{2}\right)=10^{2}(1-0.95)+[10+10 v]^{2}(0.95)(1-0.9)+\left[10+10 v+10 v^{2}\right]^{2}(0.95)(0.9)=708.27\)
\(\operatorname{Var}(X)=708.27-(26.23)^{2}=20.26\)
Standard Deviation \(=(20.26)^{1 / 2}=4.5\)
It's not significantly less work, but since everyone receives the first 10 , you could use the formula \(\operatorname{Var}(X\)-constant \()=\operatorname{Var}(X)\), and work with the three payment amounts as \(0,10 v\), \(10 v+10 v^{2}\), and you would get the same answer.
Question: . For a life annuity-due issued to (55), you are given:
i) The annuity pays an annual benefit of \(X\) through age 64
ii) Beginning at age 65 , the annuity pays \(75 \%\) of \(X\)
iii) The present value of this annuity is 250,000
iv) Mortality follows the Standard Ultimate Life Table
v) \(\quad i=0.05\)

Calculate \(X\).

Options:
A
17,400
B
17,500
C
17,600
D
17,700
E
17,800

Correct answer: E

Solution: Any of these ways of viewing the benefit structure would be fine; all give the same answer:
(i) A temporary life annuity-due of \(X\) plus a deferred life annuity-due of \(0.75 X\).
(ii) A whole life annuity-due of \(0.75 X\) plus a temporary deferred life annuity-due of \(0.25 X\)
(iii) A whole life annuity-due of \(X\) minus a deferred life annuity-due of \(0.25 X\)

This solution views it the first way.
\(X \ddot{a}_{55: \overline{10}}+0.75 X \ddot{a}_{65} \bullet{ }_{10} E_{55}=8.0192 X+(0.75 X)(0.59342)(13.5498)=14.05 X=250,000\)
\(\Rightarrow X=17,794\)
Question: . For a 30 -year temporary life annuity due of 1 per year, payable monthly, on (35), you are given:
i) \(\quad \mu_{35+t}=\frac{0.15}{75-t}, \quad 0 \leq t \leq 75\)
ii) \(\quad i=0.06\)
iii) \(\quad \ddot{a}_{35: 301}=14.2546\)

Calculate the single net premium using the three term Woolhouse formula.

Options:
A
13.792
B
13.796
C
13.862
D
13.866
E
13.870

Correct answer: D

Solution: \[
\begin{aligned}
& \ddot{a}_{35: 301}^{(12)} \approx \ddot{a}_{35: \overline{30}}-\frac{11}{24}\left(1-{ }_{30} E_{35}\right)-\frac{143}{1728}\left(\delta+\mu_{35}-{ }_{30} E_{35}\left(\delta+\mu_{65}\right)\right) \\
& { }_{30} p_{35}=e^{-\int_{0}^{30} \frac{0.15}{75-t} d t}=e^{\left.0.15 \ln (75-t)\right|_{0} ^{30}}=e^{0.15 \ln 45-0.15 \ln 75}=e^{0.15 \ln (45 / 75)}=(45 / 75)^{0.15} \\
& { }_{30} E_{35}=\frac{1}{1.06^{30}}\left(\frac{45}{75}\right)^{0.15}=0.161267 \\
& \delta=\ln (1.06)=0.058269 \\
& \mu_{35}=0.15 / 75=0.002 \\
& \mu_{65}=0.15 / 45=0.003333 \\
& \ddot{a}_{35: 301}^{(12)} \approx 14.2546-\frac{11}{24}(1-0.161267)-\frac{143}{1728}(0.058269+0.002-0.161267(0.058269+0.003333)) \\
& =13.866
\end{aligned}
\]
Question: . You are given the following information about a special fully discrete 2 -payment, 2 -year term insurance on (80):
(i) Mortality follows the Standard Ultimate Life Table
(ii) \(\quad i=0.03\)
(iii) The death benefit is 1000 plus a return of all premiums paid without interest
(iv) Level premiums are calculated using the equivalence principle Calculate the net premium for this special insurance.

