question_id stringlengths 18 25 | question_number stringlengths 1 4 | question_type stringclasses 1
value | question stringlengths 69 1.25k | options listlengths 5 5 | correct_answer stringclasses 5
values | explanation stringlengths 63 2.87k |
|---|---|---|---|---|---|---|
exam-fam-l--question-2.1 | 2.1 | multiple_choice | . 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. | [
[
"A",
"2.2"
],
[
"B",
"2.5"
],
[
"C",
"2.7"
],
[
"D",
"3.0"
],
[
"E",
"3.2"
]
] | B | 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 ... |
exam-fam-l--question-2.2 | 2.2 | multiple_choice | 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 eac... | [
[
"A",
"100"
],
[
"B",
"200"
],
[
"C",
"300"
],
[
"D",
"400"
],
[
"E",
"500"
]
] | D | 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) & \(\o... |
exam-fam-l--question-2.3 | 2.3 | multiple_choice | . You are given that mortality follows Gompertz Law with \(B=0.00027\) and \(c=1.1\). Calculate \(f_{50}(10)\). | [
[
"A",
"0.048"
],
[
"B",
"0.050"
],
[
"C",
"0.052"
],
[
"D",
"0.054"
],
[
"E",
"0.056"
]
] | A | \[
\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)... |
exam-fam-l--question-2.4 | 2.4 | multiple_choice | . You are given \({ }_{t} q_{0}=\frac{t^{2}}{10,000}\) for \(0<t<100\).
Calculate \(\stackrel{\circ}{e}_{75 \text { iol }}\). | [
[
"A",
"6.6"
],
[
"B",
"7.0"
],
[
"C",
"7.4"
],
[
"D",
"7.8"
],
[
"E",
"8.2"
]
] | E | \(\stackrel{\circ}{e}_{75: \overline{10}}=\int_{t=0}^{t=10}{ }_{t} p_{75} d t\) 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 \(0<t<100-x\)
\(=\int_{0}^{10} \frac{10000-75^{2}-150 t-t^{2}}{10000-75^{2}} d t\)... |
exam-fam-l--question-2.5 | 2.5 | multiple_choice | . 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 \(e_{41}\). | [
[
"A",
"\\(\\quad 36.1\\)"
],
[
"B",
"37.1"
],
[
"C",
"38.1"
],
[
"D",
"39.1"
],
[
"E",
"40.1"
]
] | B | \(e_{40}=e_{40: 20}+{ }_{20} p_{40} \cdot e_{60}\)
\(=18+(1-0.2)(25)\)
\(=38\)
\(e_{40}=e_{40: 11}+p_{40} \cdot e_{41}\)
\(\Rightarrow e_{41}=\frac{e_{40}-e_{40: 1}}{p_{40}}=\frac{e_{40}-p_{40}}{p_{40}}=\frac{38-0.997}{0.997}=37.11434\) |
exam-fam-l--question-2.6 | 2.6 | multiple_choice | . 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 \(1000 \mu_{35}\). | [
[
"A",
"5.6"
],
[
"B",
"6.7"
],
[
"C",
"13.3"
],
[
"D",
"16.7"
],
[
"E",
"20.1"
]
] | C | \[
\begin{aligned}
\mu_{x} & =-\frac{d}{d_{x}} \ln S_{0}(x)=-\frac{1}{3} \frac{d}{d_{x}} \ln \left(1-\frac{x}{60}\right) \\
& =\frac{1}{180}\left(1-\frac{x}{60}\right)^{-1}=\frac{1}{3(60-x)}
\end{aligned}
\]
Therefore, \(1000 \mu_{35}=(1000) \frac{1}{3(25)}=\frac{1000}{75}=13.3\). |
exam-fam-l--question-2.7 | 2.7 | multiple_choice | 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. | [
[
"A",
"0.13"
],
[
"B",
"0.15"
],
[
"C",
"0.17"
],
[
"D",
"0.19"
],
[
"E",
"0.21"
]
] | B | \[
\begin{aligned}
& { }_{20} q_{30}=\frac{S_{0}(30)-S_{0}(50)}{S_{0}(30)}=\frac{\left(1-\frac{30}{250}\right)-\left(1-\left[\frac{50}{100}\right]^{2}\right)}{1-\frac{30}{250}}=\frac{\frac{220}{250}-\frac{3}{4}}{\frac{220}{250}} \\
& =\frac{440-375}{440}=\frac{65}{440}=\frac{13}{88}=0.1477
\end{aligned}
\] |
exam-fam-l--question-2.8 | 2.8 | multiple_choice | . 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 \... | [
[
"A",
"0.89"
],
[
"B",
"0.92"
],
[
"C",
"0.94"
],
[
"D",
"0.96"
],
[
"E",
"0.99"
]
] | C | 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, re... |
exam-fam-l--question-2.9 | 2.9 | multiple_choice | . 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... | [
[
"A",
"1,317"
],
[
"B",
"1,328"
],
[
"C",
"1,339"
],
[
"D",
"1,350"
],
[
"E",
"1,361"
]
] | E | The distribution is binomial with 10,000 trials.
\(\operatorname{Var}\left[L_{15}\right]=n p q=10,000\left({ }_{15} p_{50}\right)\left({ }_{15} q_{50}\right)\)
\({ }_{15} p_{50}=e^{\left[-A(15)-\frac{B}{\ln C} * c^{50}\left(c^{15}-1\right)\right]}=0.837445\)
\({ }_{15} q_{50}=1-{ }_{15} p_{50}=0.162555\)
\(\operatornam... |
exam-fam-l--question-2.10 | 2.10 | multiple_choice | . You are given:
i) \(\mu_{x+t}=\beta t^{2}, t \geq 0\)
ii) \(l_{x}=1000\)
iii) \(l_{x+10}=400\)
Calculate \(1000 \beta\). | [
[
"A",
"2.75"
],
[
"B",
"2.80"
],
[
"C",
"2.85"
],
[
"D",
"2.90"
],
[
"E",
"2.95"
]
] | A | \({ }_{10} p_{x}=\frac{l_{x+10}}{l_{x}}=e^{-\int_{0}^{10} \mu_{x+t} \cdot d t} \Rightarrow \frac{400}{1000}=e^{-\int_{0}^{10} \beta t^{2} \cdot d t} \Rightarrow 0.4=e^{-\beta t^{3} / 3 t_{0}^{10}}\)
\(==>0.4=e^{-\beta \cdot 100^{3} / 3}=>\ln (0.4)=-\beta\left(\frac{1000}{3}\right)==>\beta=-\ln (0.4)(.003)=0.0027489\) |
exam-fam-l--question-2.11 | 2.11 | multiple_choice | . 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. | [
[
"A",
"0.0294"
],
[
"B",
"0.0296"
],
[
"C",
"0.0298"
],
[
"D",
"0.0300"
],
[
"E",
"0.0302"
]
] | C | 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 survivor... |
exam-fam-l--question-2.12 | 2.12 | multiple_choice | . 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 \(\operatorname{Var}\left[T_{0}\right]\). | [
[
"A",
"3.9"
],
[
"B",
"4.1"
],
[
"C",
"4.3"
],
[
"D",
"4.5"
],
[
"E",
"4.7"
]
] | A | \(E\left[T_{0}\right]=\int_{0}^{8}{ }_{t} p_{0} d t=\int_{0}^{8}\left(1-\frac{t^{2}+t}{72}\right) d t \rightarrow \frac{1}{72}\left[72 t-\frac{t^{3}}{3}-\frac{t^{2}}{2}\right]_{0}^{8}=5.1852\)
\(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^... |
exam-fam-l--question-3.1 | 3.1 | multiple_choice | . 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 ... | [
[
"A",
"104"
],
[
"B",
"117"
],
[
"C",
"122"
],
[
"D",
"135"
],
[
"E",
"142"
]
] | B | 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.... |
exam-fam-l--question-3.2 | 3.2 | multiple_choice | 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 ov... | [
[
"A",
"14.1"
],
[
"B",
"14.3"
],
[
"C",
"14.5"
],
[
"D",
"14.7"
],
[
"E",
"14.9"
]
] | D | \[
\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}... |
exam-fam-l--question-3.4 | 3.4 | multiple_choice | . 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 \(90 \%\). | [
[
"A",
"800"
],
[
"B",
"815"
],
[
"C",
"830"
],
[
"D",
"845"
],
[
"E",
"860"
]
] | B | 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 \(\operatorname{Var}(S)=(0.21206)(1-0.21206)(4,000)=668.36\)
\[
\operatorname{StdDev}(S)=\sqrt{668.36}=25.853
\]
T... |
exam-fam-l--question-3.5 | 3.5 | multiple_choice | . 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 dist... | [
[
"A",
"-24"
],
[
"B",
"9"
],
[
"C",
"42"
],
[
"D",
"73"
],
[
"E",
"106"
]
] | E | 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_... |
exam-fam-l--question-3.6 | 3.6 | multiple_choice | . 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... | [
[
"A",
"5.30"
],
[
"B",
"5.39"
],
[
"C",
"5.68"
],
[
"D",
"5.85"
],
[
"E",
"6.00"
]
] | D | \(e_{[61]}=e_{[61] ; 3]}+{ }_{3} p_{[61]}\left(e_{64}\right)\)
\(p_{[61]}=0.90\),
\({ }_{2} p_{[61]}=0.9(0.88)=0.792\),
\({ }_{3} p_{[61]}=0.792(0.86)=0.68112\)
\(e_{[61]: 3]}=\sum_{k=1}^{3}{ }_{k} p_{[61]}=0.9+0.792+0.68112=2.37312\)
\(e_{[61]}=2.37312+0.68112 e_{64}=2.37312+0.68112(5.10)=5.847\) |
exam-fam-l--question-3.7 | 3.7 | multiple_choice | . 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 & ... | [
[
"A",
"15.2"
],
[
"B",
"16.4"
],
[
"C",
"17.7"
],
[
"D",
"19.0"
],
[
"E",
"20.2"
]
] | B | \[
\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\} /\... |
exam-fam-l--question-3.8 | 3.8 | multiple_choice | . 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
Usi... | [
[
"A",
"1500"
],
[
"B",
"1505"
],
[
"C",
"1510"
],
[
"D",
"1515"
],
[
"E",
"1520"
]
] | B | \(E(N)=1000\left({ }_{40} p_{35}+{ }_{40} p_{45}\right)=1000\left(\frac{85,203.5}{99,556.7}+\frac{61,184.9}{99,033.9}\right)=1473.65\)
\(\operatorname{Var}(N)=1000_{40} p_{35}\left(1-{ }_{40} p_{35}\right)+1000_{40} p_{45}\left(1-{ }_{40} p_{45}\right)=359.50\)
Since \(1473.65+1.645 \sqrt{359.50}=1504.84\)
\(N=1505\) |
exam-fam-l--question-3.11 | 3.11 | multiple_choice | . 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... | [
[
"A",
"0.02"
],
[
"B",
"0.03"
],
[
"C",
"0.04"
],
[
"D",
"0.05"
],
[
"E",
"0.06"
]
] | B | \({ }_{2.5} q_{50}={ }_{2} q_{50}+{ }_{2} p_{50 \quad 0.5} q_{52}=0.02+(0.98)\left(\frac{0.5}{2}\right)(0.04)=0.0298\) |
exam-fam-l--question-3.12 | 3.12 | multiple_choice | . 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]+... | [
[
"A",
"0.035"
],
[
"B",
"0.045"
],
[
"C",
"0.055"
],
[
"D",
"0.065"
],
[
"E",
"0.075"
]
