Practice_Exam_20-Solutions

Practice_Exam_20-Solutions - PRACTICE EXAM 20 PRACTICE...

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Unformatted text preview: PRACTICE EXAM 20 PRACTICE EXAMINATION NUMBER 20 SOLUTIONS 1. May 1990 Course 110 Examination, Problem No. 49 Let X and Y be continuous random variables with joint density function xy forOSxSZandOSySl, fx.r(x’y)= 0 otherwise. What is Pr[§ SY S X)? A. B C 1 ,1 ,1 D2 53 32 8 4 8 4 Solution. The joint density is positive in the rectangle defined by the conditions 0 S x S 2 and O S y S l, but the region of probability sought is defined by the condition is y S x, as shown below. l 2 The probability sought is calculated as the integral of the joint density over the region between the lines y =25 and y = x, marked in the figure above, either as ASM Study Manual for Course Pll Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 673 - SECTION 24 2. May 1990 Course 110 Examination, Problem No. 50 A coin is twice as likely to turn up tails as heads. If the coin is tossed independently, what is the probability that the third head occurs on the fifth trial? A.—8— 3.39. c.19 13.31”— E.3 81 243 81 243 5 Solution. Because this coin is twice as likely to turn up tails as heads, we have Pr(Head) =% and Pr(Tail) = Therefore, Pr(3rd head on 5th toss) = Pr({2 heads in first 4 tosses} 0 {Head on 5th toss}) = . 4 12 2 21 8 =Pr(2heads1nfirst4tosses)-Pr(Headon5thtoss)= 2 - 3 - E We can also express the probability sought in terms of a negative binomial distribution. Let X be the number of tails that are tossed that are tossed until the third head occurs. Then X has a negative binomial distribution with p =% and r = 3, and we are looking for Pr(X = 2). This also gives us Pr<x=2>=[:J-(%T(aer- Answer A. 3. You are given that the time T until a major engine failure in a new car is exponentially distributed with mean of 8 years. You are given that if the first major engine failure occurs during the first three years of the car’s life, the cost of the repair to the car owner is $0, because the repair is covered by the manufacturer’s warranty. If the first major engine failure occurs at any later time, the cost to the repair is $8,000 (we disregard any time value of money). Calculate the expected cost of the first major engine repair that is paid by the car owner. A. $5125 B. $5498 C. $5617 D. $8000 E. $43497 Solution. The expected value sought is 3 $0 « Pr(O S T S 3) + $8000 ' Pr(T > 3) = $8000e-i = $5498.31. Answer B. 4. You are given a continuous random variable X, whose density is ASM Study Manual for Course P/1 Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 674 - PRACTICE EXAM 20 f(x)={l—le for—1<x<1, 0 otherwise. If 21:0_75 is the 75-th percentile of this random variable, find E(X| X > xo_7s A. l-—l- B.O.5284 C. l—fl D. E E.0.75 Ji 3 3 Solution. First, we find the 75-th percentile of X. We must have 30.1: 30.15 0 x1175 0.75 = fo(x)dx = j (1 -|x|)dx = {(1- (—x))dx+ I (1 —x)dx = _.. -1 -1 o 3=Io.7s = 0.5 + (x—05x2) = 0.5 + xm -O.5x:..,5. x=0 Therefore, 0.5x§_.,5 — x0” + 0.25 = 0, resulting in x035 = l i J; Because the 75-th percentile must lie between 0 and 1, we have x015, = l- J; Furthermore, fx (3‘) f l , (x)=——-—-i- “'ng Pr[X>1—J;] for l-J;<x<l, and jixx>l_‘g](x) X = 0 otherwise. This gives us l l E XX>1—\/-T- = wa—ll—dx: jx-l—xdx= 2 , 1 I l in]; Pr X>1- E H]; 4 I x=l . =4 j x (1—x)dx=4[lx2-1x3] =1—£. x=l- l 3 Answer C. 5. You are given that X and Y are independent, identically distributed, normal random variables with mean 4 and variance 12. Find Pr(]X—Y| <1). 