Exam-Style Questions on Life Table Functions: Part 2

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 S3L18-1.

Similar to Question 6 from the Casualty Actuarial Society's Spring 2007 Exam 3.

The survival function for elephantine fleas is s(x) = (1444 - x²)/1444 for 0 ≤ x ≤ 38. Find ė₂₁, the expected future lifetime of a 21-year-old elephantine flea.

Solution S3L18-1. We use the formula ė_x = ₀^∞∫_tp_xdt from Section 10. We need to find _tp₂₁ = s(21+t)/s(21) = [(1444 - (21+t)²)/1444]/[(1444 - (21)²)/1444] = (1444 - (21+t)²)/(1444 - (21)²) =

(1444 - (21+t)²)/1003 = 1444/1003 - (21+t)²/1003.

In determining the upper bound for our integral, we need to recognize that elephantine fleas do not live indefinitely. In fact, no elephantine flea lives past the age of 38, since s(38) = 0. In determining the future lifetime of a flea already aged 21, we need to take into account the fact that the most this flea will live is 38 - 21 = 17 years. Thus, the upper bound of our integral is 17.

Thus, ė₂₁ = ₀¹⁷∫(1444/1003 - (21+t)²/1003)dt = (1444t/1003 - (21+t)³/3009) │₀¹⁷ =

(1444*17/1003 - (38)³/3009) + (21)³/3009 = ė₂₁ =about 9.316384181 years.

Problem S3L18-2.

Similar to Question 31 from the Casualty Actuarial Society's Fall 2007 Exam 3.

You know the following data about orange bluebirds:

_xp₀ = 0.8345

s(x + h) = 0.6664

_h│zq_x = 0.113

Find s(x + h + z).

Solution S3L18-2. We know that _xp₀ = s(x)/s(0) = s(x) = 0.8345.

Furthermore, we know from Section 2 that _h│zq_x = [s(x + h) - s(x + h + z)]/s(x).

Thus, 0.113 = [0.6664 - s(x + h + z)]/0.8345

0.0942985 = 0.6664 - s(x + h + z)

s(x + h + z) = 0.5721015.

Problem S3L18-3.

Similar to Question 32 from the Casualty Actuarial Society's Fall 2007 Exam 3.

The lives of a group of conical armadillos are described by the following life table.

x.........l_x

76.......10000

77.......7893

78.......7765

Use the uniform distribution of deaths assumption for fractional ages to find _0.8p_76.4 for conical armadillos. Do not round your answers at any step. Fractional answers are possible.

Solution S3L18-3. _0.8p_76.4= s(77.2)/s(76.4).

By age 76.4 years, we assume that 0.4 of the armadillos to die in the 77^th year will have died. Thus, 0.4(10000 - 7893) = 842.8 armadillos have died and thus 10000 - 842.8= 9157.2 are alive at age 76.4. So s(76.4) = 9157.2.

By age 77.2 years, we assume that 0.2 of the armadillos to die in the 78^th year will have died. Thus, 0.2(7893 - 7765) = 25.6 armadillos have died and thus 7893 - 25.6 = 7867.4 are alive at age 77.2. So s(77.2) = 7867.4.

Hence, 7867.4/9157.2 = _0.8p_76.4 = about 0.8591490849.

Problem S3L18-4. Similar to Question 11 from the Casualty Actuarial Society's Fall 2006 Exam 3.

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

In the Illustrative Life Table, find the average number of complete years lived between the ages of 34 and 37.

Solution S3L18-4. We seek to find a truncated version of E[K] = e_x = _k=0^∞Σ_k+1p_x. In our case, this will be _k=0²Σ_k+1p₃₄ = ₁p₃₄ + ₂p₃₄ + ₃p₃₄. The easiest way to use the life table here entails the recognition that

_tp_x = l_x+t/l_x and that in each of our three computations for the values of the "p" functions, there will be a common denominator l_x = l₃₄ = 9438571.

Thus, ₁p₃₄ + ₂p₃₄ + ₃p₃₄ = (l₃₅ + l₃₆ + l₃₇)/l₃₄ = (9420657 + 9401688 + 9381566)/9438571 = about 2.988154755.

Problem S3L18-5. Similar to Question 11 from the Casualty Actuarial Society's Fall 2006 Exam 3.

Vercingetorix and Bob are two speckled slugs. Vercingetorix is currently 4 years old, and Bob is currently 6 years old. We know the following about speckled slugs:

s(4) = 0.89

s(6) = 0.77

_4│3q₀ = 0.28

_6│3q₀ = 0.45

Find the probability that at least one of the two speckled slugs will be alive three years from now.

Solution S3L18-5. We note that _4│3q₀ = (s(4) - s(7))/s(0) = (s(4) - s(7))

Thus, 0.28 = 0.89 - s(7) and so s(7) = 0.61.

The probability that Vercingetorix will be alive 3 years from now is s(7)/s(4) = 0.61/0.89 = 61/89.

Furthermore, _6│3q₀ = (s(6) - s(9))/s(0) = (s(6) - s(9))

Thus, 0.45 = 0.77 - s(9) and so s(9) = 0.32.

The probability that Bob will be alive 3 years from now is s(9)/s(6) = 0.32/0.77 = 32/77.

The probability that neither slug will be alive 3 years from now is (1 - 61/89)(1 - 32/77). Thus, the probability that at least one will be alive 3 years from now is 1 - (1 - 61/89)(1 - 32/77) = 799/979 = about 0.8161389173.

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