] | C | \({ }_{3.5} p_{[61]}-{ }_{3.5} p_{[60]+1}={ }_{0.5} p_{64}\left({ }_{3} p_{[61]}-{ }_{3} p_{[60]+1}\right)\)
\(=\left(\frac{l_{65}}{l_{64}}\right)^{0.5}\left(\frac{l_{64}}{l_{[61]}}-\frac{l_{64}}{l_{[60]+1}}\right)\)
\(=\left(\frac{4016}{5737}\right)^{0.5}\left(\frac{5737}{8654}-\frac{5737}{9600}\right)\)
\(=0.05466\) |
exam-fam-l--question-3.13 | 3.13 | multiple_choice | . 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 & 7,01... | [
[
"A",
"1.5"
],
[
"B",
"1.6"
],
[
"C",
"1.7"
],
[
"D",
"1.8"
],
[
"E",
"1.9"
]
] | B | \(\stackrel{\circ}{e}_{[58]+2}=e_{[58]+2}+0.5\)
\(e_{[58]+2}=p_{[58]+2}\left(1+e_{61}\right)=p_{[58]+2}\left[1+\frac{e_{60}}{p_{60}}-1\right]\)
\(=\frac{l_{61}}{l_{[58]+2}} \times \frac{e_{60}}{p_{60}}=\frac{2210}{3548} \times \frac{1}{(2210 / 3904)}=\frac{3904}{3549}=1.100338\)
\(\stackrel{e}{e}_{[58]+2}=1.100338+0.5=... |
exam-fam-l--question-3.14 | 3.14 | multiple_choice | . 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 \(l_{93}=825\)
(i... | [
[
"A",
"0.195"
],
[
"B",
"0.220"
],
[
"C",
"0.345"
],
[
"D",
"0.465"
],
[
"E",
"0.668"
]
] | C | 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}=... |
exam-fam-l--question-3.15 | 3.15 | multiple_choice | . You are given the following survival function:
\[
S_{0}(x)=\left(1-\frac{x}{100}\right)^{0.5}, 0 \leq x \leq 100
\]
Calculate \(1000 \mu_{25}\). | [
[
"A",
"6.7"
],
[
"B",
"10.0"
],
[
"C",
"13.3"
],
[
"D",
"16.7"
],
[
"E",
"20.0"
]
] | A | \(\mu_{25}=\frac{-s^{\prime}(25)}{s(25)}=\frac{0.5(0.01)(1-0.01 \times 25)^{-0.5}}{(1-0.01 \times 25)^{0.5}}=0.00667\) |
exam-fam-l--question-3.16 | 3.16 | multiple_choice | . You are given:
i) The following extract from a three-year select and ultimate table:
\begin{tabular}{|l|l|l|l|l|l|}
\hline [ \(x\) ] & \(q_{[x]}\) & \(q_{[x]+1}\) & \(q_{\{x\}+2}\) & \(q_{x+3}\) & \(x+3\) \\
\hline 50 & 0.020 & 0.031 & 0.043 & 0.056 & 53 \\
\hline 51 & 0.025 & 0.037 & 0.050 & 0.065 & 54 \\
\hline 52... | [
[
"A",
"91"
],
[
"B",
"92"
],
[
"C",
"93"
],
[
"D",
"94"
],
[
"E",
"95"
]
] | C | \begin{tabular}{|c|c|c|c|c|c|}
\hline\([x]\) & \(q_{[x]}\) & \(q_{[x]+1}\) & \(q_{[x]+2}\) & \(q_{x+3}\) & \(x+3\) \\
\hline 52 & 0.030 & 0.043 & 0.057 & 0.072 & 55 \\
\hline
\end{tabular}
\(1000\left({ }_{0.6 \mid 1.5} q_{[52]+1.7}\right)=1000 \frac{l_{[52]+2.3}-l_{[52]+3.8}}{l_{[52]+1.7}}\)
\(l_{\text {[52] }}=1000\)... |
exam-fam-l--question-4.1 | 4.1 | multiple_choice | . For a special whole life insurance policy issued on (40), you are given:
(i) Death benefits are payable at the end of the year of death
(ii) The amount of benefit is 2 if death occurs within the first 20 years and is 1 thereafter
(iii) \(\quad Z\) is the present value random variable for the payments under this insur... | [
[
"A",
"0.27"
],
[
"B",
"0.32"
],
[
"C",
"0.37"
],
[
"D",
"0.42"
],
[
"E",
"0.47"
]
] | A | \(E[Z]=2 \cdot A_{40}-{ }_{20} E_{40} A_{60}=(2)(0.36987)-(0.51276)(0.62567)=0.41892\)
\(E\left[Z^{2}\right]=0.24954\) which is given in the problem.
\(\operatorname{Var}(Z)=E\left[Z^{2}\right]-(E[Z])^{2}=0.24954-0.41892^{2}=0.07405\)
\(S D(Z)=\sqrt{0.07405}=0.27212\)
An alternative way to obtain the mean is \(E[Z]=2 ... |
exam-fam-l--question-4.2 | 4.2 | multiple_choice | . For a special 2 -year term insurance policy on \((x)\), you are given:
(i) Death benefits are payable at the end of the half-year of death
(ii) The amount of the death benefit is 300,000 for the first half-year and increases by 30,000 per half-year thereafter
(iii) \(q_{x}=0.16\) and \(q_{x+1}=0.23\)
(iv) \(\quad i^{... | [
[
"A",
"0.08"
],
[
"B",
"0.11"
],
[
"C",
"0.14"
],
[
"D",
"0.18"
],
[
"E",
"0.21"
]
] | D | \begin{tabular}{|l|l|}
\hline Half-year & PV of Benefit \\
\hline 1 & \(300,000 v^{0.5}=(300,000)(1.09)^{-1}=275,229\) \\
\hline 2 & \(330,000 v^{1}=(330,000)(1.09)^{-2}=277,754\) \\
\hline 3 & \(360,000 v^{1.5}=(360,000)(1.09)^{-3}=277,986\) \\
\hline 4 & \(390,000 v^{2}=(390,000)(1.09)^{-4}=276,286\) \\
\hline
\end{t... |
exam-fam-l--question-4.3 | 4.3 | multiple_choice | . You are given:
(i) \(\quad q_{60}=0.01\)
(ii) Using \(i=0.05, A_{60: 3 \mid}=0.86545\)
Using \(i=0.045\) calculate \(A_{60: 31} \cdot\) | [
[
"A",
"0.866"
],
[
"B",
"0.870"
],
[
"C",
"0.874"
],
[
"D",
"0.878"
],
[
"E",
"0.882"
]
] | D | \[
\begin{aligned}
& A_{60: 3 \mid}=q_{60} v+\left(1-q_{60}\right) q_{60+1} v^{2}+\left(1-q_{60}\right)\left(1-q_{60+1}\right) v^{3}=0.86545 \\
& q_{60+1}=\frac{A_{60: 31}-q_{60} v-\left(1-q_{60}\right) v^{3}}{\left(1-q_{60}\right) v^{2}-\left(1-q_{60}\right) v^{3}}=\frac{0.86545-\frac{0.01}{1.05}-\frac{0.99}{1.05^{3}}... |
exam-fam-l--question-4.4 | 4.4 | multiple_choice | For a special increasing whole life insurance on (40), payable at the moment of death, you are given:
(i) The death benefit at time \(t\) is \(b_{t}=1+0.2 t, \quad t \geq 0\)
(ii) The interest discount factor at time \(t\) is \(v(t)=(1+0.2 t)^{-2}, \quad t \geq 0\)
(iii) \(\quad{ }_{t} p_{40} \mu_{40+t}= \begin{cases}0... | [
[
"A",
"0.036"
],
[
"B",
"0.038"
],
[
"C",
"0.040"
],
[
"D",
"0.042"
],
[
"E",
"0.044"
]
] | A | \[
\begin{aligned}
\operatorname{Var}(Z) & =E\left(Z^{2}\right)-E(Z)^{2} \\
E(Z) & =E\left[(1+0.2 T)(1+0.2 T)^{-2}\right]=E\left[(1+0.2 T)^{-1}\right] \\
& =\int_{0}^{40} \frac{1}{(1+0.2 t)} f_{T}(t) d t=\frac{1}{40} \int_{0}^{40} \frac{1}{1+0.2 t} d t \\
& =\left.\frac{1}{40} \frac{1}{0.2} \ln (1+0.2 t)\right|_{0} ^{4... |
exam-fam-l--question-4.5 | 4.5 | multiple_choice | . For a 30 -year term life insurance of 100,000 on (45), you are given:
(i) The death benefit is payable at the moment of death
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\delta=0.05\)
(iv) Deaths are uniformly distributed over each year of age
Calculate the \(95{ }^{\text {th }}\) percentile of t... | [
[
"A",
"\\(\\quad 30,200\\)"
],
[
"B",
"31,200"
],
[
"C",
"35,200"
],
[
"D",
"36,200"
],
[
"E",
"37,200"
]
] | C | The earlier the death (before year 30), the larger the loss. Since we are looking for the \(95{ }^{\text {th }}\) percentile of the present value of benefits random variable, we must find the time at which \(5 \%\) of the insureds have died. The present value of the death benefit for that insured is what is being asked... |
exam-fam-l--question-4.6 | 4.6 | multiple_choice | . For a 3 -year term insurance of 1000 on (70), you are given:
(i) \(\quad q_{70+k}^{S U L T}\) is the mortality rate from the Standard Ultimate Life Table, for \(k=0,1,2\)
(ii) \(\quad q_{70+k}\) is the mortality rate used to price this insurance, for \(k=0,1,2\)
(iii) \(\quad q_{70+k}=(0.95)^{k} q_{70+k}^{\text {SULT... | [
[
"A",
"29.05"
],
[
"B",
"29.85"
],
[
"C",
"30.65"
],
[
"D",
"31.45"
],
[
"E",
"32.25"
]
] | B | \begin{tabular}{|l|l|l|l|l|}
\hline Time & Age & \(q_{x}^{\text {SULT }}\) & \begin{tabular}{l}
Improvement \\
factor
\end{tabular} & \(q_{x}\) \\
\hline 0 & 70 & 0.010413 & \(100.00 \%\) & 0.010413 \\
\hline 1 & 71 & 0.011670 & \(95.00 \%\) & 0.011087 \\
\hline 2 & 72 & 0.013081 & \(90.25 \%\) & 0.011806 \\
\hline
\e... |
exam-fam-l--question-4.7 | 4.7 | multiple_choice | . For a 25 -year pure endowment of 1 on ( \(x\) ), you are given:
(i) \(\quad Z\) is the present value random variable at issue of the benefit payment
(ii) \(\operatorname{Var}(Z)=0.10 E[Z]\)
(iii) \(\quad{ }_{25} p_{x}=0.57\)
Calculate the annual effective interest rate. | [
[
"A",
"\\(\\quad 5.8 \\%\\)"
],
[
"B",
"\\(\\quad 6.0 \\%\\)"
],
[
"C",
"\\(\\quad 6.2 \\%\\)"
],
[
"D",
"\\(\\quad 6.4 \\%\\)"
],
[
"E",
"\\(6.6 \\%\\)"
]
] | B | \[
\begin{aligned}
& \operatorname{Var}(Z)=0.10 E[Z] \Rightarrow v_{25}^{50} p_{x}\left(1-{ }_{25} p_{x}\right)=0.10 \cdot v_{25}^{25} p_{x} \\
& \Rightarrow \frac{(1-0.57)}{(1+i)^{50}}=0.10 \times \frac{1}{(1+i)^{25}} \\
& \Rightarrow(1+i)^{25}=\frac{0.43}{0.10}=4.3 \Rightarrow i=0.06
\end{aligned}
\] |
exam-fam-l--question-4.8 | 4.8 | multiple_choice | . For a whole life insurance of 1000 on (50), you are given:
(i) The death benefit is payable at the end of the year of death
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.04\) in the first year, and \(i=0.05\) in subsequent years
Calculate the actuarial present value of this insurance. | [
[
"A",
"187"
],
[
"B",
"189"
],
[
"C",
"191"
],
[
"D",
"193"
],
[
"E",
"195"
]
] | C | Let \(A_{51}^{\text {SULT }}\) designate \(A_{51}\) using the Standard Ultimate Life Table at \(5 \%\).