6%“ ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004—2008 by Knysztof Ostaszewski - 675 - SECTION 24 A. 0.5832 B. 0.5793 C. 0.5000 D. 0.3333 E. 0.1618 Solution. We have E(X—Y)= E(X)-E(Y)= 4—4 =0, and Var(X- Y) = Var(X)+Var(Y)—2Cov(X,Y) = 12+ 12— o = 24, and therefore (X—Y) ~ N(0,Jfi) = N(0,2JE). Let us write Z for a standard normal random variable, and (I) for the CDF of Z. We conclude that pr(lx_yl<1)=Pr(-1<X-Y<l)=Pr(-:Jgo ((XZ/g-O <%) = Pr[% <z <fi): ¢[fi)—¢(—E§EJ= 1 l l =<I) — - l—q) —— =24) — -1==2<I) 0.2041 —l= (2J6) [ (NED [2J6] ‘ ) = 2 '(0.41-0.5832+0.59 ~0.5793)— 1 = 0.1618. using linear hitupolan'on Answer E. 6. For cell phones produced by a certain manufacturer: (i) Lifetimes follow a Pareto distribution with a >1 and 0 = 4. (ii) The expected lifetime of a cell phone is 2 years. Calculate the probability that the lifetime of a cell phone is at least 6 years. A. 0.04 B. 0.06 C. 0.08 D. 0.10 E. 0.12 Solution. The expected value of a variable following the Pareto distribution is a6 l, and in this case 9 = 4 and the expected value is 2. This gives a = 3. The probability that the lifetime of a cell phone exceeds 6 years is 9 a 4 3 6 = — = —— =0.43=0.064. SA) (6+6) [6+4] 7. The amount of damage X in a car accident is given by the exponential distribution with mean 10,000. The insurance policy covering that damage has a deductible of 500 and a policy limit of ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 676 - PRACTICE EXAM 20 100,000. Find the coefficient of variation of the amount paid per payment (i.e., amount paid given that a payment is made by the insurance company). A. 13.43 B. 13.47 C. 14.25 D. 15.11 E. 16.00 Solution. Let us write Y for the amount paid per payment. We have Y = (x— 500| 500 < x < 100000). Note that for 0 S y S 99500, s, (y): Pr(Y > y) = Pr(X—500 > yl500 < X <100000)= Pr({X > 500 + y}n{500 < x <100000}) _ 91(500 + y < X < 100000) _ Pr(500 < x < 100000) Pr(500 < X < 100000) 1 -500» 1 -m y _ 10000 e W ' 10000 e W _ em 'e-m - e"° 1 e_% - l e_% e-ODS _ e-IO 10000 10000 and for y > 99500, 3,, (y) = 0. This implies that )l -0.05 - e , e fr (3') - W for 0 < y < 100000 and zero otherwise. Based on this, y w 995m .. e-ons.e loooo_e l0 E(Y)=I5y(y)dy= I Wdy= )7 e _— -o.05 _ -1o " e e )‘=0 "”’°°] _ 99500.24"10 _ _10000 e'°”5 99500e-°"° _ . (1 _ 8-995 ) _ _ 3-005 _ e-IO e-ons _ e-IO —0.lO 933% z 9995 .2509. e = 10000 — 8 Also, ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 677 - SECTION 24 y -005 _ ‘lomo ’ u = 3’2 V = —% 99500 -o,os 'm __ _ 2 _ 2 e -e _ e e _ )— I y .10000(e-0.05 e—lo)dy- -L — o - e-ODS . e 10000 (1 =2 (1 dv=——-—d u y y 10000(e‘°°5-e"°) y ——,—_——1 Inteynfionbypam y “99500 y 2 8-095 _ e'm 99500 e-oos . e'm = — — + J. 2y—dy = y e-ODS _ 6—10 0 e-ODS _ e-IO y=0 _.."_. u _ 2y v _ 10000e4’05 -e m — . e-lO + e-ODS _ e-IO _ " — e-oos _ e-IO , '- e-ons . e‘m du = Zdy dv= mdy e - e ———,——_—1 Intcgmtion by pans in the second integral y=99500 y . e—lO e-0.05 _ e-m 99500 e-ODS . e_mooo = —————-——+ —20000y— + j 20000———dy= e—oos _ e-lO e-oos _ e-1o 0 8-055 _ e—IO y=0 y=99500 _ 995002 ~e"° 99500-20000-e‘lo + 2 100002 em” -e '°°°° _ e-ons _ e-IO e-oos _ e-IO e-ons _ e-IO r=0 e"° -1o'° (—9.952 — 19.9 — 2)+ 2 . 10l0 ~e'°”5 = —— z 18129180349. e-oos _ e-lo Based on this Var(Y) = E(Y2)- (19(1))2 z 18029275309. The coefficient of variation is Var(Y) E(Y) Answer A. = 13.4337. 