Options:
A
32
B
33
C
34
D
35
E
36

Correct answer: D

Solution: The equation of value is given by
Actuarial Present Value of Premiums = Actuarial Present Value of Death Benefits.
The death benefit in the first year is \(1000+P\). The death benefit in the second year is \(1000+2 P\).
The formula is \(P \ddot{a}_{80: 2}=1000 A_{80: 21}^{1}+P(I A)_{80: 21}^{1}\).
Solving for P we obtain \(P=\frac{1000 A_{80: 21}^{1}}{\ddot{a}_{80: 21}-(I A)_{80: 21}^{1}}\).
\(\ddot{a}_{80: 21}=1+p_{80} v=1+\frac{0.967342}{1.03}=1.93917\)
\(1000 A_{80: 2 \mid}^{1}=1000\left(v q_{80}+v^{2} p_{80} q_{81}\right)=1000\left(\frac{0.032658}{1.03}+\frac{(0.967342)(0.036607)}{1.03^{2}}\right)=65.08552\)
(IA) \({ }_{80: 2 \mid}^{1}=v q_{80}+2 v^{2} p_{80} q_{81}=\frac{0.032658}{1.03}+(2) \frac{(0.967342)(0.036607)}{1.03^{2}}=0.09846\)
\(P=\frac{65.08552}{1.93917-0.09846}=35.36 \rightarrow D\)
Question: For a fully discrete 10 -year term life insurance policy on ( \(x\) ), you are given:
(i) Death benefits are 100,000 plus the return of all gross premiums paid without interest
(ii) Expenses are \(50 \%\) of the first year's gross premium, \(5 \%\) of renewal gross premiums and 200 per policy expenses each year
(iii) Expenses are payable at the beginning of the year
(iv) \(\quad A_{x: 10 \mid}=0.17094\)
(v) \(\quad(I A)_{x: 10 \mid}^{1}=0.96728\)
(vi) \(\quad \ddot{a}_{x: \overline{10}}=6.8865\)

Calculate the gross premium using the equivalence principle.

Options:
A
3200
B
3300
C
3400
D
3500
E
3600

Correct answer: E

Solution: \(G \ddot{a}_{x: \overline{10} \mid}=100,000 A_{x: \overline{10} \mid}^{1}+G(I A)_{x: \overline{10} \mid}^{1}+0.45 G+0.05 G \ddot{a}_{x: \overline{10} \mid}+200 \ddot{a}_{x: \overline{10} \mid}\)
\(G=\frac{(100,000)(0.17094)+200(6.8865)}{(1-0.05)(6.8865)-0.96728-0.45}=3604.23\)
Question: . \(S\), now age 65 , purchased a 20 -year deferred whole life annuity-due of 1 per year at age 45. You are given:
(i) Equal annual premiums, determined using the equivalence principle, were paid at the beginning of each year during the deferral period
(ii) Mortality at ages 65 and older follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)
(iv) \(Y\) is the present value random variable at age 65 for \(S\) 's annuity benefits Calculate the probability that \(Y\) is less than the actuarial accumulated value of \(S\) 's premiums.

Options:
A
0.35
B
0.37
C
0.39
D
0.41
E
0.43

Correct answer: C

Solution: Let \(C\) be the annual contribution, then \(C=\frac{{ }_{20} E_{45} \ddot{a}_{65}}{\ddot{a}_{45: 20}}\)
Let \(K_{65}\) be the curtate future lifetime of (65). The required probability is
\(\operatorname{Pr}\left(\frac{C \ddot{a}_{45: \overline{20}}}{{ }_{20} E_{45}}>\ddot{a}_{\overline{K_{65}+1}}\right)=\operatorname{Pr}\left(\frac{{ }_{20} E_{45} \ddot{a}_{65}}{\ddot{a}_{45: 20}} \frac{\ddot{a}_{45} \cdot \overline{20}}{{ }_{20} E_{45}}>\ddot{a}_{\overline{K_{65}+1}}\right)=\operatorname{Pr}\left(\ddot{a}_{65}>\ddot{a}_{\overline{K_{65}+1}}\right)=\operatorname{Pr}\left(13.5498>\ddot{a}_{\overline{K_{65}+1}}\right)\)

Thus, since \(\ddot{a}_{21}=13.4622\) and \(\ddot{a}_{22}=13.8212\) we have
\(\operatorname{Pr}\left(\ddot{a}_{K_{65}+1}<13.5498\right)=\operatorname{Pr}\left(K_{65}+1 \leq 21\right)=1{ }_{-21} p_{65}=1-\frac{l_{86}}{l_{65}}=1-\frac{57,656.7}{94,579.7}=0.390\)
Question: . For whole life annuities-due of 15 per month on each of 200 lives age 62 with independent future lifetimes, you are given:
(i) \(\quad i=0.06\)
(ii) \(\quad A_{62}^{(12)}=0.4075\) and \({ }^{2} A_{62}^{(12)}=0.2105\)
(iii) \(\quad \pi\) is the single premium to be paid by each of the 200 lives
(iv) \(S\) is the present value random variable at time 0 of total payments made to the 200 lives
Using the normal approximation, calculate \(\pi\) such that \(\operatorname{Pr}(200 \pi>S)=0.90\).

