Practice_Exam_3-Solutions

Practice_Exam_3-Solutions - PRACTICE EXAMINATION NO 3...

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Unformatted text preview: PRACTICE EXAMINATION NO. 3 PRACTICE EXAMINATION NUMBER 3 SOLUTIONS 1. Sample Course 1 Examination, Problem No. 15 An insurance company issues insurance contracts to two classes of independent lives, as shown below. Class Probability of Death Benefit Amount Number in Class A 0.01 200 500 B 0.05 100 300 The company wants to collect an amount, in total, equal to the 95-th percentile of the distribution of total claims. The company will collect an amount from each life insured that is proportional to that life’s expected claim. That is, the amount for life j with expected claim E (X1) would be kE(X,). Calculate k. A. 1.30 B. 1.32 C. 1.34 D. 1.36 E. 1.38 Solution. Let X be a random claim for an insured individual from class A. Then X = 200 with probability 0.01 and X = 0 with probability 0.99, i.e., X is a multiple of a Bernoulli Trial with p = 0.01. This gives E(X) = ZOO-0.01: 2 and Var(X) = 2002 001-099 = 396. Similarly, let Y be a random claim for an insured individual from class B. Then Y = 100 with probability 0.05 and Y = 0 with probability 0.95. This gives E(Y) = 100 - 0.05 = 5 and Var(l’)=1002 -0.05 - 0.95 = 475. For class A, total claims are Xl + X2 + + X500, where Xl ,...,X50° is a random sample from the distribution of X. Thus these total claims can be approximated with a normal distribution. For class B, total claims are Yl + Y2 + ...+ Y300, where l"l ,...,Y30° is a random sample from the distribution of Y. These total claims can also be approximated by a normal distribution. The sum of these two totals is the amount of total claims, and because this is a sum of two independent random variables each of which can be approximated with a normal random variable, the sum itself can be approximated by a normal random variable. Let us write T for that sum. Then E(T)= SOD-2+ 300 ~ 5 = 2500, and Var(T) = 500 - 396 + 300 - 475 = 340500. The company wants to collect the total amount of premiums 1‘ such that (note that Z, as always, denotes a standard normal random variable, and (I) is its CDF) _ < _ r T-E(T)<t-E(T) z <r—2500 _ t—2500 figs—Pr” t)_P[,/Var(T) Var(T)] “[2 J34osoo)"¢[./34osoo]' From the table, the 95-th percentile of the standard normal distribution, is approximated using linear interpolation as ASM Study Manual for Course P/I Actuarial Examination. © Copyright 20044008 by Krzysztof Ostaszewski - 179 - PRACTICE EXAMINATION N0. 3 z”. =1.64+M-(1.65—1.64)=1.645. ’ 0.9505 - 0.9495 Thus, 1.645 = t "2500 —— , or s} 340500 t = 2500 + 1.645 - \I 340500 = 34593966. Since k - E (T) = t, we conclude that proportionality constant k for the premium per insured should be k = t = 34593966 513840. E(T) 2500 Answer E. 2. Sample Course 1 Examination, Problem No. 17 An actuary is reviewing a study she performed on the size of claims made ten years ago under homeowners insurance policies. In her study, she concluded that the size of claims followed an exponential distribution and that the probability that a claim would be less than 1000 was 0.25. The actuary feels that the conclusions she reached in her study are still valid today with one exception: every claim made today would be twice the size of a similar claim made ten years ago as a result of inflation. Calculate the probability that the size of a claim made today is less than 1000. A. 0.063 B. 0.125 C. 0.134 D. 0.163 E. 0.250 Solution. Given that the claims follows the exponential distribution, if we write [1 for its mean, the PDF is X X fx (x) = leg for 0 S x < co, and zero otherwise, while the GDP is Fx (x) = l — e7 for the same p values of x. The current distribution is such that X 0.25 = Fx (2000) = 1 — e7 and this gives u = . We need to find FX (1000). Given the information we now have, we find it to be: Fx(1000)=1-e_% =1—eEgg = —\/§=1-%=0.134. Answer C. 3. Sample Course 1 Examination, Problem No. 22 A dental insurance policy covers three procedures: orthodontics, fillings and extractions. ASM Study Manual for Course Pl] Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 180 - PRACTICE EXAMINATION NO. 3 During the life of the policy, the probability that the policyholder needs: 0 Orthodontic work is 1/2. 