Free SOA Exam FAM (Fundamentals of Actuarial Mathematics) Mortality Models Practice Questions

Mortality models on SOA Exam FAM cover life tables, survival functions, force of mortality, and select-and-ultimate mortality tables. These are the foundation for all life contingency calculations in actuarial science.

125 questions 66 easy 40 medium 19 hard 2026 syllabus

Sample Questions

Question 1 Easy
In a life table, which of the following relationships is correct?
Solution
B is correct.

In a life table: lxl_x = expected number alive at exact age xx; dxd_x = expected number of deaths between ages xx and x+1. Therefore: dx=lx−lx+1d_x = l_x - l_{x+1} Equivalently: dx=lx⋅qxd_x = l_x \cdot q_x (number alive times probability of death).
Question 2 Medium
For a constant force of mortality μ=0.02\mu = 0.02, calculate exe_x, the curtate expectation of life.
Solution
B is correct.

The curtate expectation of life is:

ex=∑k=1∞kpxe_x = \sum_{k=1}^{\infty} {}_{k}p_x

With constant force μ\mu, kpx=e−μk=(e−μ)k{}_{k}p_x = e^{-\mu k} = (e^{-\mu})^k.

Let p=e−μ=e−0.02=0.98020p = e^{-\mu} = e^{-0.02} = 0.98020.

This is a geometric series:

ex=∑k=1∞pk=p1−p=0.980200.01980=49.505≈49.5e_x = \sum_{k=1}^{\infty} p^k = \frac{p}{1-p} = \frac{0.98020}{0.01980} = 49.505 \approx 49.5

Note: The complete expectation is e˚x=1/μ=50\mathring{e}_x = 1/\mu = 50, while the curtate expectation is slightly less because it only counts complete years survived. Under UDD, e˚x≈ex+0.5\mathring{e}_x \approx e_x + 0.5, consistent with 50≈49.5+0.550 \approx 49.5 + 0.5.
Question 3 Hard
For a Gompertz model μx=Bcx\mu_x = Bc^x, you are given:

10p50=0.920{}_{10}p_{50} = 0.920 and 10p60=0.850{}_{10}p_{60} = 0.850

Calculate c10c^{10}.
Solution
A is correct.

For Gompertz, the survival probability is:
10px=exp⁡(−Bcx(c10−1)ln⁡c){}_{10}p_x = \exp\left(-\frac{Bc^x(c^{10} - 1)}{\ln c}\right)

Taking logarithms of each:
ln⁡(10p50)=−Bc50(c10−1)ln⁡c\ln({}_{10}p_{50}) = -\frac{Bc^{50}(c^{10}-1)}{\ln c}
ln⁡(10p60)=−Bc60(c10−1)ln⁡c\ln({}_{10}p_{60}) = -\frac{Bc^{60}(c^{10}-1)}{\ln c}

Dividing:
ln⁡(10p60)ln⁡(10p50)=c60c50=c10\frac{\ln({}_{10}p_{60})}{\ln({}_{10}p_{50})} = \frac{c^{60}}{c^{50}} = c^{10}

ln⁡(0.920)=−0.08338\ln(0.920) = -0.08338
ln⁡(0.850)=−0.16252\ln(0.850) = -0.16252

c10=−0.16252−0.08338=1.9492≈1.949c^{10} = \frac{-0.16252}{-0.08338} = 1.9492 \approx 1.949

The answer is 1.949.
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