\[
\begin{aligned}
\text { APV (insurance) } & =1000\left(\frac{1}{1.04}\right)\left(q_{50}+p_{50} A_{51}^{\text {SULT }}\right) \\
& =1000\left(\frac{1}{1.04}\right)[0.001209+(1-0.001209)(0.19780)] \\
& =191.12
\end... |
exam-fam-l--question-4.9 | 4.9 | multiple_choice | . You are given:
(i) \(\quad A_{35: \overline{15 \mid}}=0.39\)
(ii) \(\quad A_{35: 15 \mid}^{1} 0.25\)
(iii) \(\quad A_{35}=0.32\)
Calculate \(A_{50}\). | [
[
"A",
"0.35"
],
[
"B",
"0.40"
],
[
"C",
"0.45"
],
[
"D",
"0.50"
],
[
"E",
"0.55"
]
] | D | \[
\begin{aligned}
A_{35}= & A_{35: 15 \mid}^{1}+A_{35: 15 \mid} \frac{1}{50} A_{50} \\
& 0.32=0.25+0.14 A_{50} \\
& A_{50}=\frac{0.07}{0.14}=0.50
\end{aligned}
\] |
exam-fam-l--question-4.11 | 4.11 | multiple_choice | . You are given:
(i) \(\quad Z_{1}\) is the present value random variable for an \(n\)-year term insurance of 1000 issued to ( \(x\) )
(ii) \(\quad Z_{2}\) is the present value random variable for an \(n\)-year endowment insurance of 1000 issued to ( \(x\) )
(iii) For both \(Z_{1}\) and \(Z_{2}\) the death benefit is p... | [
[
"A",
"143,400"
],
[
"B",
"177,500"
],
[
"C",
"211,200"
],
[
"D",
"245,300"
],
[
"E",
"279,300"
]
] | A | \[
\begin{aligned}
& \begin{array}{r}
\operatorname{Var}\left(Z_{2}\right)=(1000)^{2}\left[{ }^{2} A_{x: n \mid}-\left(A_{x: n \mid}\right)^{2}\right]=15,000 \\
=(1000)^{2}\left({ }^{2} A_{x: n \mid}^{1}+{ }^{2} A_{x: n}\right)-(1000)^{2}\left[A_{x: n}^{1}+A_{x n}^{1}\right]^{2}
\end{array} \\
& =(1000)^{2} A_{x: n}^{1... |
exam-fam-l--question-4.12 | 4.12 | multiple_choice | . For three fully discrete insurance products on the same ( \(x\) ), you are given:
(i) \(\quad Z_{1}\) is the present value random variable for a 20-year term insurance of 50
(ii) \(\quad Z_{2}\) is the present value random variable for a 20-year deferred whole life insurance of 100
(iii) \(\quad Z_{3}\) is the presen... | [
[
"A",
"62"
],
[
"B",
"109"
],
[
"C",
"167"
],
[
"D",
"202"
],
[
"E",
"238"
]
] | C | \[
Z_{3}=2 Z_{1}+Z_{2} \text { so that } \operatorname{Var}\left(Z_{3}\right)=4 \operatorname{Var}\left(Z_{1}\right)+\operatorname{Var}\left(Z_{2}\right)+4 \operatorname{Cov}\left(Z_{1}, Z_{2}\right)
\]
\[
\begin{aligned}
& \text { where } \operatorname{Cov}\left(Z_{1}, Z_{2}\right)=\underbrace{E\left[Z_{1} Z_{2}\right... |
exam-fam-l--question-4.13 | 4.13 | multiple_choice | . For a 2 -year deferred, 2 -year term insurance of 2000 on [65], you are given:
(i) The following select and ultimate mortality 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 65 & 0.08 & 0.10 & 0.12 & 0.14 ... | [
[
"A",
"260"
],
[
"B",
"290"
],
[
"C",
"350"
],
[
"D",
"370"
],
[
"E",
"410"
]
] | C | \[
\begin{aligned}
{ }_{2 \mid 2} A_{65}= & \underbrace{v^{3}}_{\text {payment year 3 }} \underbrace{{ }_{2} p_{[65]}}_{\text {Lives 2 years }} \times \underbrace{q_{[65]+2}}_{\text {Die year 3 }} \\
& +\underbrace{v^{4}}_{\text {payment year 4 }} \underbrace{{ }_{3} p_{[65]}}_{\text {Lives 3 years }} \times \underbrac... |
exam-fam-l--question-4.14 | 4.14 | multiple_choice | . A fund is established for the benefit of 400 workers all age 60 with independent future lifetimes. When they reach age 85 , the fund will be dissolved and distributed to the survivors.
The fund will earn interest at a rate of \(5 \%\) per year.
The initial fund balance, \(F\), is determined so that the probability t... | [
[
"A",
"350,000"
],
[
"B",
"360,000"
],
[
"C",
"370,000"
],
[
"D",
"380,000"
],
[
"E",
"390,000"
]
] | E | Out of 400 lives initially, we expect \(400_{25} p_{60}=400 \frac{l_{85}}{l_{60}}=400\left(\frac{61,184.9}{96,634.1}\right)=253.26\) survivors
The standard deviation of the number of survivors is \(\sqrt{400_{25} p_{60}\left(1-{ }_{25} p_{60}\right)}=9.639\)
To ensure \(86 \%\) funding, using the normal distribution t... |
exam-fam-l--question-4.15 | 4.15 | multiple_choice | . For a special whole life insurance on ( \(x\) ), you are given:
(i) Death benefits are payable at the moment of death
(ii) The death benefit at time \(t\) is \(b_{t}=e^{0.02 t}\), for \(t \geq 0\)
(iii) \(\quad \mu_{x+t}=0.04\), for \(t \geq 0\)
(iv) \(\delta=0.06\)
(v) \(\quad Z\) is the present value at issue rando... | [
[
"A",
"0.020"
],
[
"B",
"0.036"
],
[
"C",
"0.052"
],
[
"D",
"0.068"
],
[
"E",
"0.083"
]
] | E | \(E[Z]=\int_{0}^{\infty} b_{t} \cdot v^{t} \cdot{ }_{t} p_{x} \cdot \mu_{x+t} d t=\int_{0}^{\infty} e^{0.02 t} \cdot e^{-0.06 t} \cdot e^{-0.04 t} \cdot 0.04 d t\)
\(=0.04 \int_{0}^{\infty} e^{-0.08 t} d t=\frac{0.04}{0.08}=\frac{1}{2}\)
\(E\left[Z^{2}\right]=\int_{0}^{\infty}\left(b_{t} \cdot v^{t}\right)^{2}{ }_{t} p... |
exam-fam-l--question-4.16 | 4.16 | multiple_choice | . You are given the following extract of ultimate mortality rates from a two-year select and ultimate mortality table:
\begin{tabular}{|c|c|}
\hline\(x\) & \(q_{x}\) \\
\hline 50 & 0.045 \\
\hline 51 & 0.050 \\
\hline 52 & 0.055 \\
\hline 53 & 0.060 \\
\hline
\end{tabular}
The select mortality rates satisfy the follo... | [
[
"A",
"0.08"
],
[
"B",
"0.09"
],
[
"C",
"0.10"
],
[
"D",
"0.11"
],
[
"E",
"0.12"
]
] | D | \(A_{[50]: 31}^{1}=v q_{[50]}+v^{2} p_{[50]} q_{[50]+1}+v^{3} p_{[50]} p_{[50]+1} q_{52}\)
where: \(v=\frac{1}{1.04}\)
\(q_{[50]}=0.7(0.045)=0.0315\)
\(p_{[50]}=1-q_{[50]}=0.9685\)
\(q_{[50]+1}=0.8(0.050)=0.040\)
\(p_{[50]+1}=1-q_{[50]+1}=0.960\)
\(q_{52}=0.055\)
So: \(A_{[50]: 31}^{1}=0.1116\) |
exam-fam-l--question-4.17 | 4.17 | multiple_choice | . For a special whole life policy on (48), you are given:
(i) The policy pays 5000 if the insured's death is before the median curtate future lifetime at issue and 10,000 if death is after the median curtate future lifetime at issue
(ii) Mortality follows the Standard Ultimate Life Table
(iii) Death benefits are paid a... | [
[
"A",
"1130"
],
[
"B",
"1160"
],
[
"C",
"1190"
],
[
"D",
"1220"
],
[
"E",
"1250"
]
] | A | The median of \(K_{48}\) is the integer \(m\) for which
\[
P\left(K_{48}<m\right) \leq 0.5 \text { and } P\left(K_{48}>m\right) \leq 0.5
\]
This is equivalent to finding \(m\) for which
\(\frac{l_{48+m}}{l_{48}} \geq 0.5\) and \(\frac{l_{48+m+1}}{l_{48}} \leq 0.5\).
Based on the SULT and \(l_{48}(0.5)=(98,783.9)(0.5)=... |
exam-fam-l--question-4.18 | 4.18 | multiple_choice | . You are given that \(T\), the time to first failure of an industrial robot, has a density \(f(t)\) given by
\[
f(t)= \begin{cases}0.1, & 0 \leq t<2 \\ 0.4 t^{-2}, & 2 \leq t<10\end{cases}
\]
with \(f(t)\) undetermined on \([10, \infty)\).