0 Orthodontic work or a filling is 2/3. 0 Orthodontic work or an extraction is 3/4. 0 A filling and an extraction is 1/8. The need for orthodontic work is independent of the need for a filling and is independent of the need for an extraction. Calculate the probability that the policyholder will need a filling or an extraction during the life of the policy. A. 7/24 B. 3/8 C. 2/3 D. 17/24 E. 5/6 Solution. Let 0 be the event of orthodontic work, F be the event of a filling, E be the event an extraction. We are given that Pr(0)=%, Pr(OUF)=§-, Pr(OUE)=-:-:-, and Pr(FnE)=%. Since 0 and F are independent, Pr(O n F) = Pr(O) - Pr(F), and since 0 and E are independent, Pr(O n E) = Pr(O) - Therefore, §=Pr(OUE)=Pr(0)+Pr(E)—Pr(OnE)=%+Pr(E)-%Pr(E), and from that éPr(E)=%, resulting in Pr(E)= Also, 2 l 3:Pr(OUF)=Pr(0)+Pr(F)—Pr(OflF)=§+Pr(F)‘%Pr(F)’ so that %m(r)=%, and pr(p)=§. Finally, 11117 PF E=PF+PE—PFE=— —--=—. (u) () () (n)3+2824 Answer D. 4. Sample Course 1 Examination, Problem No. 23 The value, v, of an appliance is based on the number of years since purchase, t, as follows: v(t) = e(7'°2'). If the appliance fails within seven years of purchase, a warranty pays the owner the value of the appliance. After seven years, the warranty pays nothing. The time until failure of the appliance failing has an exponential distribution with mean 10. Calculate the expected payment from the warranty. A. 98.70 B. 109.66 C. 270.43 D. 320.78 B. 352.16 Solution. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski ~ 18] o PRACTICE EXAMINATION N 0. 3 If T is the time to failure, then we are given that f, (t) = 0.1e'°"‘ . The expected payment is 7 7 1 _L 1 'I 1 'l Iv(t)fr (Odt = J’e7-02: _ _ _ ,0 dt = . e'l-Olr _ e-o.n dt = _l_0 _J‘e1-03: dt = o o o 10 10 o 1 -1 "’ = — - e7’03' = —- -(e7‘2" — e7) = -e —-e = 320.7811. 10 0.3 mo 3 3 3 You can also use the Darth Vader Rule, but you have to apply it to the payment amount, not T. We define a new random variable describing the payment amount: {8-0-22 0 < z s 7, 1 1114.9 U: 0, t>7. It is clear that U is nonnegative, as it is a value of the exponential function, or 0. We have Pr(U = 0) = Pr(T > 7) = e"'°-‘ = e'°'7. Note also that the value function v(t) = en'om is decreasing for 0 < t < 7, with its largest value being v(0) = e7 and its smallest value being v( 7) = e”. This means that s” (u) = 1— e""7 for u S e“. Recall that FT (t) = Pr(T S t) = l — e'°‘" for t > 0. For u > e” we have s” (u) = Pr(U 2 u) = Pr(e7'°lr 2 u) = Pr(7 - 0.2T Z lnu) = = Pr(T S 35 _ Slnu)=1_ e-0.l(35-Slnu) = l _ e-as _ €0.51“ =1_ e-as _ J; as long as u S e7. and s” (u) = 0 for u > e’. Therefore E(U)= T(l—e'°'7)du+ j. (l—e'”~/H)du= =(e7 —e4‘9)-§(e7 - e4-9)= %e7 —%e43 z 320.781126. In this case, the Darth Vader Rule is actually the harder way to do this problem, unusually so. Answer D. 5. Sample Course 1 Examination, Problem No. 24 An automobile insurance company divides its policyholders into two groups: good drivers and bad drivers. For the good drivers, the amount of an average claim is 1400, with a variance of 40,000. For the bad drivers, the amount of an average claim is 2000, with a variance of 250,000. Sixty percent of the policyholders are classified as good drivers. Calculate the variance of the amount of a claim for a policyholder. A. 124,000 B.145,000 C. 166,000 D.210,400 E.235,000 ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004—2008 by Krzysztof Osmszewski - I82 - PRACTICE EXAMINATION N 0. 