8. You are given that X and Y are jointly uniformly in the region defined by the conditions 0<x<2 and x2 <y<4. Find Var(X|Y=3). ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 678 - PRACTICE EXAM 20 1 1 A c. — D. — . 2 J2 12 ul— 1 '4 Solution. The relationship x2 < y < 4 can be rewritten as x < J; < 2. Therefore, for a fixed value of y, x varies between 0 and J}. As the joint distribution is uniform, the conditional distribution of X given that Y: 3 is also uniform. Being uniform on [0,6] = [Op/3], this conditional distribution has its variance equal to (ii-or l 12 4 ' Answer A. 9. Four actuarial students are taking Course P examination at the same time. For two of them, the probability of passing the examination is 0.6, while for the other two, that probability is 0.9. Their performances are independent of each other. Find the probability that at least one of them passes the examination. A. 0.9999 B. 0.9984 C. 0.9084 D. 0.8667 E. 0.8298 Solution. Let El, E2, E3, and E4 be the events of passing Course P examination by students 1, 2, 3, and 4, respectively. The probability that at least one of them passes the examination is Pr(E, uEz uE, UE4)=1— Pr((E, UEZ uE, UE4)C)= =1— Pr(E,cr1Efn Ef n Ef)=1-Pr(Ef)Pr(Ef)Pr(E§)Pr(Ef)= =1—(1-0.6)-(1—0.6)-(1—0.9)-(l—0.9)= 0.9984. Answer B. 10. The joint density of X and Y is «(y-x) xe forOSxSlandy>x, fxy (xvy) = . 0 othermse. Find Pr(Y > E(Y|X = x)|X = x). A. 0.25 B. e" c. 0.75 D. (1+ e)" E. 2(1+ e)" Solution. Note that ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 679 - SECTION 24 2 +03 M x fx (x) = l “""""dy = l xe"""’dy = xe" I e“"’dy = — xe (3'8? ally x x x so that X is uniform on the interval [0, 1], Based on this f)’ = X) = fx'y (x’y) = xe'XU-x) y-Heo _2 )=exze x =1, i=3 fx ('7‘) for y > x. Furthermore +°° ‘“ u - v - -le"’ E(Y|X = x) = Ixye‘("’)dy = Jcex2 Iye'”dy = y _ x = x X du = dy dv = e'wdy \__v____—J integrationbypam y—Hw m = xe“2 ~ (—ze’”] + are“2 - Ilery = x F: x x )-H«7 _xz =xe‘1-e"2+£-e‘2-(—le"’) =x+e‘2-£——=x+l. x x y“ x x This results in My > E(Y|X= x)|x=x)= Pr[[Y >x+l X “D: x +0- 1_x = I xe-XU-xldy = x .(_l)e‘x(f'1) P": = ( x ] = 3-1 1 x F"; X+- 1 Answer B. 11. Study Note P-09-08, Problem No. 131 Let N1 and N2 represent the numbers of claims submitted to a life insurance company in April and May, respectively. The joint probability function of N1 and N2 is Ill—l N,.N1 ’ 2 — 0 otherwise. Calculate the expected number of claims that will be submitted to the company in May if exactly 2 claims were submitted in April. A. 3(e2 -l) B. 3e2 C. i D. e2 -1 E. e2 16 16 4—e Solution. We begin by finding the conditional probability function of N2 given N I = nI , i.e., ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Ktzysztof Ostaszewski - 680 - PRACTICE EXAM 20 AW(MMJ Nl = =.._'_L—. fN, "1) fN‘ (n!) We have ~ 3 1”fl ” _ - M4 fN. ("1): mew, ("19”2)='Z[Z) ' 2e "l (l—e "1) = "2:1 '11:! geometric series with ratio l-:"" and first term 2"“ 3(1)“ ‘4 4 1—0—ew)'4 4 ' Therefore fN N ("1312) _ _ nz-l f1n2N=nl =—'*—’—=e"‘l—e"' . N ( I l ) le (n1) ( ) This is actually the probability function of a geometric random variable with parameter p = e'"'. The mean of this distribution is i = 117 = e"' , and that equals e2 when n1 = 2. P e ‘ Answer E. 12. Study Note P-09-08, Problem No. 132 A store has 80 modems in its inventory, 30 coming from Source A and the remainder from Source B. Of the modems