Consider a supplemental warranty on this robot that pays 100,000 at the time \(... | [
[
"A",
"\\(\\quad 82,000\\)"
],
[
"B",
"84,000"
],
[
"C",
"87,000"
],
[
"D",
"91,000"
],
[
"E",
"95,000"
]
] | A | The present value random variable \(\mathrm{PV}=1,000,000 e^{-0.05 T}, 2 \leq T \leq 10\) is a decreasing function of \(T\) so that its \(90^{\text {th }}\) percentile is
\(1,000,000 e^{-0.05 p}\) where \(\boldsymbol{p}\) is the solution to \(\int_{2}^{p} 0.4 t^{-2} d t=0.10\).
\[
\begin{aligned}
& \int_{2}^{p} 0.4 t^{... |
exam-fam-l--question-4.19 | 4.19 | multiple_choice | . (80) purchases a whole life insurance policy of 100,000 .
You are given:
(i) The policy is priced with a select period of one year
(ii) The select mortality rate equals \(80 \%\) of the mortality rate from the Standard Ultimate Life Table
(iii) Ultimate mortality follows the Standard Ultimate Life Table
(iv) \(\quad... | [
[
"A",
"\\(\\quad 58,950\\)"
],
[
"B",
"59,050"
],
[
"C",
"\\(\\quad 59,150\\)"
],
[
"D",
"59,250"
],
[
"E",
"59,350"
]
] | B | Superscript SULT refers to values from the SULT. Values without superscripts refer to this select life.
\[
\begin{aligned}
q_{80} & =0.8 q_{80}^{S U L T}=0.0261264 \Rightarrow p_{80}=0.9738736 \\
A_{80} & =v q_{80}+v p_{80} A_{81}^{S U L T} \\
& =(1.05)^{-1}(0.0261264)+(1.05)^{-1}(0.9738736)(0.60984)=0.59051
\end{align... |
exam-fam-l--question-4.20 | 4.20 | multiple_choice | . For a special fully continuous whole life insurance on ( \(x\) ), you are given:
i) \(\quad \mu_{x+t}=0.03, t \geq 0\)
ii) \(\delta=0.06\)
iii) The death benefit at time \(t\) is \(b_{t}=e^{0.05 t}, t \geq 0\)
iv) \(\quad Z\) is the present value random variable at issue for this insurance
Calculate \(\operatorname{... | [
[
"A",
"0.0300"
],
[
"B",
"0.0325"
],
[
"C",
"0.0350"
],
[
"D",
"0.0375"
],
[
"E",
"0.0400"
]
] | D | \[
\begin{aligned}
E(Z) & =\int_{0}^{\infty} p_{x} \times \mu \times e^{0.05 \times t} e^{-\delta \times t} d t \\
& =\int_{0}^{\infty} e^{-0.03 \times t} \times 0.03 \times e^{0.05 \times t} e^{-0.06 \times t} d t \\
& =\frac{0.03}{0.04} \times\left. e^{-0.04 t}\right|_{0} ^{\infty}=0.75 \\
E\left(Z^{2}\right) & =\int... |
exam-fam-l--question-4.21 | 4.21 | multiple_choice | . For a two-year term insurance of 1 on ( \(x\) ) payable at the moment of death, you are given:
i) \(q_{x}=0.04\)
ii) \(\quad q_{x+1}=0.06\)
iii) Deaths are uniformly distributed over each year of age
iv) \(i=0.04\)
v) \(\quad Z\) is the present value random variable for this insurance
Calculate Var[Z]. | [
[
"A",
"0.065"
],
[
"B",
"0.069"
],
[
"C",
"0.073"
],
[
"D",
"0.077"
],
[
"E",
"0.081"
]
] | E | \[
\begin{aligned}
& E(Z)=\frac{i}{\delta}\left[v q_{x}+v^{2} p_{x} q_{x+1}\right]=\frac{0.04}{\ln (1.04)}\left[\frac{1}{1.04}(0.04)+\frac{1}{1.04^{2}}(0.96)(0.06)\right]=0.09353831 \\
& E\left(Z^{2}\right)=\frac{(1+i)^{2}-1}{2 \delta}\left[v^{2} q_{x}+v^{4} p_{x} q_{x+1}\right]=\frac{\left(1.04^{2}-1\right)}{2 \ln (1.... |
exam-fam-l--question-4.22 | 4.22 | multiple_choice | . (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 fir... | [
[
"A",
"43,000"
],
[
"B",
"44,000"
],
[
"C",
"45,000"
],
[
"D",
"46,000"
],
[
"E",
"47,000"
]
] | A | 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}... |
exam-fam-l--question-5.1 | 5.1 | multiple_choice | 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 \(\operatorname{Pr}(Y>\mathrm{E}[Y])\). | [
[
"A",
"0.705"
],
[
"B",
"0.710"
],
[
"C",
"0.715"
],
[
"D",
"0.720"
],
[
"E",
"0.725"
]
] | A | \(E(Y)=\bar{a}_{\overline{10} \mid}+e^{-\delta(10)} e^{-\mu(10)} \bar{a}_{x+10}\)
\(=\frac{\left(1-e^{-0.6}\right)}{0.06}+e^{-0.7} \frac{1}{0.07}\)
=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... |
exam-fam-l--question-5.2 | 5.2 | multiple_choice | 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 \(a_{x: n} \cdot\) | [
[
"A",
"9.3"
],
[
"B",
"9.6"
],
[
"C",
"9.8"
],
[
"D",
"10.0"
],
[
"E",
"10.3"
]
] | B | \(A_{x: \frac{1}{n}}={ }_{n} E_{x}\)
\(A_{x}=A_{x: n \mid}^{1}+{ }_{n} E_{x} A_{x+n}\)
\(0.3=A_{x: n \mid}^{1}+(0.35)(0.4) \Rightarrow A_{x: n \mid}^{1}=0.16\)
\(A_{x: n \mid}=A_{x: n \mid}^{1}+{ }_{n} E_{x}=0.16+0.35=0.51\)
\(\ddot{a}_{x: \bar{n}}=\frac{1-A_{x: \bar{n}}}{d}=\frac{1-0.51}{(0.05 / 1.05)}=10.29\)
\(a_{x:... |
exam-fam-l--question-5.3 | 5.3 | multiple_choice | . 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 \(t=10.5\). | [
[
"A",
"5.8"
],
[
"B",
"6.0"
],
[
"C",
"6.2"
],
[
"D",
"6.4"
],
[
"E",
"\\(\\quad 6.6\\)"
]
] | C | \((\bar{I} \bar{a})_{40: \bar{t}}=\int_{0}^{t} s_{s} p_{40} v^{s} d s \Rightarrow \frac{d(\bar{I} \bar{a})_{40: \bar{t}}}{d t}=t_{t} p_{40} v^{t}\)
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)(0.9759... |
exam-fam-l--question-5.4 | 5.4 | multiple_choice | (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) \(\delt... | [
[
"A",
"214"
],
[
"B",
"216"
],
[
"C",
"218"
],
[
"D",
"220"
],
[
"E",
"222"
]
] | A | \(\stackrel{\circ}{e}_{40}=\frac{1}{\mu}=50\) 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\... |
exam-fam-l--question-5.5 | 5.5 | multiple_choice | . 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 substandar... | [
[
"A",
"1700"
],
[
"B",
"1710"
],
[
"C",
"1720"
],
[
"D",
"1730"
],
[
"E",
"1740"
]
] | A | \(\ddot{a}_{45}^{S}=1+v p_{45}^{S} \ddot{a}_{46}^{S U L T}\)
\(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}\rig... |
exam-fam-l--question-5.6 | 5.6 | multiple_choice | 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 nor... | [
[
"A",
"1170"
],
[
"B",
"1180"
],
[
"C",
"1190"
],
[
"D",
"1200"
],
[
"E",
"1210"
]
] | D | 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 \(Y=\sum_{i=1}^{100} Y_{i}\) ... |
exam-fam-l--question-5.7 | 5.7 | multiple_choice | 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. | [
[
"A",
"\\(\\quad 17,060\\)"
],
[
"B",
"17,310"
],
[
"C",
"17,380"
],
[
"D",
"17,490"
],
[
"E",
"17,530"
]
] | C | \[
\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}}... |
exam-fam-l--question-5.8 | 5.8 | multiple_choice | . 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 ... | [
[
"A",
"0.88"
],
[
"B",
"0.90"
],
[
"C",
"0.92"
],
[
"D",
"0.94"
],
[
"E",
"\\(\\quad 0.96\\)"
]
] | C | 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 ... |
exam-fam-l--question-5.9 | 5.9 | multiple_choice | . 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 \(k\). | [
[
"A",
"0.005"
],
[
"B",
"0.010"
],
[
"C",
"0.015"
],
[
"D",
"0.020"
],
[
"E",
"0.025"
]
] | C | \[
\ddot{a}_{[x]: \bar{n}]}=1+v p_{[x]} \ddot{a}_{x+1: \overline{n-1}]}=1+(1+k)\left(v p_{x} \ddot{a}_{x+1: \overline{n-1}}\right)=1+(1+k)\left(\ddot{a}_{x: \bar{n}]}-1\right)
\]
Therefore, we have
\[
k=\frac{\ddot{a}_{[x]: n]}-1}{\ddot{a}_{x: n]}-1}-1=\frac{21.167}{20.854}-1=0.015
\] |
exam-fam-l--question-5.10 | 5.10 | multiple_choice | . 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 annu... | [
[
"A",
"0.826"
],
[
"B",
"0.836"
],
[
"C",
"\\(\\mathbf{0 . 8 4 6}\\)"
],
[
"D",
"0.856"
],
[
"E",
"\\(\\quad 0.866\\)"
]
] | C | \[
\ddot{a}_{\overline{65: \overline{10}}}=\ddot{a}_{10 \mid}+{ }_{10} E_{65} \times \ddot{a}_{75}=8.10782+0.55305 \times 10.3178=13.8141
\]
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 requi... |
exam-fam-l--question-5.11 | 5.11 | multiple_choice | . 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. | [
[
"A",
"\\(\\quad 4.4\\)"
],
[
"B",
"\\(\\quad 4.5\\)"
],
[
"C",
"\\(\\quad 4.6\\)"
],
[
"D",
"4.7"
],
[
"E",
"\\(\\quad 4.8\\)"
]
] | B | \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](... |
exam-fam-l--question-5.12 | 5.12 | multiple_choice | . 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\... | [
[
"A",
"17,400"
],
[
"B",
"17,500"
],
[
"C",
"17,600"
],
[
"D",
"17,700"
],
[
"E",
"17,800"
]
] | E | 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... |
exam-fam-l--question-5.13 | 5.13 | multiple_choice | . 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. | [
[
"A",
"13.792"
],
[
"B",
"13.796"
],
[
"C",
"13.862"
],
[
"D",
"13.866"
],
[
"E",
"13.870"
]
] | D | \[
\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}... |
exam-fam-l--question-6.1 | 6.1 | multiple_choice | . 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 t... | [
[
"A",
"32"
],
[
"B",
"33"
],
[
"C",
"34"
],
[
"D",
"35"
],
[
"E",
"36"
]
] | D | 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 obta... |
exam-fam-l--question-6.2 | 6.2 | multiple_choice | 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) E... | [
[
"A",
"3200"
],
[
"B",
"3300"
],
[
"C",
"3400"
],
[
"D",
"3500"
],
[
"E",
"3600"
]
] | E | \(G \ddot{a}_{x: \overline{10} \mid}=100,000 A_{x: \overline{10} \mid}^{1}+G(I A)_{x: \overline{10} \mid}^{1}+0.45 G+0.05 G \ddot{a}_{x: \overline{10} \mid}+200 \ddot{a}_{x: \overline{10} \mid}\)
\(G=\frac{(100,000)(0.17094)+200(6.8865)}{(1-0.05)(6.8865)-0.96728-0.45}=3604.23\) |
exam-fam-l--question-6.3 | 6.3 | multiple_choice | . \(S\), 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 L... | [
[
"A",
"0.35"
],
[
"B",
"0.37"
],
[
"C",
"0.39"
],
[
"D",
"0.41"
],
[
"E",
"0.43"
]
] | C | 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{... |
exam-fam-l--question-6.4 | 6.4 | multiple_choice | . 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 prese... | [
[
"A",
"1850"
],
[
"B",
"1860"
],
[
"C",
"1870"
],
[
"D",
"1880"
],
[
"E",
"1890"
]
] | E | 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.