3 Solution. The distribution of the random amount of the claim X can be viewed as a mixture of two distributions: 0 X1 : random claim amount for good drivers, with probability weight of 0.60, and 0 2 : random claim amount for bad drivers, with probability weight of 0.40. We have: E(X,) = 1400, Var(X,) = 40,000, E(X2) = 2000, Var(X2) = 250,000, so that by using the weights of 60% for good drivers and for 40% bad drivers, we get E(X)=0.6.E(X1)+0.4-E(X2)= 0.6-1400+0.4-2000 =1640, E(X2)=0.6-E(X3)+0.4-E(x§)= V3131) (50(1))2 “WM (50‘: i): = 1200000 + 1700000 = 2900000. Var(X) = [13(X2)-(E(X))2 = 2900000 —16402 = 210400. Answer D. _ . 2 . 2 = —0.6 [40,000+1400 ]+0.4 [250,000+2000 ] 6. Sample Course 1 Examination, Problem No. 26 2 + e' 9 Let X be a random variable with moment generating function M (t) =[ J . Calculate the variance ofX. A.2 B.3 C8 D9 E.11 Solution. Recall that the MGF of a Bernoulli Trial is qte°"+p-e” =(1-p)+pe‘. I 2 + e = %+ ée‘ is the MGF ofa Bernoulli Trial with p = Hence Therefore, 9 l M (t) = [2 T: J is the MGF of a sum of 9 independent identically distributed Bernoulli Trials 1 with p = That sum is, of course, the binomial distribution with n = 9 and p = %, whose l 2 variance is n =9-—-—=2. P4 3 3 Answer A. Alternatively, you can recall this key property of moment generating functions: ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 183 - PRACTICE EXAMINATION NO. 3 E<X">=“""fz“’|.- d Here, I 9 l 8 I E(X)=dM(t) =_d_ 2+e =9 2+e e_ :3, dt (=0 dt 3 3 3 l=0 t=0 d2M(t) d 2+e' " 2+e' ‘ 2+e' 7 3X2 = ——3' = 3' +8' ' =11. ()dt2,_odte[3] 33 33 e - [=0 l=0 Finally, Var(X)= E(X2)-(E(X))2 =11- 32 = 2. AnswerA,again. 7. Sample Course 1 Examination, Problem No. 29 An insurance company designates 10% of its customers as high risk and 90% as low risk. The number of claims made by a customer in a calendar year is Poisson distributed with mean 9 and is independent of the number of claims made by a customer in the previous calendar year. For high risk customers 9 = 0.6, while low risk customers 0 = 0.1 . Calculate the expected number of claims made in calendar year 1998 by a customer who made one claim in calendar year 1997. A. 0.15B. 0.18 C. 0.24 D. 0.30E. 0.40 Solution. You might think at first that since the number of claims in a year is independent of number of claims in another year, you can disregard the information that a customer made one claim in 1997, because you are working on the expected number of claims in 1998. However, the probability of being a high risk or being a low risk customer is not independent of the number of claims in any year, and the fact that a customer made a claim in 1997 affects the probability that this specific customer is high risk or low risk. So the question is really asking this: What is the expected number of claims by a customer of whom we do not know whwther he/she is high risk or low risk, but we know that the last time we ran this experiment of counting claims of a customer in a year, this customer had one claim. The prior probability of being high risk is 10% and the prior probability of being low risk is 90%. Given one claim in a year, what are the posterior probabilities of being high risk and low risk, and based on that, what is the expected number of claims this customer makes? Once you phrase this problem in the way just described, you can see that this is a Bayes Theorem problem. Let us now label the events: S is the sample space of all customers, H