Options:
A
1850
B
1860
C
1870
D
1880
E
1890

Correct answer: E

Solution: Let \(X_{i}\) be the present value of a life annuity of \(1 / 12\) per month on life \(i\) for \(i=1,2, \ldots, 200\).
Let \(S=\sum_{i=1}^{200} X_{i}\) be the present value of all the annuity payments.
\(E\left[X_{i}\right]=\ddot{a}_{62}^{(12)}=\frac{1-A_{62}^{(12)}}{d^{(12)}}=\frac{1-0.4075}{0.05813}=10.19267\)
\(\operatorname{Var}\left(X_{i}\right)=\frac{{ }^{2} A_{62}^{(12)}-\left(A_{62}^{(12)}\right)^{2}}{\left(d^{(12)}\right)^{2}}=\frac{0.2105-(0.4075)^{2}}{(0.05813)^{2}}=13.15255\)
\(E[S]=(200)(180)(10.19267)=366,936.12\)
\(\operatorname{Var}(S)=(200)(180)^{2}(13.15255)=85,228,524\)
With the normal approximation, for \(\operatorname{Pr}(S \leq M)=0.90\)
\(M=E[S]+1.282 \sqrt{\operatorname{Var}(S)}=366,936.12+1.282 \sqrt{85,228,524}=378,771.45\)
So \(\pi=\frac{378,771.45}{200}=1893.86\)
Question: . For a fully discrete whole life insurance of 1000 on (30), you are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) \(\quad i=0.05\)
(iii) The premium is the net premium

Calculate the first year for which the expected present value at issue of that year's premium is less than the expected present value at issue of that year's benefit.

Options:
A
21
B
25
C
29
D
33
E
37

Correct answer: D

Solution: Let \(k\) be the policy year, so that the mortality rate during that year is \(q_{30+k-1}\). The objective is to determine the smallest value of \(k\) such that
\(v^{k-1}\left(_{k-1} p_{30}\right)\left(1000 P_{30}\right)<v^{k}\left({ }_{k-1} p_{30}\right) q_{30+k-1}(1000)\)
\(P_{30}<v q_{30+k-1}\)
\(\frac{0.07698}{19.3834}<\frac{q_{29+k}}{1.05}\)
\(q_{29+k}>0.00417\)
\(29+k>61 \Rightarrow k>32\)
Therefore, the smallest value that meets the condition is 33.
Question: . For fully discrete whole life insurance policies of 10,000 issued on 600 lives with independent future lifetimes, each age 62, you are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) \(\quad i=0.05\)
(iii) Expenses of \(5 \%\) of the first year gross premium are incurred at issue
(iv) Expenses of 5 per policy are incurred at the beginning of each policy year
(v) The gross premium is \(103 \%\) of the net premium.
(vi) \(\quad{ }_{0} L\) is the aggregate present value of future loss at issue random variable Calculate \(\operatorname{Pr}\left({ }_{0} L<40,000\right)\), using the normal approximation.

Options:
A
0.75
B
0.79
C
0.83
D
0.87
E
\(\quad 0.91\)

Correct answer: B

Solution: Net Premium \(=10,000 A_{62} / \ddot{a}_{62}=10,000(0.31495) / 14.3861=218.93\)
\(G=218.93(1.03)=225.50\)
Let \({ }_{0} L^{*}\) be the present value of future loss at issue for one policy.
\[
\begin{aligned}
& \begin{aligned}
{ }_{0} L^{*} & =10,000 v^{K+1}-(G-5) \ddot{a}_{\overline{K+1}}+0.05 G \\
& =10,000 v^{K+1}-(225.50-5) \frac{1-v^{K+1}}{d}+0.05(225.50) \\
& =(10,000+4630.50) v^{K+1}-4630.50+11.28 \\
& =14,630.50 v^{K+1}-4619.22
\end{aligned} \\
& E\left({ }_{0} L^{*}\right)=14,630.50 A_{62}-4619.22=14,630.50(0.31495)-4619.22=-11.34 \\
& \operatorname{Var}\left({ }_{0} L^{*}\right)=(14,630.50)^{2}\left({ }^{2} A_{62}-A_{62}^{2}\right)=(14,630.50)^{2}\left(0.12506-0.31495^{2}\right)=5,536,763
\end{aligned}
\]