\(E\left[X_{i}\right]=\ddot{a}_{62}^{(12)}=\frac{1-A_{62}^{(12)}}{d^{(12)}}=\frac{1-0.4075}{0.05813}=10.19267\)
\(\ope... |
exam-fam-l--question-6.5 | 6.5 | multiple_choice | . 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 v... | [
[
"A",
"21"
],
[
"B",
"25"
],
[
"C",
"29"
],
[
"D",
"33"
],
[
"E",
"37"
]
] | D | 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}{19.3... |
exam-fam-l--question-6.6 | 6.6 | multiple_choice | . 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... | [
[
"A",
"0.75"
],
[
"B",
"0.79"
],
[
"C",
"0.83"
],
[
"D",
"0.87"
],
[
"E",
"\\(\\quad 0.91\\)"
]
] | B | 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}-(2... |
exam-fam-l--question-6.7 | 6.7 | multiple_choice | . For a special fully discrete 20 -year endowment insurance on (40), you are given:
(i) The only death benefit is the return of annual net premiums accumulated with interest at \(5 \%\) to the end of the year of death
(ii) The endowment benefit is 100,000
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(... | [
[
"A",
"2680"
],
[
"B",
"2780"
],
[
"C",
"2880"
],
[
"D",
"2980"
],
[
"E",
"3080"
]
] | C | There are four ways to approach this problem. In all cases, let \(\pi\) denote the net premium.
The first approach is an intuitive result. The key is that in addition to the pure endowment, there is a benefit equal in value to a temporary interest only annuity due with annual payment \(\pi\). However, if the insured su... |
exam-fam-l--question-6.8 | 6.8 | multiple_choice | . 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
... | [
[
"A",
"7.5"
],
[
"B",
"9.5"
],
[
"C",
"11.5"
],
[
"D",
"13.5"
],
[
"E",
"15.5"
]
] | B | \(\ddot{a}_{60: \overline{10}}=7.9555\)
\(\ddot{a}_{60: 201}=12.3816\)
Annual level amount \(=\frac{40+5 \ddot{a}_{60: \overline{10}}+5 \ddot{a}_{60: \overline{20}}}{\ddot{a}_{60}}=\frac{141.686}{14.9041}=9.51\) |
exam-fam-l--question-6.9 | 6.9 | multiple_choice | . 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 & ... | [
[
"A",
"617"
],
[
"B",
"627"
],
[
"C",
"637"
],
[
"D",
"647"
],
[
"E",
"657"
]
] | D | \(\ddot{a}_{50: \overline{10} \mid}=8.0550\)
\(A_{50: 20 \mid}^{1}=A_{50: 20 \mid}-{ }_{20} E_{50}=0.38844-0.34824=0.04020\)
\(\ddot{a}_{50: 20 \mid}=12.8428\)
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{... |
exam-fam-l--question-6.10 | 6.10 | multiple_choice | . 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 \(p_{x+2}\). | [
[
"A",
"0.88"
],
[
"B",
"0.89"
],
[
"C",
"0.90"
],
[
"D",
"0.91"
],
[
"E",
"0.92"
]
] | D | \(\ddot{a}_{x: 3 \mid}=\frac{\text { Actuarial PV of the benefit }}{\text { Level Annual Premium }}=\frac{152.85}{56.05}=2.727\)
\(\ddot{a}_{x: 31}=1+\frac{0.975}{1.06}+\frac{0.975\left(p_{x+1}\right)}{(1.06)^{2}}=2.727\)
\(\Rightarrow p_{x+1}=0.93\)
Actuarial PV of the benefit \(=\)
\(152.85=1,000\left[\frac{0.025}{1.... |
exam-fam-l--question-6.12 | 6.12 | multiple_choice | . For a fully discrete whole life insurance of 1000 on ( \(x\) ), you are given:
(i) The following expenses are incurred at the beginning of each year:
\begin{tabular}{|l|c|c|}
\hline & Year 1 & Years 2+ \\
\hline Percent of premium & \(75 \%\) & \(10 \%\) \\
\hline Maintenance expenses & 10 & 2 \\
\hline
\end{tabular... | [
[
"A",
"14,600"
],
[
"B",
"33,100"
],
[
"C",
"51,700"
],
[
"D",
"70,300"
],
[
"E",
"\\(\\quad 88,900\\)"
]
] | E | 1,020 in the solution is the 1,000 death benefit plus the 20 death benefit claim expense.
\(A_{x}=1-d \ddot{a}_{x}=1-d(12.0)=0.320755\)
\(G \ddot{a}_{x}=1,020 A_{x}+0.65 G+0.10 G \ddot{a}_{x}+8+2 \ddot{a}_{x}\)
\(G=\frac{1,020 A_{x}+8+2 \ddot{a}_{x}}{\ddot{a}_{x}-0.65-0.10 \ddot{a}_{x}}=\frac{1,020(0.320755)+8+2(12.0)}... |
exam-fam-l--question-6.13 | 6.13 | multiple_choice | . For a fully discrete whole life insurance of 10,000 on (45), you are given:
(i) Commissions are \(80 \%\) of the first year premium and \(10 \%\) of subsequent premiums. There are no other expenses
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)
(iv) \(\quad{ }_{0} L\) denotes the loss ... | [
[
"A",
"-580"
],
[
"B",
"-520"
],
[
"C",
"-460"
],
[
"D",
"\\(\\quad-400\\)"
],
[
"E",
"\\(\\quad-340\\)"
]
] | D | If \(T_{45}=10.5\), then \(K_{45}=10\) and \(K_{45}+1=11\).
\[
\begin{aligned}
& { }_{0} L=10,000 v^{K_{45}+1}-G(1-0.10) \ddot{a}_{\overline{K_{45}+1}}+G(0.80-0.10)=10,000 v^{11}-0.9 G \ddot{a}_{\overline{11}}+0.7 G \\
& 4953=10,000(0.58468)-0.9 G(8.72173)+0.7 G \\
& G=(5846.8-4953) /(7.14956)=125.01 \\
& E\left({ }_{0... |
exam-fam-l--question-6.15 | 6.15 | multiple_choice | . For a fully discrete whole life insurance of 1000 on \((x)\) with net premiums payable quarterly, you are given:
(i) \(\quad i=0.05\)
(ii) \(\quad \ddot{a}_{x}=3.4611\)
(iii) \(\quad P^{(W)}\) and \(P^{(U D D)}\) are the annualized net premiums calculated using the 2-term Woolhouse ( \(W\) ) and the uniform distribut... | [
[
"A",
"1.000"
],
[
"B",
"1.002"
],
[
"C",
"1.004"
],
[
"D",
"1.006"
],
[
"E",
"1.008"
]
] | B | Woolhouse: \(\quad{ }^{W} \ddot{a}_{x}^{(4)}=3.4611-\frac{3}{8}=3.0861\)
\[
{ }^{U D D} \ddot{a}_{x}^{(4)}=\alpha(4) \ddot{a}_{x}-\beta(4)
\]
UDD:
\[
\begin{aligned}
& =1.00019(3.4611)-0.38272 \\
& =3.0790
\end{aligned}
\]
and
\[
A_{x}=1-d \ddot{a}_{x}=1-(0.04762)(3.4611)=0.83518
\]
\(P^{(W)}=\frac{1000(0.83518)}{3.08... |
exam-fam-l--question-6.16 | 6.16 | multiple_choice | . For a fully discrete 20 -year endowment insurance of 100,000 on (30), you are given:
(i) \(\quad d=0.05\)
(ii) Expenses, payable at the beginning of each year, are:
\begin{tabular}{|l|c|c|c|c|}
\hline & \multicolumn{2}{|c|}{ First Year } & \multicolumn{2}{c|}{ Renewal Years } \\
\hline & \begin{tabular}{c}
Percent ... | [
[
"A",
"2410"
],
[
"B",
"2530"
],
[
"C",
"2800"
],
[
"D",
"3130"
],
[
"E",
"3280"
]
] | A | \(P_{30: 201}=\frac{1}{\ddot{a}_{30: 201}}-d \Rightarrow \frac{2,143}{100,000}+0.05=\frac{1}{\ddot{a}_{30: 201}} \Rightarrow \ddot{a}_{30: 201}=14\)
\(A_{30: \overline{20}}=1-d \ddot{a}_{30: \overline{20}}=1-0.05(14)=0.3\)
\(G \ddot{a}_{30: \overline{20}}=100,000 A_{30: \overline{20}}+\left(200+50 \ddot{a}_{30: \overli... |
exam-fam-l--question-6.17 | 6.17 | multiple_choice | . An insurance company sells special fully discrete two-year endowment insurance policies to smokers (S) and non-smokers (NS) age \(x\). You are given:
(i) The death benefit is 100,000 ; the maturity benefit is 30,000
(ii) The level annual premium for non-smoker policies is determined by the equivalence principle
(iii)... | [
[
"A",
"\\(-30,000\\)"
],
[
"B",
"\\(-29,000\\)"
],
[
"C",
"\\(\\quad-28.000\\)"
],
[
"D",
"\\(\\quad-27.000\\)"
],
[
"E",
"-26.000"
]
] | A | \[
q_{x}^{\mathrm{NS}}=q_{x+1}^{\mathrm{NS}}=1-e^{-0.1}=0.095
\]
Then the annual premium for the non-smoker policies is \(P^{\mathrm{NS}}\), where
\[
\begin{aligned}
P^{\mathrm{NS}}\left(1+v p_{x}^{\mathrm{NS}}\right) & =100,000 v q_{x}^{\mathrm{NS}}+100,000 v^{2} p_{x}^{\mathrm{NS}} q_{x+1}^{\mathrm{NS}}+30,000 v^{2}... |
exam-fam-l--question-6.18 | 6.18 | multiple_choice | . For a 20 -year deferred whole life annuity-due with annual payments of 30,000 on (40), you are given:
(i) The single net premium is refunded without interest at the end of the year of death if death occurs during the deferral period
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)
Calcu... | [
[
"A",
"162,000"
],
[
"B",
"164,000"
],
[
"C",
"165,200"
],
[