is the subset of S consisting of high-risk customers, L is the subset of S consisting of low-risk customers, and N y is the number of claims in year y for a given customer. We are asked to find E (Nlml N19,, = 1). For a distribution of the number of claims that is Poisson, its parameter (typically denoted by A) is the expected value. For high risk customers, ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 184 ‘ PRACTICE EXAMINATION NO. 3 this is 0.6, and for low risk customers this is 0.1. For a mixed group of customers, the resulting distribution is a mixture of two Poisson distributions, and the unconditional mean would be simply 10% -0.6 + 90%-0.1 = 0.06 + 0.09 = 0.15. But the customer we are interested in is known to have had a claim in 1997, and this affects his/her probability of being high risk or low risk. Therefore, the desired mean will be calculated as: E(N,,,,|N,,,, = 1) = Pr(H|N,,,, =1).0.6+Pr(L|Is/,,.,7 =1)-0.1. k -9 We know that Pr(N = k) = 9 e for k = 0,1,2,3... and 0 is the mean of NW. That mean of 1997 N1997 can be either 0.6, or 0.1 . Based on this we get: 0.6e'05 Pr(N,997 =1|H)= Pr(N,,,, =1|9 = 0.6) = = 0.3293, and 0 le-OJ Pr(N,99., =1|L)= Pr(N,9,, = 119 = 0.1) = ' l = 0.0905. Applying Bayes’ Theorem, we conclude that: Pr(H|N.,,, = 1) = Pr(9 = 0.6|N,,.,7 = 1) = Pr(N,99, =1|H)-Pr(H) = Pr(N,.,97 = 1|H)-Pr(H)+Pr(N,,,, =1]L)-Pr(L) _ 0.1-0.3293 ' 0.1~0.3293+ 0.9 0.0905 = 0.2879, and Pr(L|N,,,, = 1) = 1 — Pr(H|N,,,, = 1) = 0.7121. Finally, the revised (posterior) calculation of the mean as a weighted average is: E(N,993) = 0.2879 -0.6 + 0.7121 01 = 0.2440. Answer C. 8. Sample Course 1 Examination, Problem No. 31 Let X and Ybe random losses with joint density function fx_y (x, y) = 2x for 0 < x <1 and 0 < )2 <1. An insurance policy is written to cover the loss X + Y. The policy has a deductible of 1. Calculate the expected payment under the policy. A. 1/4 B. 1/3 C. 1/2 D. 7/12 E. 5/6 Solution. Consider the region where the joint density is positive, i.e., 0 < x <1 and 0 < y < 1, and inside of ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 185 - PRACTICE EXAMINATION NO. 3 it the region where x + y > 1, i.e., y >1— I, so that a positive payment is made under this policy. y 1 "_. ’ The expected payment is the integral of the payment amount x + y —1 (this is what is paid after the deductible is applied) over the shaded region, i.e., the region where the payment after deductible is positive: 1 1 l 1 1 3 "1 II(x+y—l)-2xdxdy=II(2x2+2xy-2x)dxdy=‘[[%+x2y—x2] dy= Ol-y Ol-y o “1-, ' 2 “' ' 1 2 1 =11x21-w-ID war—1w("o-aw o 3 My 0 3 3 ' 1 1 3 3,13 ' y3 -y‘ y3 ’=' 1 = ——+——— +— —- d = ——+ 2 d = —+— =—. {[3, 3 3 3y 3y 33'ij 3 y y 12 3 F0 4 Answer A. 9. Sample Course 1 Examination, Problem No. 34 Under a group insurance policy, an insurer agrees to pay 100% of the medical bills incurred during the year by employees of a small company, up to a maximum total of one million dollars. The total amount of bills incurred, X, has probability density function x(4— x) fx (x) = 9 , 0, otherwise. where x is measured in millions. Calculate the total amount, in millions of dollars, the insurer would expect to pay under this policy. for0<x<3, A. 0.120 B. 0.301 C. 0.935 D. 2.338 E. 3.495 Solution. Because of the policy limit of 1 million, we simply need to calculated the expected value of min(X,1), i.e., ASM Study Manual for Course Pl] Actuarial Examination. (9 Copyright 2004-2008 by Knysztof Ostaszewski - 186 - PRACTICE EXAMINATION NO. 3 l x(4-x) 3 x(4—x) l4262—3:3 3425-3:2 E X,1 = - dx+ l- dx= dx+ dx= tmww 49 £9! 3 4‘ 2 3 3 = 4i—x— + i—x— =-4—-—1-—+2—1—E+-1—=0.9352. 27 36 o 18 271 27 36 9 27 You could also use version of Darth Vader Rule for E (min(X ,1)), i.e., . ‘ '3u(4—") l 2 2 1 3 "=3 E(rmn(X,l))=£sx(x)dx=‘£,I 9 dudx=o(§u __27u luck: I x=l (fie—ea = o 9 :80 =1__2-+-—1—=0.9352. 