# Practice Exam 1. x l x d x

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1 Practice Eam. You are given: (i) The following life table. (ii) 2q l d Determine d 5. (A) 2 (B) 2 (C) 22 (D) 24 (E) For a Continuing Care Retirement Community, you are given the following annual transition probability matri: Independent Temporary Permanent Living Nursing Nursing From Unit Facility Facility Gone Independent Living Unit.7..2 Temporary Nursing Facility.7... Permanent Nursing Facility.8.2 Gone You are given the following costs per year in each unit, payable at the beginning of the year. Independent Living Unit 5 Temporary Nursing Facility Permanent Nursing Facility 2 Determine the actuarial present value at 6% of 3 years of costs for someone currently in the Temporary Nursing Facility. (A) 84 (B) 2 (C) 22 (D) 234 (E) 292 To SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM 9 Eercises continue on the net page...

2 92 PART IV. PRACTICE EXAMS 3. For a fully discrete 2-year term insurance on (6) with benefit b payable at the end of the year of death, you are given (i) t p 6+t (ii) The annual benefit premium is (iii) i.5. Determine the revised annual benefit premium if an interest rate of i.4 is used. (A) (B) (C) 25.7 (D) (E) You are given: (i) t p () /( + t ) (ii) t p (2) /( + t ) 3 Determine q (). (A).68 (B).74 (C).79 (D).83 (E).9 5. For a special whole life insurance paying at the moment of death, you are given: (i) If death occurs in the first years, the benefit is the refund of the single benefit premium with interest at a rate of δ.3. (ii) If death occurs after the first years, the benefit is only.. t (iii) µ (t ).2 t > (iv) δ.6 Determine the single benefit premium. (A) 3 (B) 32 (C) 33 (D) 34 (E) A drill has two gears, and fails only if both gears fail. The gears are independent, and time to failure of each one is uniformly distributed over (, 2]. A 3-year warranty covers the drill. Determine the average amount of time to the earlier of failure of the drill or epiration of the warranty. (A) (B) (C) (D) (E) A continuous deferred life annuity pays per year starting at time. If death occurs before time, the single benefit premium is refunded without interest at the moment of death. You are given: (i) µ.2 (ii) δ.5 Calculate the single benefit premium for this annuity. (A) 8.9 (B) 8.29 (C) 8.38 (D) 8.48 (E) 8.58 SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM Eercises continue on the net page...

3 PRACTICE EXAM You are given that A.4 +. for < 6. Calculate 2 V 3. (A) 4 (B) 3 (C) 2 (D) 2 3 (E) Tais arrive in a Poisson process at a hotel at a rate of 2 per minute. Fares paid by riders follow a gamma distribution with probability density function f () 4 e /2 768 Fares are independent of each other and of the number of tais. Calculate the variance of aggregate fares collected by all tais in an hour. (A) 3,6 (B) 7,2 (C),8 (D) 4,4 (E) 8,. A special 9-year term insurance pays the following benefit at the end of the year of death: Year of death t Benefit b t You are given the following values for increasing and decreasing term insurances: n (I A) :n (DA) :n Determine the actuarial present value of the term insurance. (A).7 (B).8 (C).3 (D).4 (E).7. Cars arrive at a toll booth in a Poisson process at the rate of 6 per minute. Determine the probability that the third car will arrive between 3 and 4 seconds from now. (A).6 (B).9 (C).24 (D).28 (E) Each night, your supper has one of four main courses: beef, chicken, fish, or pasta. Your main course is always different from the one of the previous night. Each of the 3 main courses not eaten the previous night are equally likely. Determine the probability that you will have pasta three nights from now, given that you had beef tonight. (A) /4 (B) 7/27 (C) 5/9 (D) 4/5 (E) /3 3. You are given (i) e 85:2 8 (ii) e 84:2 8.4 (iii) Uniform distribution of deaths is assumed between integral ages. Determine m 84. (A).33 (B).36 (C).38 (D).32 (E).323 SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM Eercises continue on the net page...