"D",
"166,400"
],
[
"E",
"168,800"
]
] | D | \[
\begin{aligned}
P & =30,000_{20 \mid} \ddot{a}_{40}+P A_{40: 20 \mid}^{1} \\
\Rightarrow \quad P & =30,000_{20 \mid} \ddot{a}_{40} /\left(1-A_{40: 20 \mid}^{1}\right) \\
& =30,000(5.46429) /(1-0.0146346)=166,363
\end{aligned}
\] |
exam-fam-l--question-6.19 | 6.19 | multiple_choice | . For a fully discrete whole life insurance of 1 on (50), you are given:
(i) Expenses of 0.20 at the start of the first year and 0.01 at the start of each renewal year are incurred
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)
(iv) Gross premiums are determined using the equivalence pri... | [
[
"A",
"0.023"
],
[
"B",
"0.028"
],
[
"C",
"0.033"
],
[
"D",
"0.038"
],
[
"E",
"0.043"
]
] | C | Let \(\pi\) be the annual premium, so that \(\pi \ddot{a}_{50}=A_{50}+0.01 \ddot{a}_{50}+0.19\)
\(\Rightarrow \pi=\frac{A_{50}+0.19}{\ddot{a}_{50}}+0.01=\frac{0.18931+0.19}{17.0245}+0.01=0.03228\)
Loss at issue: \(L_{0}=v^{k+1}-(\pi-0.01) \ddot{a}_{\overline{k+1}}\left(1-v^{k+1}\right) / d+0.19\)
\[
\begin{aligned}
\Ri... |
exam-fam-l--question-6.20 | 6.20 | multiple_choice | . For a special fully discrete 3 -year term insurance on (75), you are given:
(i) The death benefit during the first two years is the sum of the net premiums paid without interest
(ii) The death benefit in the third year is 10,000
(iii)
\begin{tabular}{|c|c|}
\hline\(x\) & \(p_{x}\) \\
\hline 75 & 0.90 \\
76 & 0.88 \\... | [
[
"A",
"449"
],
[
"B",
"459"
],
[
"C",
"469"
],
[
"D",
"479"
],
[
"E",
"489"
]
] | B | EPV \((\) premiums \()=\) EPV \((\) benefits \()\)
\(P\left(1+v p_{x}+v^{2}{ }_{2} p_{x}\right)=P\left(v q_{x}+2 v^{2} p_{x} q_{x+1}\right)+10000\left(v^{3}{ }_{2} p_{x} q_{x+2}\right)\)
\(P\left(1+\frac{0.9}{1.04}+\frac{0.9 \times 0.88}{1.04^{2}}\right)=P\left(\frac{0.1}{1.04}+\frac{2 \times 0.9 \times 0.12}{1.04^{2}}... |
exam-fam-l--question-6.21 | 6.21 | multiple_choice | . For a special fully discrete 15 -year endowment insurance on (75), you are given:
(i) The death benefit is 1000
(ii) The endowment benefit is the sum of the net premiums paid without interest
(iii) \(\quad d=0.04\)
(iv) \(\quad A_{75: \overline{15} \mid}=0.70\)
(v) \(\quad A_{75: 15} \frac{1}{15}=0.11\)
Calculate th... | [
[
"A",
"80"
],
[
"B",
"90"
],
[
"C",
"100"
],
[
"D",
"110"
],
[
"E",
"120"
]
] | C | \(P \times \ddot{a}_{75: 15 \mid}=1000\left(A_{75: 15 \mid}^{1}+15 \times P \times A_{75: 15 \mid}\right) \rightarrow P=\frac{1000 A_{75: 15 \mid}^{1}}{\ddot{a}_{75: 15 \mid}-15 \times A_{75: 15 \mid} \frac{1}{15 \mid}}\)
\(A_{75: \overline{15} \mid}^{1}=A_{75: \overline{15} \mid}-A_{75: \overline{15} \mid}=0.7-0.11=0.... |
exam-fam-l--question-6.22 | 6.22 | multiple_choice | . For a whole life insurance of 100,000 on (45) with premiums payable monthly for a period of 20 years, you are given:
(i) The death benefit is paid immediately upon death
(ii) Mortality follows the Standard Ultimate Life Table
(iii) Deaths are uniformly distributed over each year of age
(iv) \(\quad i=0.05\)
Calculat... | [
[
"A",
"98"
],
[
"B",
"100"
],
[
"C",
"102"
],
[
"D",
"104"
],
[
"E",
"106"
]
] | C | Let the monthly net premium \(=\pi\)
\[
12 \pi=\frac{100,000 \bar{A}_{45}}{\ddot{a}_{45: 20}^{(12)}} \quad \begin{aligned}
& \alpha(12)=1.00020 \\
& \beta(12)=0.46651 \\
& \frac{i}{\delta}=1.02480
\end{aligned}
\]
\[
\begin{aligned}
& 100,000 \bar{A}_{45}=100,000 \frac{i}{\delta} A_{45}=(1.02480)(15,161)=15,536.99 \\
&... |
exam-fam-l--question-6.23 | 6.23 | multiple_choice | . For fully discrete 30 -payment whole life insurance policies on ( \(x\) ), you are given:
(i) The following expenses payable at the beginning of the year:
\begin{tabular}{|l|c|c|c|c|}
\hline & \(1^{\text {st }}\) Year & \begin{tabular}{c}
Years \\
\(2-15\)
\end{tabular} & \begin{tabular}{c}
Years \\
\(16-30\)
\end... | [
[
"A",
"30.3"
],
[
"B",
"35.1"
],
[
"C",
"39.9"
],
[
"D",
"44.7"
],
[
"E",
"49.5"
]
] | D | \[
\begin{aligned}
& G \ddot{a}_{x: 301}=\operatorname{APV}[\text { gross premium }]=\operatorname{APV}[\text { Benefits }+ \text { expenses }] \\
& \quad=F A_{x}+\left(30+30 \ddot{a}_{x}\right)+G\left(0.6+0.10 \ddot{a}_{x: 301}+0.10 \ddot{a}_{x: 151}\right) \\
& \begin{aligned}
G & =\frac{F A_{x}+30+30 \ddot{a}_{x}}{\... |
exam-fam-l--question-6.24 | 6.24 | multiple_choice | . For a fully continuous whole life insurance of 1 on ( \(x\) ), you are given:
(i) \(\quad L\) is the present value of the loss at issue random variable if the premium rate is determined by the equivalence principle
(ii) \(\quad L^{*}\) is the present value of the loss at issue random variable if the premium rate is 0... | [
[
"A",
"0.18"
],
[
"B",
"0.21"
],
[
"C",
"0.24"
],
[
"D",
"0.27"
],
[
"E",
"0.30"
]
] | E | In general, the loss at issue random variable can be expressed as:
\(L=\bar{Z}_{x}-P \times \bar{Y}_{x}=\bar{Z}_{x}-P \times\left(\frac{1-\bar{Z}_{x}}{\delta}\right)=\bar{Z}_{x} \times\left(1+\frac{P}{\delta}\right)-\frac{P}{\delta}\)
Using actuarial equivalence to determine the premium rate:
\[
\begin{aligned}
P & =\f... |
exam-fam-l--question-6.25 | 6.25 | multiple_choice | . For a fully discrete 10 -year deferred whole life annuity-due of 1000 per month on (55), you are given:
(i) The premium, \(G\), will be paid annually at the beginning of each year during the deferral period
(ii) Expenses are expected to be 300 per year for all years, payable at the beginning of the year
(iii) Mortali... | [
[
"A",
"\\(\\quad 12,110\\)"
],
[
"B",
"12,220"
],
[
"C",
"12,330"
],
[
"D",
"12,440"
],
[
"E",
"12,550"
]
] | C | Need EPV(Ben + Exp) - EPV(Prem) \(=-800\)
\(\operatorname{EPV}(\) Prem \()=G \ddot{a}_{55: \overline{10}}=8.0192 G\)
\(\operatorname{EPV}(B e n+\operatorname{Exp})=12,000_{10} \ddot{a}_{55}^{(12)}+300 \ddot{a}_{55}\)
\[
\begin{aligned}
& =12,000_{10} E_{55} \ddot{a}_{65}^{(12)}+300 \ddot{a}_{55} \\
& =12,000_{10} E_{55... |
exam-fam-l--question-6.26 | 6.26 | multiple_choice | . For a special fully discrete whole life insurance policy of 1000 on (90), you are given:
(i) The first year premium is 0
(ii) \(P\) is the renewal premium
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)
(v) Premiums are calculated using the equivalence principle Calculate \(P\). | [
[
"A",
"150"
],
[
"B",
"160"
],
[
"C",
"170"
],
[
"D",
"180"
],
[
"E",
"190"
]
] | D | EPV \((\) Premiums \()=P a_{90}=P\left(\ddot{a}_{90}-1\right)=(4.1835) P\)
\(\operatorname{EPV}(\) Benefits \()=1000 A_{90}=1000(0.75317)=753.17\)
Therefore,
\(P=\frac{753.17}{4.1835}=180.03\) |
exam-fam-l--question-6.28 | 6.28 | multiple_choice | . For a fully discrete 5 -payment whole life insurance of 1000 on (40), you are given:
(i) Expenses incurred at the beginning of the first five policy years are as follows:
\begin{tabular}{|l|c|c|c|c|}
\hline & \multicolumn{2}{|c|}{ Year 1 } & \multicolumn{2}{c|}{ Years 2-5 } \\
\hline & \begin{tabular}{c}
Percent \\... | [
[
"A",
"31"
],
[
"B",
"36"
],
[
"C",
"41"
],
[
"D",
"46"
],
[
"E",
"51"
]
] | B | \(G \ddot{a}_{40: 51}=1000 A_{40}+0.15 G+0.05 G \ddot{a}_{40: 51}+5+5 \ddot{a}_{40: 51}\)
\(\ddot{a}_{40: 5 \mid}=\ddot{a}_{40}-{ }_{5} E_{40} \bullet \ddot{a}_{45}=18.4578-(0.78113)(17.8162)=4.5410\)
\(G=\frac{121.06+5+5(4.5410)}{-0.15+0.95(4.5410)}=35.73\) |
exam-fam-l--question-6.29 | 6.29 | multiple_choice | . (35) purchases a fully discrete whole life insurance policy of 100,000 .
