Exam-Style Questions on Life Table Functions

The problems in this section were designed to be similar to problems from past versions of the Casualty Actuarial Society's Exam 3L and the Society of Actuaries' Exam MLC. They use original exam questions as their inspiration - and the specific inspiration for each problem is cited so as to give students a chance to see the original. All of the original problems are publicly available, and students are encouraged to refer to them. But all of the values, names, conditions, and calculations in the problems here are the original work of Mr. Stolyarov.

Original Problems and Solutions from The Actuary's Free Study Guide

Problem S3L17-1.

Similar to Question 13 from the Casualty Actuarial Society's Spring 2008 Exam 3L.

The future lifetimes of bald chimpanzees can be modeled by the survival function

s(x) = ((94 - x)/94)^3.1. Find ė₄₄, the expected future lifetime of a 44-year-old bald chimpanzee.

Solution S3L17-1. We use the formula ė_x = ₀^∞∫_tp_xdt from Section 10. We need to find _tp₄₄ = s(44+t)/s(44) = ((94 - 44 - t)/94)^3.1/((94 - 44)/94)^3.1 = ((50 - t)/94)^3.1/((50)/94)^3.1 = ((50 - t)/50)^3.1.

In determining the upper bound for our integral, we need to recognize that bald chimpanzees do not live indefinitely. In fact, no bald chimpanzee lives past the age of 94, since s(94) = 0. In determining the future lifetime of a chimpanzee already aged 44, we need to take into account the fact that the most this chimpanzee will live is 94 - 44 = 50 years. Thus, the upper bound of our integral is 50.

Thus, ė₄₄ = ₀⁵⁰∫((50 - t)/50)^3.1dt = (50 - t)^4.1/[4.1(50)^3.1]│₀⁵⁰ = (50)^4.1/[4.1(50)^3.1] = 50/4.1 = ė₄₄ = about 12.19512195 years.

Problem S3L17-2.

Similar to Question 14 from the Casualty Actuarial Society's Spring 2008 Exam 3L.

This question uses the Illustrative Life Table, which can be found here.

In the Illustrative Life Table, the following adjustments are made, whereby every adjusted value of q_x is replaced by the value of q_x*.

q₂₄* = 3q₂₄

q₂₅* = (1/2)q₂₅

q₂₆* = q₂₄

q₂₇* = 8q₅

q_x* = (1/2)q_xfor 28 ≤ x ≤ 30.

Find _2│5q₂₄ using the modified values (q_x*).

Solution S3L17-2. From the Illustrative Life Table, we know the following:

q₅ = 0.00098

q₂₄ = 0.00118

q₂₅ = 0.00122

q₂₈ = 0.00139

q₂₉ = 0.00146

q₃₀ = 0.00153

Thus,

q₂₄* = 3q₂₄ = 0.00354

q₂₅* = (1/2)q₂₅ = 0.00061

q₂₆* = q₂₄ = 0.00118

q₂₇* = 8q₅ = 0.00784

q₂₈* = (1/2)q₂₈ = 0.000695

q₂₉* = (1/2)q₂₉ = 0.00073

q₃₀* = (1/2) q₃₀ = 0.000765

We need to figure out _2│5q₂₄ = [s(26)-s(31)]/s(24) = [s(26│24) - s(31│24)]/s(24│24).

(That is, we can modify our survival function to s(x│24). This is the survival function for all lives that have survived to age 24. So s(24│24) = 1.)

Moreover, s(26│24) and s(31│24) are simple to find, since we already know the probabilities q_x*.

s(26│24) = (1 - q₂₄*)(1 - q₂₅*) = (1 - 0.00354)(1 - 0.00061) = s(26│24) = 0.9958521594

s(31│24) = (1 - q₂₄*)(1 - q₂₅*)(1 - q₂₆*)(1 - q₂₇*)(1 - q₂₈*)(1 - q₂₉*)(1 - q₃₀*) =

0.9958521594(1 - 0.00118)(1 - 0.00784)(1 - 0.000695)(1 - 0.00073)(1 - 0.000765) =

s(31│24) = 0.9847190973.