27 108 AnswerC. 10. Sample Course 1 Examination, Problem No. 35 Suppose the remaining lifetimes of a husband and a wife are independent and uniformly distributed on the interval (0, 40). An insurance company offers two products to married couples: 0 One which pays when the husband dies; and 0 One which pays when both the husband and wife have died. Calculate the covariance of the two payment times. A. 0.0 B. 44.4 C. 66.7 D. 200.0 B. 466.7 Solution. Let H be the random time to death of the husband, Wbe the time to death of the wife, and X be the time to the second death of the two. Clearly, X = max(H ,W). We have fa(h)=fw(w)=% for 0 S h S 40. and o s w s 40. Thus E(H) = 5(w) = 20.1:unhemore, FX(X)= S.x)=Pr(max(H’W)Sx)= x x x =Pr HanWSx =PrHS -Pr S =—-—= . ({ }{ }) ( x) (Wx 40401600 Thisimpliesthat x2 sx(x)=l —1600 for OSxS40, and ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 187 - PRACTICE EXAMINATION NO. 3 40 2 3 E(X)=j[1- x )dx=4O-£[email protected]fl=8—Ou O 1600 4800 3 3 3 In order to find covariance, we also need to find E (XH ) = E (H max(H,W)). We separate the double integral into two parts: one based on the region where the wife lives longer and one based on the region where the husband lives longer, as illustrated in the graph below 1 402 + -404 = 600. 1 3' 12800 11. Sample Course 1 Examination, Problem No. 37 An insurance contract reimburses a family’s automobile accident losses up to a maximum of two accidents per year. The joint probability distribution for the number of accidents of a three- person family (X, Y, Z) is p(x,y,z) = k(x+ 2y+ z), where x = 0,1, y = 0,1,2, 2 = 0,1,2, and x, y and z are the number of accidents incurred by X, Y and Z, respectively. Determine the expected number of unreimbursed accident losses given that X is not involved in any accidents. A. 5/21 B. 1/3 C. 5/9 D. 46/63 E. 7/9 ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostasaewski - 188 - PRACTICE EXAMINATION N0. 3 Solution. This discrete distribution is fully described as follows: Pr(X=O,Y=O,Z=O)=O, Pr(X=l,Y=0,Z=0)=k, Pr(X=0,Y =0,z=1)= k, Pr(X= 1,Y=0,Z=1)= 2k, Pr(X=0,Y=0,Z=2)=2k, n(x=1,r=o,z=2)= 3k, Pr(X=0,Y=1,Z=O)=2k, Pr(X=1,Y=1,Z=0)=3k, Pr(X=0,Y=1,Z=l)=3k, Pr(X=l,Y=1,Z=l)=4k, Pr(X=0,Y=l,Z=2)=4k, Pr(X=1,Y=1,Z=2)=5k, Pr(X=0,Y=2,Z=0)=4k, Pr(X=1,Y=2,Z=0)=5k, Pr(X=O,Y = 2,z=1)= 5k, Pr(X=1,Y = 2,2 =1): 6k, Pr(X=O,Y= 2,Z=2)=6k, Pr(X= 1,1' = 2,z=2)= 7k. Adding all of these probabilities gives 63k = 1, so that k = Furthermore, Pr(X=0)=£+i+£+l+i+i+i+i+£=a, 63 63 63 63 63 63 63 63 63 63 and using the definition of conditional probability, we get Pr(Y=0,Z=OlX=0)=0, Pr(Y=0,Z=l|X=0)=%, Pr(Y=o,z=2|X=o)=l, Pr(Y=1,z=olx=o)=l, 27 27 3 4 Pr(Y—1,Z—l|X—0)—E7, Pr(Y—l,Z—2|X—O)_E, Pr(Y=2,z=o|X=o)=i, Pr(Y=2,Z=1IX=0)=i. 27 27 Pr(Y=2,Z=2|X=0)=%. Let us write W for the number of unreimbursed losses given that X is not involved in accidents. We are looking for E But 0, (Y+zsz)|x=o, w: 1, (Y+Z=3)|X=0, =(max(Y+Z,2)-2)|X=O. 2, (Y+Z=4)|X=0, Therefore E(W)=1-Pr(Y=l,Z=2|X=0)+l-Pr(Y=2,Z=l|X=0)+2«Pr(Y=2,Z=2|X=0)= =1.[i+i)+2.£=l. 27 9 ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Knyszlof Ostaszewski - 189 - PRACTICE EXAMINATION N0. 3 12. Sample Course 1 Examination, Problem No. 39 The loss amount, X, for a medical insurance policy has cumulative distribution function 0, x<0, 3 Fx(x)= %[2x2—%), 03x33 1, x>3. Calculate the mode of the distribution. A. 2- B. l C. 2 D. 2 E. 3 3 2 Solution. The mode is the point where the PDF (for continuous distributions) or the probability function (for discrete distributions) is maximized. This is a continuous distribution with PDF: 4x x2 x = F’ x = __ _. ._ . m ) .