You are given:
(i) The annual gross premium, calculated using the equivalence principle, is 1770
(ii) The expenses in policy year 1 are \(50 \%\) of premium and 200 per policy
(iii) The expenses in policy years 2 and later are \(10 \%\) of prem... | [
[
"A",
"20.0"
],
[
"B",
"20.5"
],
[
"C",
"21.0"
],
[
"D",
"21.5"
],
[
"E",
"22.0"
]
] | B | Per equivalence Principle:
\[
\begin{aligned}
G \ddot{a}_{35} & =100,000 A_{35}+0.4 G+150+0.1 G \ddot{a}_{35}+50 \ddot{a}_{35} \\
1770 \ddot{a}_{35} & =100,000\left(1-d \ddot{a}_{35}\right)+0.4(1770)+150+0.1(1770) \ddot{a}_{35}+50 \ddot{a}_{35} \\
1770 \ddot{a}_{35} & =100,000+708+150+\ddot{a}_{35}\left(177+50-100,000\... |
exam-fam-l--question-6.30 | 6.30 | multiple_choice | . For a fully discrete whole life insurance of 100 on \((x)\), you are given:
(i) The first year expense is \(10 \%\) of the gross annual premium
(ii) Expenses in subsequent years are \(5 \%\) of the gross annual premium
(iii) The gross premium calculated using the equivalence principle is 2.338
(iv) \(\quad i=0.04\)
(... | [
[
"A",
"900"
],
[
"B",
"1200"
],
[
"C",
"1500"
],
[
"D",
"1800"
],
[
"E",
"2100"
]
] | A | The loss at issue is given by:
\[
\begin{aligned}
& L_{0}=100 v^{K+1}+0.05 G+0.05 G \ddot{a}_{\overline{K+1}}-G \ddot{a}_{\overline{K+1}} \\
& =100 v^{K+1}+0.05 G-0.95 G\left(\frac{1-v^{K+1}}{d}\right) \\
& =\left(100+\frac{0.95 G}{d}\right) v^{K+1}+0.05 G-0.95 \frac{G}{d}
\end{aligned}
\]
Thus, the variance is
\[
\be... |
exam-fam-l--question-6.31 | 6.31 | multiple_choice | . For a fully continuous whole life insurance policy of 100,000 on (35), you are given:
(i) The density function of the future lifetime of a newborn:
\[
f(t)= \begin{cases}0.01 e^{-0.01 t}, & 0 \leq t<70 \\ g(t), & t \geq 70\end{cases}
\]
(ii) \(\delta=0.05\)
(iii) \(\quad \bar{A}_{70}=0.51791\)
Calculate the annual n... | [
[
"A",
"1000"
],
[
"B",
"1110"
],
[
"C",
"1220"
],
[
"D",
"1330"
],
[
"E",
"1440"
]
] | D | \[
\begin{aligned}
& \bar{A}_{35}=\left(1-e^{-35(\mu+\delta)}\right) \times\left(\frac{\mu}{\mu+\delta}\right)+e^{-35(\mu+\delta)} \bar{A}_{70}=0.063421+0.146257=0.209679 \\
& \bar{a}_{35}=\frac{1-\bar{A}_{35}}{\delta}=\frac{1-0.209679}{0.05}=15.80642 \\
& \bar{P}_{35}=\frac{\bar{A}_{35}}{\bar{a}_{35}}=\frac{0.209679}{... |
exam-fam-l--question-6.33 | 6.33 | multiple_choice | . An insurance company sells 15 -year pure endowments of 10,000 to 500 lives, each age \(x\), with independent future lifetimes. The single premium for each pure endowment is determined by the equivalence principle.
(i) You are given:
(ii) \(\quad i=0.03\)
(iii) \(\quad \mu_{x}(t)=0.02 t, \quad t \geq 0\)
(iv) \({ }_{0... | [
[
"A",
"0.08"
],
[
"B",
"0.13"
],
[
"C",
"0.18"
],
[
"D",
"0.23"
],
[
"E",
"0.28"
]
] | B | The probability that the endowment payment will be made for a given contract is:
\[
\begin{aligned}
15 p_{x} & =\exp \left(-\int_{0}^{15} 0.02 t d t\right) \\
& =\exp \left(-\left.0.01 t^{2}\right|_{0} ^{15}\right) \\
& =\exp \left(-0.01(15)^{2}\right) \\
& =0.1054
\end{aligned}
\]
Because the premium is set by the eq... |
exam-fam-l--question-6.34 | 6.34 | multiple_choice | . For a fully discrete whole life insurance policy on (61), you are given:
(i) The annual gross premium using the equivalence principle is 500
(ii) Initial expenses, incurred at policy issue, are \(15 \%\) of the premium
(iii) Renewal expenses, incurred at the beginning of each year after the first, are \(3 \%\) of the... | [
[
"A",
"\\(\\quad 23,300\\)"
],
[
"B",
"23,400"
],
[
"C",
"\\(\\quad 23,500\\)"
],
[
"D",
"23,600"
],
[
"E",
"\\(\\quad 23,700\\)"
]
] | A | Let \(B\) be the amount of death benefit.
\(\operatorname{EPV}(\) Premiums \()=500 \ddot{a}_{61}=500(14.6491)=7324.55\)
EPV(Benefits) \(=\mathrm{B} \cdot A_{61}=(0.30243) \mathrm{B}\)
\(\operatorname{EPV}(\) Expenses \()=(0.12)(500)+(0.03)(500) \ddot{a}_{61}=(0.12)(500)+(0.03)(7324.55)=279.74\)
EPV(Premiums) = EPV(Bene... |
exam-fam-l--question-6.35 | 6.35 | multiple_choice | . For a fully discrete whole life insurance policy of 100,000 on (35), you are given:
(i) First year commissions are \(19 \%\) of the annual gross premium
(ii) Renewal year commissions are \(4 \%\) of the annual gross premium
(iii) Mortality follows the Standard Ultimate Life Table
(iv) \(\quad i=0.05\)
Calculate the ... | [
[
"A",
"410"
],
[
"B",
"450"
],
[
"C",
"490"
],
[
"D",
"530"
],
[
"E",
"570"
]
] | D | Let \(G\) be the annual gross premium. By the equivalence principle, we have
\(G \ddot{a}_{35}=100,000 A_{35}+0.15 G+0.04 G \ddot{a}_{35}\)
so that
\(G=\frac{100,000 A_{35}}{0.96 \ddot{a}_{35}-0.15}=\frac{100,000(0.09653)}{0.96(18.9728)-0.15}=534.38\) |
exam-fam-l--question-6.36 | 6.36 | multiple_choice | . For a fully continuous 20 -year term insurance policy of 100,000 on (50), you are given:
(i) Gross premiums, calculated using the equivalence principle, are payable at an annual rate of 4500
(ii) Expenses at an annual rate of \(R\) are payable continuously throughout the life of the policy
(iii) \(\quad \mu_{50+t}=0.... | [
[
"A",
"400"
],
[
"B",
"500"
],
[
"C",
"600"
],
[
"D",
"700"
],
[
"E",
"800"
]
] | B | By the equivalence principle,
\[
4500 \bar{a}_{x: 20 \mid}=100,000 \bar{A}_{x: 20 \mid}^{1}+R \bar{a}_{x: 20 \mid}
\]
where
\[
\begin{aligned}
& \bar{A}_{x: 201}^{1}=\frac{\mu}{\mu+\delta}\left(1-e^{-20(\mu+\delta)}\right)=\frac{0.04}{0.12}\left(1-e^{-20(0.12)}\right)=0.3031 \\
& \bar{a}_{x: \overline{201}}=\frac{1-e^{... |
exam-fam-l--question-6.37 | 6.37 | multiple_choice | . For a fully discrete whole life insurance policy of 50,000 on (35), with premiums payable for a maximum of 10 years, you are given:
(i) Expenses of 100 are payable at the end of each year including the year of death
(ii) Mortality follows the Standard Ultimate Life Table
(iii) \(\quad i=0.05\)
Calculate the annual g... | [
[
"A",
"790"
],
[
"B",
"800"
],
[
"C",
"810"
],
[
"D",
"820"
],
[
"E",
"830"
]
] | D | By the equivalence principle, we have
\[
G \ddot{a}_{35: \overline{10}}=50,000 A_{35}+100 a_{35}+100 A_{35}
\]
so
\[
G=\frac{50,100 A_{35}+100\left(\ddot{a}_{35}-1\right)}{\ddot{a}_{35: \overline{10}}}=\frac{50,100(0.09653)+100(17.9728)}{8.0926}=819.69
\] |
exam-fam-l--question-6.38 | 6.38 | multiple_choice | . For an \(n\)-year endowment insurance of 1000 on \((x)\), you are given:
(i) Death benefits are payable at the moment of death
(ii) Premiums are payable annually at the beginning of each year
(iii) Deaths are uniformly distributed over each year of age
(iv) \(\quad i=0.05\)
(v) \(\quad{ }_{n} E_{x}=0.172\)
(vi) \(\qu... | [
[
"A",
"10.1"
],
[
"B",
"11.3"
],
[
"C",
"12.5"
],
[
"D",
"13.7"
],
[
"E",
"14.9"
]
] | B | Let \(P\) be the annual net premium
\[
P=\frac{1000 \bar{A}_{x: \bar{n}}}{\ddot{a}_{x: n}}=\frac{1000(0.192)}{\ddot{a}_{x: \bar{n}}}
\]
where
\[
\begin{aligned}
& \ddot{a}_{x: \bar{n}}=\frac{1-A_{x: \bar{n}}}{d}=\frac{(1.05)}{(0.05)}\left(1-A_{x: \bar{n}}^{1}-A_{x: \bar{n}}\right) \\
& A_{x: \bar{n}}=\frac{i}{\delta}\l... |
exam-fam-l--question-6.39 | 6.39 | multiple_choice | . XYZ Insurance writes 10,000 fully discrete whole life insurance policies of 1000 on lives age 40 and an additional 10,000 fully discrete whole life policies of 1000 on lives age 80.
XYZ used the following assumptions to determine the net premiums for these policies:
(i) Mortality follows the Standard Ultimate Life Ta... | [
[
"A",
"29"
],
[
"B",
"32"
],
[
"C",
"35"
],
[
"D",
"38"
],
[
"E",
"41"
]
] | A | Premium at issue for (40): \(\frac{1000 A_{40}}{\ddot{a}_{40}}=\frac{121.06}{18.4578}=6.5587\)
Premium at issue for (80): \(\frac{1000 A_{80}}{\ddot{a}_{80}}=\frac{592.93}{8.5484}=69.3615\)
Lives in force after ten years:
Issued at age 40: \(10,000_{10} p_{40}=10,000 \times \frac{98,576.4}{99,338.3}=9923.30\)
Issued at... |
exam-fam-l--question-6.40 | 6.40 | multiple_choice | . For a special fully discrete whole life insurance, you are given:
(i) The death benefit is \(1000(1.03)^{k}\) for death in policy year \(k\), for \(k=1,2,3 \ldots\)
(ii) \(q_{x}=0.05\)
(iii) \(\quad i=0.06\)
(iv) \(\quad \ddot{a}_{x+1}=7.00\)
(v) The annual net premium for this insurance at issue age \(x\) is 110
Ca... | [
[
"A",
"110"
],
[
"B",
"112"
],
[
"C",
"116"
],
[
"D",
"120"
],
[
"E",
"122"
]
] | C | Let \(P\) be the annual net premium at \(x+1\). Also, let \(A_{y}^{*}\) be the expected present value for the special insurance described in the problem issued to \((y)\).