from Source A, 20% are defective. Of the modems from Source B, 8% are defective. Calculate the probability that exactly two out of a random sample of five modems from the store’s inventory are defective. A. 0.010 B. 0.078 C. 0.102 D. 0.105 E. 0.125 Solution: The number of defective modems is 20% - 30 + 8% - 50 = 10. The probability that exactly two of a random sample of five are defective is (85°) Answer C. 13. Study Note P-09-08, Problem No. 133 A man purchases a life insurance policy on his 40-th birthday. The policy will pay 5000 only if he dies before his 50-th birthday and will pay 0 otherwise. The length of lifetime, in years, of a male born the same year as the insured has the cumulative distribution function ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 681 - SECTION 24 0 for t S 0 [31(1): 1-1.1' l- em for t > 0. Calculate the expected payment to the man under this policy. A. 333 B. 348 C. 421 D. 549 E. 574 Solution. H! " We have Pr(Man dies before age 50) = Pr(T < 50|T > 40) = w = Pr(T < 40) ( ) ( ) l-l.l‘° l—l.l’° (I I” n”) F50-F40 e‘°°°-e'°°° '-' =W=T=l—e woo e IMO Therefore, the expected benefit is 5000 Pr(Man dies before age 50) = 5000 00696 = 347.96. Answer B. 14. Study Note P-09-08, Problem No. 134 A mattress store sells only king, queen and twin-size mattresses. Sales records at the store indicate that one-fourth as many queen-size mattresses are sold as king and twin-size mattresses combined. Records also indicate that three times as many king-size mattresses are sold as twin- size mattresses. Calculate the probability that the next mattress sold is either king or queen-size. A.0.12 B.0.15 C.0.80 D.0.85 E. 0.95 Solution. Let t denote the relative frequency at which twin-sized mattresses are sold. We know that the relative frequency at which king-sized mattresses are sold is 3t and the relative frequency at 3t +t which queen-sized mattresses are sold is = t. Since the sum of all relative frequencies must be 1, we have 1+ 3t + t: 1, and therefore t = 0.2. The probability we seek is 3t + t = 0.80. Answer C. 15. Fall 1979 Part 2 Examination, Problem No. 7 Sixty percent of new drivers have had driver education. During their first year, new drivers without driver education have probability 0.08 of having an accident, but new drivers with driver education have only a 0.05 probability of an accident. What is the probability a new driver has had driver education, given that the driver has had no accidents the first year? ASM Study Manual for Course P/l Actuarial Examination. ‘9 Copyright 2004-2008 by Knysztof Ostaszewski - 682 - PRACTICE EXAM 20 A. 0.05 -0.6 . 0.92.0.4 C. 0.95 -0.4 +092 .04 0.05 +0.6 0.6 +0.92-0.4 0.95 ~0.6 0.95 - 0.4 0.95 - 0.6 D. ——————- E. —— 0.95 - 0.4 + 0.92 - 0.4 +0.92 - 0.6 0.95 -0.6 + 0.92 - 0.4 Solution. Let E be the event of a driver having had driver’s education, and let A be the event of having an accident. We are given that Pr(E) = 0.60, and that Pr(A|E) = 0.05 and Pr(A|Ec) = 0.08. Using the Bayes’ Theorem, we get Pr(A°|E) - Pr(E) Pr(Ac|E) . Pr(E)+ Pr(AC|EC)-Pr(Ec) _ (1 — Pr(A|E))~Pr(E) _ 0.95 060 _ (1- Pr(A|E))-Pr(E)+(l — Pr(A|E‘)). 13456) ' 0.95 . 0.60 +0.92 - 0.40 ' Pr(E|AC) = Answer E. 16. May 1979 Part 2 Examination, Problem No. 2 Suppose the random variable X has mean [.1 and variance 0'2 > 0. What are the values of the numbers a and b such that a + bX has mean zero and variance 1? A.a=—£ and b=i o- 0' 13.a=—i anal):fl a a C. a=—[.t and b=0' D.a=/1 and 12:0 E. a=/,l and b=i 0' Solution. We have E(a+bX)=a+bE(X)=a+bp=0 so that p = —g. Also, Var(a+bX)= bZVar(X) =b20'2 =1 1 :1 sothatb=— and a=--. O' 0' Answer A. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 683 - SECTION 24 17. Study Note P-09-08, Problem No. 135 The number of workplace injuries, N, occurring in a factory on any given day is Poisson distributed with mean ,1. The parameter ,1 is a random variable that is determined by the level of activity in the factory, and is uniformly distributed on the interval [0, 3]. Calculate Var(N). A. it B. 2 )1 C. 0.75 D. 1.50 E. 2.25 Solution. We have Var(N) = E(Var(1v|,t))+ Var(E(N|/l)) = E(A)+ Var(zi.) = Answer E. 18. A dental insurance policy provides reimbursements for up to three fillings and one crown per policy period. The numbers of fillings needed by the insured in a policy period follows a Poisson distribution with mean 0.7, and is independent of the number of crowns needed by the insured, which in turn follows a Poisson distribution with mean 0.2. Calculate the probability that there is maximum possible reimbursement under this policy in a single policy period. A. 0.0046 B. 0.0062 C. 0.0279 D. 0.0332 E. 0.0569 Solution. Let N be the number of fillings needed by the insured in a single policy period, and M be the number of crowns needed by the insured in the same single policy period. Note that the random variables N and M are independent. The reimbursement for under this policy is maximized if the insured simultaneously needs at least three fillings and at least one crown, i.e., if N 2 3 and M 21. The probability of that is Pr({1‘123}o{M 21})=Pr(N23)-Pr(M 21)= =(1'P1'(N=0)-PI(N=l)—Pr(N=2))-(l—Pr(M=O))= =[l_e.u,0.7°_e..1,fl_e.o.7.o.7z]. 1_e..2,0.7°]= o! 1! 2! 0! = (1 -e'°'7 -0.7e'°'7 —0.245e‘°‘7)-(1—e'°2)= 0.0062. 19. 1986 Casualty Actuarial Society Part 4 Examination, Problem No. 31 Which of the following statements are true about the distribution of a random variable X? 1. If X is discrete, the value of X, which occurs most frequently is the mode. 2. If X is continuous, the expected value of X is equal to the mode of X. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 684 - PRACTICE EXAM 20 3.The median of X is the value of X , which divides the distribution in half. A.I B.l,2 C.l,3 D.2,3 E.1,2,3 Solution. Statement 1 is true, as this is the definition of the mode for a discrete distribution. Statement 2 is not true: for example, exponential distribution’s mode is 0, while its mean is positive. Statement 3 is a form of the definition of the median, so it is true. Answer C. 20. November 1986 Course 110 Examination, Problem No. 13 The following table re resents the relative fre uenc of accidents er day in a city: E-I- mm Which of the following statements are true? I. The mean and modal number of accidents are equal. II. The mean and median number of accidents are equal. III. The median and modal number of accidents are equal. A. I B. II C. 111 D. I, II, IH E. None of these answers are correct Solution. The mean of the distribution is greater than 0055 +1-0.20+2 -0.10+ 30.10 = 0.70. The mode is zero. The median is also zero because Pr(X = 0) = 0.55. Thus only the median and the mode are equal. Answer C. 21. 