( ) 9 9 In order to maximize fx ,we take its derivative and set it equal to zero: 4 2x_ f§(x)=§—?-0’ resulting in 2x = 4 and x = 2. To the left of O and to the right of 3, fx is identically zero. Also, fx (0) = 0, fx (3) = g, and fx(2) = Thus, the maximum is attained atx = 2. Answer D. 13. Sample Course 1 Examination, Problem No. 40 A small commuter plane has 30 seats. The probability that any particular passenger will not show up for a flight is 0.10, independent of other passengers. The airline sells 32 tickets for the flight. Calculate the probability that more passengers show up for the flight than there are seats available. A. 0.0042 B. 0.0343 C. 0.0382 D. 0.1221 E. 0.1564 Solution. Let X be the random number of passengers that show up for a flight. We want to find the sum of two probabilities: Pr(X = 31) and Pr(X = 32). We can treat each passenger arrival as a Bernoulli Trial with p = 0.90. Then X has binomial distribution with n = 32 and p = 0.90, and 32 Pr(X = x) =[ x ]-0.90* . 0.103“. The probability desired is: ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Kmysztof Osmszewski - 190 - PRACTICE EXAMINATION NO. 3 Pr(X=3l)+Pr(X=32)=( ].o.9o3‘-o.10‘+[ ]—0.90”-0.1o°= = 3.2 . 0.9031 + 0.9032 = 0.1564. Answer E. 14. May 1988 Course 110 Examination, Problem No. 6 Let X and Yhave a bivariate normal distribution with means [1, = fly = 0, variances of =1, 0': = 2 and correlation pxy = What is the conditional variance of Y, given X = x? A. l B. 3 C. 1 D. 3 E. 2 2 4 2 Solution. Recall that _ x-Mx 2 2 (YIX—x)~N[-uy+poy' O. ’(l—pty)oy]’ and therefore, Var(Y|X=x)=(l-p2)0'2= 1—l 2=-3-. “y y 4 2 AnswerD. 15. May 1988 Course 110 Examination, Problem No. 12 Let X be a continuous random variable with density function 1x’, for0<x<2, fx ('7‘): 4 0, otherwise. What is the density function of Y = 8 — X3 for 0 < y < 8? A. lea—yr? B. 1(s—yfi c. ins-y)i D. i(s—y)% E. 1(s—y) 3 4 6 12 4 Solution. 1 Since Y: 8 — X3,we have X3 = 8-Y, and X = (8-Y)3. Therefore, 2 £=—l(s—y)-s. dY 3 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 191 - PRACTICE EXAMINATION NO. 3 This implies that dx 1 1 3 1 -2 1 1 fr (y) = fx(x(y))|d—y| = 3((8 - yr) 3(8 -y) 3 = 5(8 —y)3. Answer D. 16. May 1988 Course 110 Examination, Problem No. 1 In a dice game, the player independently rolls a fair red die and a fair green die. The player wins ifand only ifthe red die shows a 1, 2, or 3, or ifthe total on the 2 dice is 11. What is the probability the player will win? A. l B. i C. 1—9- D. i E. 2—9- 36 9 36 9 36 Solution. The total number of possible outcomes is 36 (6 on each die). There are six ways to get a l on the red die (six outcomes of the green die accompanying the l on the red die), six ways to get a 2 on the red die, and 6 ways to get a 3 on the red die. The total of the two dice can be 11 only two ways: 5 and 6, or 6 and 5. The total number of favorable outcomes is therefore: 6 + 6 + 6 + 2 = 20. The answer is a = 36 9 Answer D. 17. May 1988 Course 110 Examination, Problem No. 2 In a large population of people, 50% are married, 20% are divorced, and 30% are single (never married). In a random sample of 4 people, what is the probability that exactly 3 are married? T [4] 4 4 3 A. (3)053 -0.2-o.3 B. (3)054 c. 0.53 D. 0.54 E. Solution. Treat each person picking as a Bernoulli Trial with picking a married person being a success. Then n = 4,12 = 0.50, and the probability of three successes is [ : ]O.S3 05' =[ g ]-0.5‘. Answer B. 18. May 1988 Course 110 Examination, Problem No. 4 Let X and Y be random variables whose joint distribution is uniform over the half-disc: ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 192 - PRACTICE EXAMINATION NO. 3 {(xv)’)|x2 + y2 51 and x ..>. 