Let \({ }_{0} L\) be the aggregate loss for 600 such policies.
\(E\left({ }_{0} L\right)=600 E\left({ }_{0} L^{*}\right)=600(-11.34)=-6804\)
\(\operatorname{Var}\left({ }_{0} L\right)=600 \operatorname{Var}\left({ }_{0} L^{*}\right)=600(5,536,763)=3,322,057,800\)
\(\operatorname{StdDev}\left({ }_{0} L\right)=3,322,057,800^{0.5}=57,637\)
\(\operatorname{Pr}\left({ }_{0} L<40,000\right)=\Phi\left(\frac{40,000+6804}{57,637}\right)=\Phi(0.81)=0.7910\)
Question: . For a fully discrete whole life insurance on (60), you are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) \(\quad i=0.05\)
(iii) The expected company expenses, payable at the beginning of the year, are:
- 50 in the first year
- 10 in years 2 through 10
- 5 in years 11 through 20
- 0 after year 20

Calculate the level annual amount that is actuarially equivalent to the expected company expenses.

Options:
A
7.5
B
9.5
C
11.5
D
13.5
E
15.5

Correct answer: B

Solution: \(\ddot{a}_{60: \overline{10}}=7.9555\)
\(\ddot{a}_{60: 201}=12.3816\)
Annual level amount \(=\frac{40+5 \ddot{a}_{60: \overline{10}}+5 \ddot{a}_{60: \overline{20}}}{\ddot{a}_{60}}=\frac{141.686}{14.9041}=9.51\)
Question: . For a fully discrete 20 -year term insurance of 100,000 on (50), you are given:
(i) Gross premiums are payable for 10 years
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)
(iv) Expenses are incurred at the beginning of each year as follows:

\begin{tabular}{|l|c|c|c|}
\hline & Year 1 & Years 2-10 & Years 11-20 \\
\hline Commission as \% of premium & \(40 \%\) & \(10 \%\) & Not applicable \\
\hline Premium taxes as \% of premium & \(2 \%\) & \(2 \%\) & Not applicable \\
\hline Maintenance expenses & 75 & 25 & 25 \\
\hline
\end{tabular}
(v) Gross premiums are calculated using the equivalence principle Calculate the gross premium for this insurance.

Options:
A
617
B
627
C
637
D
647
E
657

Correct answer: D

Solution: \(\ddot{a}_{50: \overline{10} \mid}=8.0550\)
\(A_{50: 20 \mid}^{1}=A_{50: 20 \mid}-{ }_{20} E_{50}=0.38844-0.34824=0.04020\)
\(\ddot{a}_{50: 20 \mid}=12.8428\)
APV of Premiums = APV Death Benefit + APV Commission and Taxes + APV Maintenance \(G \ddot{a}_{50: \overline{10} \mid}=100,000 A_{50: 20 \mid}^{1}+0.12 G \ddot{a}_{50: \overline{10} \mid}+0.3 G+25 \ddot{a}_{50: \overline{20} \mid}+50\)
\(8.0550 G=4020+1.2666 G+371.07\)
\(6.7883 G=4391.07\)
\[
\Rightarrow G=646.86
\]
Question: . For a fully discrete 3 -year term insurance of 1000 on \((x)\), you are given:
(i) \(\quad p_{x}=0.975\)
(ii) \(\quad i=0.06\)
(iii) The actuarial present value of the death benefit is 152.85
(iv) The annual net premium is 56.05

Calculate \(p_{x+2}\).

Options:
A
0.88
B
0.89
C
0.90
D
0.91
E
0.92

Correct answer: D

Solution: \(\ddot{a}_{x: 3 \mid}=\frac{\text { Actuarial PV of the benefit }}{\text { Level Annual Premium }}=\frac{152.85}{56.05}=2.727\)
\(\ddot{a}_{x: 31}=1+\frac{0.975}{1.06}+\frac{0.975\left(p_{x+1}\right)}{(1.06)^{2}}=2.727\)
\(\Rightarrow p_{x+1}=0.93\)
Actuarial PV of the benefit \(=\)
\(152.85=1,000\left[\frac{0.025}{1.06}+\frac{0.975(1-0.93)}{(1.06)^{2}}+\frac{0.975(0.93)\left(q_{x+2}\right)}{(1.06)^{3}}\right]\)
\(\Rightarrow q_{x+2}=0.09 \Rightarrow p_{x+2}=0.91\)