Thus, [s(26│24) - s(31│24)]/s(24│24) = (0.9958521594 - 0.9847190973)/1 = _2│5q₂₄ =about 0.0111330621

Problem S3L17-3.

Similar to Question 15 from the Casualty Actuarial Society's Spring 2008 Exam 3L.

Fuzzy cucumber plants have the following survival function:

s(x) = exp[-99x¹³⁰]. Find the force of mortality µ_x for fuzzy cucumber plants.

Solution S3L17-3. We use the formula µ_x = -s'(x)/s(x).

-s'(x) = -[-99x¹³⁰]'exp[-99x¹³⁰] = 99*130x¹²⁹exp[-99x¹³⁰] = 12870x¹²⁹exp[-99x¹³⁰]

Thus, µ_x = -s'(x)/s(x) = (12870x¹²⁹exp[-99x¹³⁰])/exp[-99x¹³⁰] = µ_x = 12870x¹²⁹

Problem S3L17-4.

Similar to Question 16 from the Casualty Actuarial Society's Spring 2008 Exam 3L.

Ordinary talking phytoplankton have the following life table associated with them:

x.........l_x

53.......10000

54.......9954

55.......8312

Use the uniform distribution of deaths assumption for fractional ages to find _0.3│0.9q_53.6 for ordinary talking phytoplankton. Do not round your answers at any step. Fractional answers are possible.

Solution S3L17-4. We use the formula _t│uq_x = (s(x + t) - s(x + t + u))/s(x) from Section 2.

So _0.3│0.9q_53.6 = (s(53.9) - s(54.8))/s(53.6)

By age 53.6 years, we assume that 0.6 of the phytoplankton to die in the 54^th year will have died. Thus, 0.6(10000 - 9954) = 27.6 phytoplankton have died and thus 10000 - 27.6 = 9972.4 are alive at age 53.6. So s(53.6) = 9972.4.

By age 53.9 years, we assume that 0.9 of the phytoplankton to die in the 54^th year will have died.

Thus, 0.9(10000 - 9954) = 41.4 phytoplankton have died and thus 10000 - 41.4 = 9958.6 are alive at age 53.9. So s(53.9) = 9958.6.

By age 54.8 years, we assume that 0.8 of the phytoplankton to die in the 55^th year will have died.

Thus, 0.8(9954 - 8312) = 1313.6 phytoplankton have died and thus 9954 - 1313.6 = 8640.4 are still alive at age 54.8. So s(54.8) = 8640.4.

Hence, _0.3│0.9q_53.6 = (s(53.9) - s(54.8))/s(53.6) = (9958.6 - 8640.4)/9972.4 = about 0.1321848301.

Problem S3L17-5.

Similar to Question 5 from the Casualty Actuarial Society's Spring 2007 Exam 3.

This question uses the Illustrative Life Table, which can be found here.

In the Illustrative Life Table, find 10000_4│1q₃₅.

Solution S3L17-5. We use the formula _t│uq_x = (s(x + t) - s(x + t + u))/s(x) from Section 2.

So _4│1q₃₅ = (s(39) - s(40))/s(35) = (s(39│35) - s(40│35))/s(35│35)

s(35│35) = 1

s(39│35) = (1 - q₃₅)(1 - q₃₆)(1 - q₃₇)(1 - q₃₈) =

(1 - 0.00201)(1 - 0.00214)(1 - 0.00228)(1 - 0.00243) = 0.9911693541

s(40│35) = (1 - q₃₅)(1 - q₃₆)(1 - q₃₇)(1 - q₃₈)(1 - q₃₉) = 0.9911693541*(1 - 0.0026) = 0.9885923048.

Thus, _4│1q₃₅ = 0.9911693541 - 0.9885923048 = about 0.0025770403.

See other sections of The Actuary's Free Study Guide for Exam 3L.