0}. What is the marginal density function of X for 0 S x S 1? 4 l 2 1 3 _ 2 1 3 A. ;(1—x1)z B. 2(1-x)2 C. n(1 x)2 D.” E.l Solution. The area of a half of the unit disc is Therefore the joint density of X and Y is constant and equal to The marginal density of X is l-xz 2 4 fx(x)= I gram; 1-x2 43 Answer A 19. May 1988 Course 110 Examination, Problem No. 7 An urn contains 10 balls: 5 are white. 3 are red, and 2 are black. Three balls are drawn at random, with replacement, from the urn. What is the probability that all 3 balls are different colors? E. 0.84 C. 0.18 D. 0.40 A. 0.03 B. 0.09 Solution. There are 103 ways to pick three balls (with replacement) from the um, and this is the total number of possible outcomes. There are 5 ways to pick one white ball, 3 ways to pick one red ball, and 2 ways to pick one black ball, and there are 3! ways to arrange the order of the picking of the three colors. Thus the probability desired is: 5-3-2-2o3_ 3~2-3_ 18 1010-10 10-10 100' Answer C. 20. May 1988 Course 110 Examination, Problem No. 8 Three fair dice are tossed independently. Let E; denote the event that the i-th die results in a 6. What is Pr(E, or;2 UE3)? 1 5 91 1 125 A. — B. — C. — D. — E. — 216 12 216 2 216 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof 0staszewski - 193 - PRACTICE EXAMINATION NO. 3 Solution. The event El u E2 u E3 is: getting at least one 6 in three simultaneously tossed dice. There are 6-6-6 total outcomes possible and there are 5-5-5 ways of not getting a six. Thus the probability desired is: 131-031 0132 UE3)=1 Answer C. _£__91. 216 216' 21. May 1988 Course 110 Examination, Problem No. 10 KB and F are events for which Pr(EU F) = 1, then Pr(Ec ch)must equal A.0 B. Pr(EC)+Pr(FC)—Pr(EC)Pr(FC) D. Pr(EC)+Pr(FC)—1 E] c. Pr(E°)+Pt(FC) Solution. The condition Pr(E u F) = 1 implies that o = Pr((E u Ff) = Pr(EC n Fe). Hence Pr(EC UFC)=Pr(EC)+Pr(FC)—Pr(EC nFC)=Pr(Ec)+Pr(FC). Answer C. 22. May 1988 Course 110 Examination, Problem No. 16 What is the probability that a hand of 5 cards chosen randomly and without replacement from a stande deck of 52 cards contains the king of spades, exactly 1 other king, and exactly 2 queens? [55] F?) [5:] (5:) (a Solution. 52 . There are 5 to choose five cards, and that’s the total number of outcomes. There 13 one way 3 4 to choose the king of spades, there are [1) to choose exactly one other king, [2] ways to ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 194 - PRACTICE EXAMINATION NO. 3 choose exactly two queens, and ( 1 J ways to choose one card (remaining) that is not a king, 3 4 44 and not a queen. This implies that the total number of favorable outcomes is ( 1 21 1 Answer B . 23. May 1988 Course 110 Examination, Problem No. 18 Let X and Y be continuous random variables with joint density function 6xy+3x2, for0<x<y<l, xx.,(x.y)={ 0, otherwise. What is E(X|Y = y)? 2 2 3 A_ 3y4 3_ 2y5 C. m D. w 3 fl 4y 4y 16 Solution. We will be working with fx 1' (Ivy) f x Y = y = --———. X( I ) fr (Y) so we start by finding x=y = 3y3 +y3 = 4y3’ x=0 f}, (y) = R61)! + 3x2)dx = (3x2y+x3) for 0 < y <1. Therefore, f x, 6 +3):2 3 _ 3 _ fx(xly=y)= x.r( 3’): x)’ 3 =_xy2+_x2y3. My) 4y 2 4 Finally, ’ 3 3 1 3 ’ 1 3 11 EXY= = x- — '2+—x2 ‘3]dx=(—x3 '2+— ‘ ’3] =— +— =— . (ly)£(2w4y 2yl6xy02y16yl6y AnswerE. 24. May 1988 Course 110 Examination, Problem No. 22 2 Let X be a discrete random variable with probability function Pr(X = x) = g, for x = l,2,3,.... What is the probability that X is even? ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof 0staszcwski - 195 - PRACTICE EXAMINATION N0. 3 A.1 3.3 c.1 13.3 13.3 4 7 3 3 4 Solution. Pr(X is even):Pr(X:2)+Pr(X=4)+Pr(X=6)+...= 222221[1)2(1]3 21 =—2+—4+7+—8...=—- l+—+ — + — +... =—-—=—. 3 3 3 3 9 9 9 9 91_1 9 AnswerA. 