1988 Casualty Actuarial Society Part 4 Examination, Problem No. 31 The following table represents data observed for a certain class of insureds. The regional claims office is being set u - to service a grou - of 10,000 olicyholders from this class. Number oflaims (n) ' ' E-l- Probabili of a olicyholder makin n claims in a car If each claims examiner can service a maximum of 500 claims in a year, and you want to staff the office so that you can handle a number of claims equal to two standard deviations more than the mean, how many examiners do you need? A.S5 B.6 C.7 D.8 E.29 Solution. Let X be the random individual claim. The mean of that distribution is ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Oslaszewski - 685 - SECTION 24 0-0.84+1-0.07+2-0.05+3-0.04 =0.29. The second moment is E(X2) = o2 084 +12 .007 + 22 .0.05 + 31-004 = 0.63. The variance is Var(X) = E(X2)- (13(X))2 = 0.63 — 0.292 = 0.5459. For 10000 policyholders, the aggregate claim mean is 10000 - 0.29 = 2900, and the aggregate variance is 10000 05459 = 5459. The aggregate standard deviation is J5459 = 73.89. The needed number of examiners is Aggregate Mean + 2 - (Aggregate standard deviation) _ 2900 + 2 - 73.89 H 500 500 Because the number of examiners must be an integer, the smallest number is 7. Answer C. 6.1. 22. February 1996 Course 110 Examination, Problem No. 25 The sum of the sample mean and median for ten distinct data points is equal to 20. The largest data point is equal to 15. Calculate the sum of the sample mean and median if the largest data point were replaced by 25. A. 20 B.21 C. 22 D. 30 E.31 Solution Since both the current largest value and 25 are greater than the median, replacing one by the other does not change the median. But increasing the largest value by 10 increases the sample mean by —10— = 1, and increases the sum of the sample mean and median by 1. Number of points Thus the new sum equals 20 + 1 = 21. Answer B. 23. Let N be the number of rolls of a fair die before getting a l, and let M be the number of rolls before getting the first odd number. Find E (N | M = 4) . A.6 3% CS DE Solution. The event {M = 4} means that the die was rolled four times before the first odd number appeared, and that first odd number appeared on the fifth roll. Given that, l was rolled on the fifth roll or later. There are three possible odd numbers, 1, 3, or 5, that can show up on the fifth roll, so that ASM Study Manual for Course Pll Actuarial Examination. (9 Copyright 2004-2008 by Knysztof Ostaszewski - 686 - PRACTICE EXAM 20 Pr(N=4|M=4)=-§-. On the other hand, for n > 4 we have Pr({M = 4}o {N = n}) Pf N: M=4 =——= ( "I ) Pr(M=4) 0:1de mm m Kent-£23m 33%1'38 munfilmm‘ EnTlsznmn [1)“ a . . 1 =_2__u___6_=1.[5]"“ (1)“ 1 9 6 .2 . .2. mm“ Notethat ii.[2)""_13°°(2)‘_1.;_2 "=59 6 9 kgo 6 9 3’ 6 as it should be, because all probabilities must add up to one. Therefore E(N|M_4)_l.4+in.l.(§]n-s—i+l+fl5.[§)n-s+l§(n_5).[§)n-s_ 3 ",5 9 6 3 9“, 6 9,.=5 6 Write n=5+(n-5) +00 k 4... k = k=n—s =13 (2) .9. 2mm = 3 9k=0 6 ‘2. k=0 6 6 hW—I W 2 ‘—’—’. 3:232: :3- Expected valueofageomlemc 5“” m randomvmiablewith =—-. oontingthenumberofp 4 5 l 2 4 10 10 =—+—-—+—-5=—-+— —=8 9 1_§ 3 3 3 3 6 Answer C. 24. A certain genetic disease is passed on to a child from the child’s father with probability p = 0.50, with occurrence of the disease being independent for each child. In a particular family, a father has the disease, and has 6 children. It is discovered that three of his children do not have the disease, and one has it. Let X be the random number of children among the remaining 2 that have the disease. Find the expected value of X. A. l B.1 C.l D.l.5 E.2 3 2 Solution. ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20042008 by Kraysztof Ostaszewski - 687 - SECTION 24 Because occurrence of the disease is independent for each child, known occurrences of the disease in four children has no effect on the remaining two children. For each of the remaining two children, occurrence of the disease is a Bernoulli Trial with probability of “success” (disease occurring) of p = 0.50. With two Bernoulli Trials, the expected value of the total number of successes is 1. Answer C. 25. One in one thousand people in a certain country have a certain disease. A test is developed, which has the probability of correctly diagnosing the disease when a person has it of 0.85, and a probability of 0.75 of correctly diagnosing a person as free of disease when a person does not have it. Find the probability that a person has the disease given that a person is diagnosed as having it by the test. A. 0.0020 B. 0.0034 C. 0.3333 D. 0.8500 E. 0.9966 Solution. Let D be the event of a random person having a disease, and E be the event of being diagnosed as having the disease. We are given that Pr(D) = 0.001, Pr(EID) = 0.85, and Pr(ECI DC) = 0.75. We need to find Pr(DIE The flip-flop of conditional probabilities tells us that we are dealing with a Bayes’ Theorem problem. Using the Bayes Theorem, we obtain _ Pr(EID)-Pr(D) _ “(DIE)” Pr(E|D)-Pr(D)+Pr(E|DC)-PI(DC) ' _ Pr(Elm-Pr(D) = Pr(ElD)-Pr(D)+(l—Pr(ECIDC))-(l—Pr(D)) 0.85 0.001 0.00085 = —— = — = 0.0034 0.85 ~0.001+(l— 0.75)~(1— 0.001) 0.00085 + 0.24975 Answer B. 26. You are given that X is an exponential random variable with mean 1 and that Y is an exponential random variable with hazard rate 4, with X and Y being independent. Find Var(max(X,Y)|X = x) as a function ofx. l A. 16 B. Jot-i6“ C. x2 +-;-xe"‘ +%e"" D. i E. —e"“-——e‘8‘ Solution. Let Z = max(X,Y). We have ASM Study Manual for Course Pll Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 688 - PRACTICE EXAM 20 x if Y < x, (ZIX = x) = (max(X'Y)|X = x) = {Y otherwise. Therefore E(zlx= x) = x-Pr(Y < x)+ nyy (y)dy=x(1- e“9+ 14y”de = _ __l -4 *"’ = " ‘ 4” v‘ 4 e y = x(1— e“)+(‘ye“’|::)+ led’dfi du = 4dy dv = e'4’dy ’ $3333?” = x(l- e'“)+xe"‘ — le‘” PM = x+le"‘ 4 ,3, Furthermore E(ZZIX= x) = x2 -Pr(Y <x)+Ty2f,, (y)dy= x2(1—e"‘)+T4y2e"’dy= .4 u=y2 v=—e ’ du = 2ydy dv = 4e""dy —v—a = x2(1- e'“ ) + (—yze'” IN”) + The-Md), = Y=x Integmtionbypans forlhemtegml _ _ l 4y 9"“ r” = ""2” V' 48 =x2+[-%Ye'4’ ]+j%e“’dy= du=2dy dv=e4ydy y=x x 3.2m y—Hoo =x2+lxe'“+ —le"” =x2+lxe'“+le"’ 2 8 y” 2 8 Finally 2 1 —4x 1 -4x 1 -4x 2 Var(Z|X=x)= x +—xe +—e - x+—e = 2 8 4 =x2 +lxe4x+le41 _x2 _lxe-4x_ie-8x =_1_e-4x __1_e-8x. 2 2 16 8 l6 AnswerE. 27. Two independent random variables X and Y are distributed uniformly on the interval [0, 40]. Determine the moment generating function of X + Y. ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 689 - SECTION 24 801 _ 401 80: _ 40: 401 _ 40: _ 80: _ e 2e2+1 Bce e2+l C.e l D. e l E.e l 1600t 1600t 40: 40t 80t Solution. For each of the two variables, its MGF is e40-t _ e01 e401 _1 M t = M t = —-—-——- = . "0 ’0 (40-0); 4» Since X and Y are independent Mx+r(t)=E(e""*”)=E(e“)E(e”)=M.(r)-M.