\(P \ddot{a}_{x+1}=1000 \sum_{k=0}^{\infty}(1.03)^{k+1} v^{k+1}{ }_{k \mid} q_{x+1}=1000 A_{x+1}^{*}\)
We are given
\(110 \ddot{a}_{x}=1000 \sum_{k=... |
exam-fam-l--question-6.41 | 6.41 | multiple_choice | . For a special fully discrete 2 -year term insurance on \((x)\), you are given:
(i) \(\quad q_{x}=0.01\)
(ii) \(\quad q_{x+1}=0.02\)
(iii) \(\quad i=0.05\)
(iv) The death benefit in the first year is 100,000
(v) Both the benefits and premiums increase by \(1 \%\) in the second year Calculate the annual net premium in ... | [
[
"A",
"1410"
],
[
"B",
"1417"
],
[
"C",
"1424"
],
[
"D",
"1431"
],
[
"E",
"1438"
]
] | B | Let \(P\) be the net premium for year 1 .
Then:
\(P+1.01 P v p_{x}=100,000 v q_{x}+(1.01)(100,000) v^{2} p_{x} q_{x+1}\)
\(P\left[1+\frac{1.01}{1.05} 0.99\right]=100,000\left(\frac{0.01}{1.05}+\frac{(1.01)(0.99)(0.02)}{(1.05)^{2}}\right) \Rightarrow P=1416.93\) |
exam-fam-l--question-6.43 | 6.43 | multiple_choice | . For a fully discrete, 5 -payment 10 -year term insurance of 200,000 on ( 30 ), you are given:
(i) Mortality follows the Standard Ultimate Life Table
(ii) The following expenses are incurred at the beginning of each respective year:
\begin{tabular}{lc|c|cc}
& \multicolumn{2}{c|}{ Year 1 } & \multicolumn{2}{c}{ Years... | [
[
"A",
"150"
],
[
"B",
"160"
],
[
"C",
"170"
],
[
"D",
"180"
],
[
"E",
"190"
]
] | C | \(A P V(\) expenses \()=0.35 G+8+0.15 G a_{30: \overline{4}}+4 a_{30: \overline{9}}\)
\(=0.20 G+4+0.15 G \ddot{a}_{30: 5 \mid}+4 \ddot{a}_{30: 10 \mid}\)
\(G \ddot{a}_{30: 51}=0.20 G+4+0.15 G \ddot{a}_{30: 51}+4 \ddot{a}_{30: \overline{10} 1}+200,000 A_{30: \overline{10} 1}^{1}\)
\(G=\frac{200,000 A_{30: \overline{10} ... |
exam-fam-l--question-6.44 | 6.44 | multiple_choice | . For a special fully discrete 10 -year deferred whole life insurance of 100 on (50), you are given:
(i) Premiums are payable annually, at the beginning of each year, only during the deferral period
(ii) For deaths during the deferral period, the benefit is equal to the return of all premiums paid, without interest
(ii... | [
[
"A",
"1.3"
],
[
"B",
"1.6"
],
[
"C",
"1.9"
],
[
"D",
"2.2"
],
[
"E",
"2.5"
]
] | D | Let \(P\) be the premium per 1 of insurance.
\[
\begin{aligned}
& P \ddot{a}_{50: \overline{10}}=P(I A)_{50: \overline{10}}^{1}+{ }_{10} E_{50} A_{60} \\
& \ddot{a}_{50: \overline{10}}=\ddot{a}_{50}-{ }_{10} E_{50} \ddot{a}_{60}=17.0-0.60 \times 15.0=8 \\
& A_{60}=1-d \ddot{a}_{60}=1-\left(\frac{0.05}{1.05}\right) 15=0... |
exam-fam-l--question-6.45 | 6.45 | multiple_choice | . For a fully continuous whole life insurance of 100,000 on (35), you are given:
(i) The annual rate of premium is 560
(ii) Mortality follows the Standard Ultimate Life Table
(iii) Deaths are uniformly distributed over each year of age
(iv) \(\quad i=0.05\)
Calculate the \(75{ }^{\text {th }}\) percentile of the loss ... | [
[
"A",
"610"
],
[
"B",
"630"
],
[
"C",
"650"
],
[
"D",
"670"
],
[
"E",
"690"
]
] | E | \(L_{0}=100,000 v^{T}-560 \bar{a}_{\bar{T}}=\left(100,000+\frac{560}{\delta}\right) e^{-\delta T}-\frac{560}{\delta}\)
Since \(L_{0}\) is a decreasing function of \(T\), the \(75^{\text {th }}\) percentile of \(L_{0}\) is \(L_{0}(t)\) where \(t\) is such that \(\operatorname{Pr}\left[T_{35}>t\right]=0.75\).
\(\frac{l_{... |
exam-fam-l--question-6.47 | 6.47 | multiple_choice | . For a 10 -year deferred whole life annuity-due with payments of 100,000 per year on ( 70 ), you are given:
(i) Annual gross premiums of \(G\) are payable for 10 years
(ii) First year expenses are \(75 \%\) of premium
(iii) Renewal expenses for years 2 and later are \(5 \%\) of premium during the premium paying period... | [
[
"A",
"\\(\\quad 64,900\\)"
],
[
"B",
"65,400"
],
[
"C",
"\\(\\quad 65,900\\)"
],
[
"D",
"66,400"
],
[
"E",
"\\(\\quad 66,900\\)"
]
] | D | \(G \ddot{a}_{70: \overline{10}}=100,000_{10} E_{70} \ddot{a}_{80}+0.05 G \ddot{a}_{70: \overline{10}}+0.7 G\)
\(7.6491 G=(100,000)(0.50994)(8.5484)+0.05 G(7.6491)+0.7 G\)
\(\Rightarrow G=66,383.54\) |
End of preview. Expand in Data Studio
ActuarialMathBench is a domain-specific benchmark dataset designed to evaluate the mathematical reasoning capabilities of Large Language Models (LLMs) within the actuarial domain. The dataset consists of 750 question-answer pairs derived from Society of Actuaries (SOA) sample exams.
Dataset Statistics
Each exam subject is part of the curriculum to obtain the designation Associate of the Society of Actuaries (ASA). The % Pass Mark is the 90th percentile pass mark percentage over the last 10 exams for the exam subject (retrieved around August 2025) sourced from Actuarial Lookup. # Merge Qa, # Filter QA, # Open-Format QA represent the data preprocessing pipeline. The dataset uploaded to huggingface only contains subsample of 150 questions per exam. For exam SRM and FAM-S older questions where used for different exams to reach 150 questions. The % Numeric Value indicate how much of the answers are numeric (e.g. 100, 100.1, 0.1, etc.) and % Numeric Range indiacte how many answers are ranges (e.g. (0, 2)).
| Exam Subject | Exam | % Pass Mark | # Merge QA | # Filter QA | # Open-Format QA | % Numeric Value | % Numeric Range |
|---|---|---|---|---|---|---|---|
| Financial Mathematics | FM | 71.00 | 462 | 461 | 442 | 85.97 | 0.00 |
| Probability Theory | P | 71.00 | 627 | 608 | 599 | 90.98 | 0.17 |
| Statistics for Risk Modelling | SRM | 63.20 | 74 | 59 | 53 | 22.64 | 0.00 |
| C | -- | 307 | 274 | 260 | 67.69 | 23.46 | |
| Short-Term Actuarial Mathematics | FAM-S | 68.65 | 107 | 99 | 98 | 84.69 | 11.22 |
| STAM | -- | 328 | 132 | 122 | 65.57 | 26.23 | |
| Long-Term Actuarial Mathematics | FAM-L | 74.65 | 163 | 163 | 161 | 98.14 | 0.00 |
Example Question
The question stem and the multiple-choice options are seperated. You can use the exams in either multiple-choice format or as open-ended questions.
{
"question_id": "exam-fam-s--question-2",
"question_number": "2",
"question_type": "multiple_choice",
"question": "You are given:\n(i) Losses follow a single-parameter Pareto distribution with density function:\n\[\nf(x)=\frac{\alpha}{x^{\alpha+1}}, \quad x>1, \quad 0<\alpha<\infty\n\]\n(ii) A random sample of size five produced three losses with values 3,6 and 14 , and two losses exceeding 25.\n\nCalculate the maximum likelihood estimate of \(\alpha\).",
"options": [
[
"A",
"0.25"
],
[
"B",
"0.30"
],
[
"C",
"0.34"
],
[
"D",
"0.38"
],
[
"E",
"\(\quad 0.42\)"
]
],
"correct_answer": "A",
"explanation": "The distribution function is \(F(x)=\int_{1}^{x} \alpha t^{-\alpha-1} d t=-\left.t^{-\alpha}\right|_{1} ^{x}=1-x^{-\alpha}\). The likelihood function is\n\[\n\begin{aligned}\nL & =f(3) f(6) f(14)[1-F(25)]^{2} \\\n& =\alpha 3^{-\alpha-1} \alpha 6^{-\alpha-1} \alpha 14^{-\alpha-1}\left(25^{-\alpha}\right)^{2} \\\n& \propto \alpha^{3}[3(6)(14)(625)]^{-\alpha}\n\end{aligned}\n\]\n\nTaking logs, differentiating, setting equal to zero, and solving:\n\(\ln L=3 \ln \alpha-\alpha \ln 157,500\) plus a constant\n\(d \ln L / d \alpha=3 \alpha^{-1}-\ln 157,500=0\)\n\(\hat{\alpha}=3 / \ln 157,500=0.2507\)."
}
Data Quality
The data source where PDF containing sample questions sourced from soa.org. The sample questions are typically question from previous exams that align with the current syllabus. The conversion from PDF to .md was done using Mathpix, further preprocessing is done with custom scripts.
The FAM-L exams contains questions with conversion errors related to Actuarial Notation (estimated impact: 23 out of 150 questions). In addition, FAM-L exams depend on the Standard Ultimate Life Table (SULT) (estimated impact: 61 out of 150 questions)
Additional Information
The methodology and results for Gemini 2.5 pro, Claude Sonnet 4, and OpenAI o3 are published
Citation
@misc{bjorn_van_braak_2026,
author = { Bjorn van Braak },
title = { ActuarialMathBench },
year = 2026,
url = { https://huggingface.co/datasets/BjornvanBraak/ActuarialMathBench },
doi = { 10.57967/hf/7465 },
publisher = { Hugging Face }
}
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