25. May 1988 Course 110 Examination, Problem No. 23 Let X1 ,X2 , and X 3 be independent continuous random variables each with density function x/E—x, for0<x<\/E, f (x)= . 0, otherwrse. What is the probability that exactly 2 of the 3 random variables exceed 1? A. g—JE B. 3—2J2 c. 3(~/§—1)(2—~/§)2 D. [g—JEHf—é] E.3[-:-- 41/13 Solution. The probability of exceeding 1 for any of these three random variables is: .15 2 l 1 1 3 J21 dx= 2 J2—1 -[— 2] =2-J2—[—-2-—)=——~/§. J: ( x) ‘F( ) 2 x 2 2 2 Treat each “draw” of these three random variable as a Bernoulli Trial, and exceeding 1 as a success. We want exactly two successes in three experiments and that probability is: [Elli-filll-G-fil}384020-3- Answer E. x: 1:1 26. May 1988 Course 110 Examination, Problem No. 25 Let the random variable X have moment generating function M (t) = em": . What is E (X 2 )? A.1 B.2 C.3 D.9 E.11 Solution. ASM Study Manual for Course Pll Actuarial Examination. (9 Copyright 2004-2008 by Knysztof Ostaszewski - 196 - PRACTICE EXAMINATION NO. 3 d2 d2 2 d 2 2 =_ =_ 3l+t =_ 3 2 31+: = E(X) dtz Mx(t)“=0 dtze mo dt(( + t)e )mo =(2e’”"+(3+2t)1e3‘*‘2) o=2+9=11. I: Answer E. 27. May 1988 Course 110 Examination, Problem No. 31 Defective items on an assembly line occur independently with probability 0.05. A random sample of 100 items is taken. What is the probability that the first item sampled is not defective, given that at least 99 of the sampled items are not defective? A. —5 B. E C. 9-8- D. g E. -—5'90 5.90 100 99 100 5.95 Solution. We start by labeling the events. Let E be the event that the first item sampled is not defective, and let F be the event that at least 99 of the items sampled are not defective. Then _ Pr(En F) _ Pr({l defective} n E )+ Pr({0 defective}nE ) ———————— — Pr E F ( l ) Pr(F) Pr(l defective or 0 defective) 99 0.95 l ]-0.9598 -o.05 +0.95 0.95” Divide top & bottom = 100 [ 1 ]0.95” -0.05+o.95‘°° by 0.9599 99-o.os+o.95 _ 5.90 100-0.05+o.95 595' Answer E. 28. May 1988 Course 110 Examination, Problem No. 32 Let X and Y be independent continuous random variables with common density function 1 for0<x<1 fx(x)={o, What is Pr(x2 2Y3)? otherwise. A.1 B.2 Cg DE El 3 5 5 3 Solution. The joint distribution of X and Yis uniform on the unit square, so probability of any event is the corresponding area within the unit square. Thus Pr(X2 2 Y 3) is the area of the region in the unit ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 197 - PRACTICE EXAMINATION NO. 3 2 square where x2 2 ya , or y S x3. That area is l 2 Ix’dx =21:3 = o 5 5 x=l x=0 Answer C. 29. May 1988 Course 110 Examination, Problem No. 33 Let X have a binomial disuibution with parameters n and p, and let the conditional distribution of Y given X = x be Poisson with mean x. What is the variance of Y? A.x B. np c. np(1 - p) D. np2 E. np(2— p) Solution. Recall the formula: Var(Y) = E(Var(Y|X))+ Var(E(Y|X)). Here, (Y [X = x) is Poisson with mean x. Therefore, E (Y IX) = X, and Var(Y |X) = X. Based on this, Var(Y) = E(Var(Y|X)) + Var(E(Y|X)) = E(X)+Var(X) = =np+npq=np(1+q)=np(2-p)- Answer E. 30. May 1988 Course 110 Examination, Problem No. 34 A sample of size 3 is drawn at random and without replacement from population {1, 2, 3, 4, 5}. What is the probability that the range of the sample is equal to 3? A. i B. 31 c. l D. l E. 3 15 125 10 10 5 Solution. There are ( 3) ways to choose such samples of size 3 from a population of size 5. How many of these samples have the range of exactly 3? These are {1,2,4},{1,3,4} ,{2,3,5},{2,4,5}, so there are four of them. The probability desired is [i = i] ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004—2008 by Knysztof Ostaszewski - 198 - Answer E. ...
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