(t)=[ Answer A. e‘o'—l 2 ew’-2e‘°'+1 401 J ‘ 1600:2 ' 28. An insurance policy has a deductible of 4 with a maximum payment of 5. When a positive loss occurs, the amount of the loss follows an exponential distribution with the mean of 8. Calculate the expected payment made by this policy, given the occurrence of a positive loss. A. 2.04 B. 2.50 C. 3.01 D. 3.27 E. 3.98 Solution. Let X be the amount of the loss, let Y be an exponential random variable with mean 8 equal to the amount of the loss given that the loss is positive, and let W be the payment made by the policy. We have 0 whenY<4, W= Y—4 when4SY<9 5 whenYZS. Therefore ‘l-ly 9 1-ly ’1-1y E(w)=j0-§e 8 dy+j(y-4) —e 8 dy+j5-—e 8 dy= 0 4 4 1 "Y .=9 “=Y‘4 v=_e 8 -1y 3 9 -1y 59 ~ly = 1 A), =0+ (y-4)[—e 3] +Ie 3 dy+§Ie 3 dy= du=dy dv=§e3dy y=4 4 4 lntegmtionbypansintheseoondimegml -2 13 9 -1, -2 -1, "9 -2 -1 =—5e 3 +— e 3 dy=—5e 3+ [-l3e 8] =—18e 3 +13e 3 =2.0412. 8 4 ’.=4 Answer A. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 690 - PRACTICE EXAM 20 29. An insurer provides earthquake insurance to a manufacturer. Given that a claim is filed, let X represent the total amount of the claim and Y represent the part of the claim arising from damages to the manufacture’s building. The joint density of X and Y is given by: i fX.Y (LY): 9 0 otherwise, where the values of both variables are in millions of dollars. Calculate the conditional variance of the total claim (in millions of dollars squared), given that part of the claim arising from damages to the manufacture’s building exceeds one million dollars. for0<y<x<3, A. 1.21 B. 1.49 C. 1.68 D. 1.87 E. 2.00 Solution. The question asks for Var(X|Y > 1). We have Fx(x|Y>1)=Pr(XSx|Y>l)=—=—=——= " 1 1 s(s—l)ds [—s3——s2] l 3-1 2-1 l _I____ 3 2 s=l_3x 2x 3 2__1_ 3_ix2 i —3 ldx‘ 13 12 “3 9_2_1+1 14 28 28 L4“) 3" 7X 2 3 2 1 x=l for I < x < 3. Therefore sx(x|Y>1)=-318—x3+-238x2+-:% and 3 x=3 l 3 27 l 1 27 E(X|Y>l)=I(—§x3+gx2+§)dx=(—mx4+Ex3+ng = l x=l [8127 81)(1 127) 80 265415 = -—+-—+— — ——+—+— =——+—+—=—. 112 28 28 112 28 28 112 28 28 7 Also d 3 3 fx(x|Y>1):;Fx(x|Y>l)=fixz-Ex. Therefore ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Knysztof Ostaszewski - 691 - SECTION 24 E(X2|Y >1) :[xz {$.73 —1—:-x)dx= Hfix‘ -%x3)dx= (7—36}? —%x‘] 363450-213 35 x=3 x=l 7o 56 7o 56 7 35 ‘3' Finally 2 Var(x|Y>1)=£_[E) =§3_£=—_‘491‘”25 =fi=1.4939. 35 7 35 49 245 245 Answer B. 30. A company buys an insurance policy to cover fire losses that happen to the company's buildings. The policy has no deductible, but it does not start payment until the 4th fire loss and thereafter. The company is responsible for the loss for their first 3 fires. Occurrences of fires in a given year for all company’s buildings combined follow a Poisson distribution with a mean of 3. Each building the company owns is worth ten million dollars, and a fire destroys it completely. What is the expected payment for fire losses from the insurance policy to the company in a given policy year, in millions of dollars? A. 2.7533 B. 4.4808 C. 6.7213 D. 8.9617 E. 16.7213 Solution. Let N be the total number of fires, and M be the number of fires covered by the insurance policy. Then N - 3 otherwise. The actual payment by the insurance policy is 10M (in millions of dollars). The expected payment, in millions of dollars, is E(10M)=ilo(n—3).pr(~=n)=mom—3))—ilo(n—3)-pr(~=n)= n=4 n=0 {0 ifNSB, M: =10[E(N)—3]+(30-Pr(N=0)+20-Pr(N=l)+10-Pr(N=2))= =3 0 l 2 = 30-e'3 -3—+20-e'3 ~1-+10-e'3 -3—= l35e'3 = 6.7213. 0! l! 2! Answer C. ASM Study Manual for Course P/l Actuarial Examination. ® Copyright 2004-2008 by Knysztof Ostaszewski - 692 - ...
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