4 94 PART IV. PRACTICE EXAMS 4. You are given: (i) The probability that a milk carton on the shelf is purchased on any day is 2%. (ii) Milk cartons are discarded after being on the shelf for 7 days. Determine the average number of full days a purchased milk carton is on the shelf. (A).69 (B).73 (C) 2.4 (D) 2.8 (E) For a fully discrete whole life insurance of with annual premiums payable for life, you are given: (i) 8V 2. (ii) 9V (iii) V (iv) q +8 q +9 (v) q +.q +9 (vi) i.4 Determine V. (A) (B) (C) (D) (E) You are given: (i) q 8. (ii) q 8.2 (iii) The force of mortality is constant between integral ages. Calculate e 8.5:. (A).93 (B).94 (C).95 (D).96 (E) For a fully continuous whole life insurance of : (i) The contract premium is 25. (ii) The variance of loss at issue is 2,,. (iii) δ.6 Employees are able to obtain this insurance for a 2% discount. Determine the variance of loss at issue for insurance sold to employees. (A),28,533 (B),295,44 (C),77,626 (D),777,778 (E),825,3 8. You are given: (i) c 4 (ii) c vq + v p ( + c + ) Determine c 3 (A) ä 3: + A 3: (B) ä 3: + A 3: (C) a 3: + A 3: (D) a 3: + A 3: (E) a 3: + A 3: SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM Eercises continue on the net page...

6 96 PART IV. PRACTICE EXAMS 24. For a fully continuous whole life insurance of on 2 independent lives () and (y ), you are given (i) (ii) Benefits are payable at the moment of the second death. Premiums are payable while both are alive. (iii) µ (t ).2 for all t. (iv) µ y (t ).3 for all t. (v) δ.5 Determine the annual benefit premium. (A) 9.57 (B) 2.5 (C) 6.7 (D) 8.37 (E) For a fully discrete whole life insurance of on (), you are given: (i) s :2 36 (ii) 2k. (iii) i.4 (iv) (v) (vi) Percent of premium epenses are 5% in all years. Per epenses are in all years. There are no withdrawals. Determine the contract premium for which 2 AS is 3% of the accumulated contract premiums. (A) 3.98 (B) 4.7 (C) 4. (D) 4.8 (E) For two lives (5) and (6) with independent future lifetimes: (i) µ 5 (t ).2t (ii) µ 6 (t ).3t Calculate 2 q 5:6 2 q 2 5:6. (A).7 (B).8 (C).3 (D).3 (E) You are given that µ Calculate 5 q 2. (A).42 (B).44 (C).46 (D).48 (E) In a three-decrement model, decrements () and (2) are uniformly distributed between integral ages in the associated single decrement tables. Decrement (3) occurs only at the end of a year. You are given: (i) l (τ) (ii) d () 9 (iii) q (2) 2q () (iv) q (3) 3q () Determine d (3). (A) 24 (B) 26 (C) 24 (D) 27 (E) 288 SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM Eercises continue on the net page...

7 PRACTICE EXAM For a double decrement model with decrements from death () and withdrawal (2), you are given: (i) The following rates in the double-decrement table for (): Death Withdrawal t q () +t q (2) +t a.5 3 2a. (ii) 3 q () Determine a. (A).4 (B).5 (C).6 (D).7 (E).8 3. A special -year term insurance pays at the moment of death. It also refunds the single benefit premium without interest if the insured survives years. You are given: (i) µ. (ii) δ.6 Calculate the single benefit premium. (A).2 (B).3 (C).4 (D).5 (E).6 Solutions to the above questions begin on page 7. SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM

8 Appendi A. Solutions to the Practice Eams Answer Key for Practice Eam B B 2 D 2 D 2 B 22 E 3 C 3 D 23 B 4 C 4 C 24 C 5 E 5 D 25 C 6 E 6 B 26 B 7 B 7 C 27 A 8 D 8 D 28 B 9 D 9 D 29 D D 2 C 3 C Practice Eam. [Lesson 3] q 52 (d 52 + d 53 )/l 52 72/l 52, so l 52 72/ But l 52 l 5 d 5 d 5 2 d 5, so d 5 2. (B) 2. [Lesson 5] The state vector after year is The state vector after 2 years is Thus the epected cost of the second year (before taking present values) is and the epected cost of the third year is The actuarial present value of 3 years of costs is.7(5) +.() +.(2) 65.56(5) +.8() +.23(2) (D) [Lesson 22] We determine b. ä 6: A 6: (.98)(.4) b A 6 :2 ä 6:2 SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM 7

9 72 PRACTICE EXAM, ANSWERS TO QUESTIONS 4 5 Now we recalculate at 4%. b 25.4(.93333).5463 ä 6: A 6: (.98)(.4) P 6: (C) [Lesson 43] 3 4 t p (τ) + t + t + t (t ) d ln t p () dt µ () q () t p (τ) µ() + t + t (t )dt 4 + t 4 dt ( + t ) ( + t ) (C) dt 5. [Lesson ] The actuarial present value of one unit of a -year deferred whole life insurance is E Ā +. The force of mortality is constant after years, so Ā + µ µ + δ The pure endowment factor E is computed using the mortality rate in effect for the first ten years, so it is e (.+.6)() e.7. Therefore, the APV of the -year deferred whole life insurance is Ā 25e.7 Let P be the single benefit premium. The δ.3 interest for the benefit in the first ten years partially offsets the δ.6 discount factor, so the APV of the first ten years of insurance is Pµ e (µ+δ δ ).P e.4 µ + δ δ.4 SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM

10 PRACTICE EXAM, ANSWERS TO QUESTIONS We now solve for P. P 25e.7 + P.25( e.4 ) P P (E) [Lesson 36] We want the 3-year temporary future lifetime of the last survivor of the two gears. e :: t p : dt t q 2 dt t 2 dt 2 3 t 3 3 3(2 2 ) (E) 2 7. [Lesson 7] We equate the single benefit premium P and the value of the sum of the annuity and the insurance for the first years. P ā + PĀ : e.7t dt + P e.7t.2 dt e P e P (.53447) P.7 7 P (B) [Lesson 29] By the insurance ratio formula (29.2), A (3).7 A (5).9 2V 3 A 5 A A (D) 9. [Lesson 55] Let X be fare per rider, S aggregate fares. The Poisson parameter per hour is 2. The gamma distribution has α 5, θ 2, mean αθ, and variance αθ 2 2, and therefore second moment By the formula for the variance of a compound Poisson distribution, equation (55.2), Var(S) λ E[X 2 ] (2)(2) 4,4 (D) SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM

11 74 PRACTICE EXAM, ANSWERS TO QUESTIONS 3. [Lesson 4] The benefits are a 9-year decreasing insurance minus twice a 4-year decreasing insurance (.7).4. (D). [Lesson 5] The probability that the third car will arrive in the interval (3, 4) is the probability of at least 3 cars in 4 seconds minus the probability of at least 3 cars in 3 seconds. For 4 seconds, the Poisson parameter is 4 and the probability is e For 3 seconds, the Poisson parameter is 3 and the probability is e The difference is (B) 2. [Lesson 49] This is a Markov chain. The transition matri going in the beef, chicken, fish, pasta order (actually the order doesn t matter) is The probabilities of fish, chicken, pasta are /3 apiece on the first night, making the state vector (,,, ) Multiplying this by the transition matri, we get for the second night s state vector For the third night s pasta, we multiply the state vector by the last column of the transition matri (in order to get the last entry of the third state vector): () (B) 3. [Lesson 8] By the recursive formula (7.5), e 84:2 p 84 ( + e 85:2 ) 8.4 p 84 (9) p q q 84 m 84.6/9.329 (D).5q /9 SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM

12 PRACTICE EXAM, ANSWERS TO QUESTIONS [Lesson 6] The average number of full days that a milk carton survives on the shelf is e :7. The t -day survival probability is t p.8 t for t 7. e :7 7.8 k k Now we remove the contribution of cartons on the shelf 7 full days, 7(.8 7 ), and divide by the probability of being purchased, (.8 7 ) (C) 5. [Lesson 3] The two recursions for the reserve from time 8 to time 9 and time 9 to time are, with the common mortality rate denoted by q: (2. + P)(.4) q ( q) ( P)(.4) q 255.4( q) We ll solve for q and for P. Subtracting the first equation from the second, Now we calculate V. 22.2(.4) q (22.2)(.4) 23.8 q (.4) + P(.4) ( ) V P(.4) (.4) 9.59 P q +.( ).8346 ( )(.4) (.8346) (D) [Section 8.2] e 8.5: e 8.5: p 8.5 e 8:.5.5.9t dt t dt ln ln ( )(.4736).9359 (B) 7. [Lesson 23] The variance of loss at issue for a contract premium of 25 is 2,, Var v T Var v T (2,6,944) SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM

13 76 PRACTICE EXAM, ANSWERS TO QUESTIONS 8 9 If we replace 25 with 2 (for a 2% discount) in the above formula, it becomes Var( L) Var v T Var v T (,777,778) We see that this is,777,778/2,6,944 times the given variance, so the final answer is Var( L),777,778 (2,,),77,626 (C) 2,6, [Lesson 9] Since each c pays () a one-year insurance (vq ), (2) a payment of one year from now (v p ), and (3) net year s c +, we need an insurance for the insurance recursion (vq ) and an annuity for the annuity recursion (v p ). Since c 4 rather than, it looks like the insurance is an endowment insurance rather than a term insurance. So let s guess that c a :4 + A :4 (*) and prove it by induction. First of all, c 4 a 4: + A 4: insurance pays immediately. Then we need works, since a -term annuity is worth nothing and a -term endowment c vq + v p ( + c + ) and epanding c with equation (*), we need However, by insurance recursion a :4 + A :4 vq + v p + v p a +:39 + v p A +:39 (**) A :4 vq + v p A +:39 and by annuity recursion a :4 v p + v p a +:39 so (**) is true. For 3, epression (*) is (D) 9. [Lesson 52] The Poisson parameter at time t for this non-homogeneous Poisson process is λ We want the probability of to be.95.5, so t du 2 u t e t.5 t ln.5 ln 2 t (ln 2) (D) SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM

14 PRACTICE EXAM, ANSWERS TO QUESTIONS [Lesson 46] Let G be the epense-loaded premium. G ä : 8ä : +.5G ä : + ä.95g 8 + ä ä : But since the benefit premium is ( ä /ä : ) 8, the last term must be 8. G (C) [Lesson 2] Let P be the premium. We need Pā 5 Ā + (P)(Ī Ā) :5 ā µ + δ µe 5(µ+δ) 5 Ā (Ī Ā) :5 5 µ + δ e t e.6t.2dt Using equation (.4), you can immediately evaluate this integral as (5) e (.6)(5). As an alternative, (Ī Ā) :5 (Ī Ā) 5 E (Ī Ā) A +5 since the two subtracted items cancel out the increasing insurance after the fifth year. For eponential mortality, (Ī Ā) µt e (µ+δ)t µ dt (µ + δ) 2 by equation (.2). Here, µ.2 and δ.4, so (Ī Ā).2/.6 2. And Ā µ/(µ + δ).2/.6. Also, 5E e.6(5). So (Ī Ā) : e.6(5) (.2) The premium is 22. [Section 8.3] P 5 Ā ā (Ī Ā) : (D) l 5 l 5 2.5() +.5l 5 5l l 5 55(2) 5 32 (E) SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM

15 78 PRACTICE EXAM, ANSWERS TO QUESTIONS [Lesson 22] Let P be the annual benefit premium. A 45:2 d ä 45:2 2 (2.6).4 :2 A 45:2 2 E A 45 2 A 45 A 45 A 45:2.2.. P A 45 : A 45 ä 45: (B) [Lesson 37] Ā y Ā + Ā y Ā y ā y The annual benefit premium is (.67/) 6.7. (C) 25. [Lesson 48] Let G be the contract premium. Then the accumulated premiums after percent of premium epenses are.95 s :2 G 34.2G, the accumulated per epenses are s :2 36, and the accumulated benefits are 2 k, so we equate 34.2G 36.3(36)G 33.2G 36 G (C) [Section 39.] 2 q 5:6 2 q 2 5:6 2 q 5 2 p 6, and 2 2q 5 ep.2t dt e.(2) p 6 ep.3t dt e.5(2) q 5 2 p 6 (.32968)(.54882).8932 (B) 27. [Lesson 4] 5 q 2 s (25) s (26) /s (2), so we will calculate these three values of s (). (Equivalently, one could calculate 5 p 2 and 6 p 2 and take the difference.) The integral of µ is.2u 2 µ u du +.5u SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM

16 PRACTICE EXAM, ANSWERS TO QUESTIONS so and the answer is s (2) ep.(2 2 ) +.5(2) ep(.5).6653 s (25) ep.(25 2 ) +.5(25) ep(.75) s (26) ep.(26 2 ) +.5(26) ep(.86) q (A) [Lesson 44] From d () 9, q () 9.9. Then q () q () q (2) 2.9 q () q () q () 2 q () +.9 The solution.9 is rejected since then q (2) >. q () ±.64.,.9 2 q (2).2 q (3).3 Since (3) occurs at the end of the year, only l (τ) p () p (2) ()(.9)(.8) 72 lives are subject to it. So d (3) 72(.3) 26. (B) 29. [Lesson 4] We have 2 q () 3 q () We set up an equation for 2 q () q () and solve Also, p (τ) q () + 2 q () 2 q () (.797)(a ) + (.797)( a.5)(2a ) a a.594a a a a (D) The other solution to the quadratic is rejected since it is greater than. 3. [Lesson ] Let P be the single benefit premium. P P e.7 7 e.7t.dt + Pe.7 e.7 P (C) SOA MLC Study Manual th edition 2nd printing Copyright 2 ASM

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