Practice_Exam_15-Solutions

Practice_Exam_15-Solutions - PRACTICE EXAM 15 PRACTICE...

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Unformatted text preview: PRACTICE EXAM 15 PRACTICE EXAMINATION NUMBER 15 SOLUTIONS 1. May 1985 Course 110 Examination, Problem No. 13 l (HY for t< 1. Find Let the random variable X have moment generating function M x (t) = E(X’). A. —24 B. 0 C. D. 24 E. Cannot be determined from the given information Solution. We have M;(t)=(-2)(-l)(l-t)". M; (t)=(-2)(-3)(-1)’(1-t)", M?’(t)=(-2)(-3)(-4)(-l)’(1-t)". and E(X’)=M§’(0)=(-2)(-3)(-4)(-1)’ = 24- Answer D. 2. May 1985 Course 110 Examination, Problem No. 16 In a certain communication system, there is an average of 1 transmission error per 10 seconds. Let the distribution of transmission errors be Poisson. What is the probability of more than 1 error in a communication one-half minute in duration? A. 1—2e’1 B. l—e'l C. l-4e'3 D. 1-3e’3 E. l-e’3 Solution. Since the distribution of the number of errors in 10 seconds is Poisson with mean 1, the distribution of errors in a half-minute (i .e., 30 seconds) time interval is Poisson with mean 3 (as there are three non-overlapping 10 second time intervals in the 30 second period). Let N be the random number of errors in a half-minute time interval. We have I Pr(N>l)=1—Pr(N=O)—Pr(N=1)=l—e'3—e‘3 ~3—=1-4e'3. l ! Answer C. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 531 - SECTION 19 3. Mr. Warrick Beige is gambling in a Las Vegas casino. Because of his emotional attitude about the process of gambling, his two consecutive gambles are dependent. Given that Mr. Beige wins in his first gamble, the conditional probability of winning in the second gamble is If the probability of winning in any gamble is %, what is the probability that Mr. Beige wins in both the first and second gamble? A B C D E.— l l l l ' 2 ' 3 ‘ 4 ‘ 6 12 Solution. Let A be the event of winning in the first gamble, and B be the event of winning in the second gamble. We are given that Pr(A) = %, Pr(B) = g, and Pr(B|A) = Therefore Answer E. 4. You are the actuary for the Glorious Property and Casualty Insurance Company. You are insuring the hips of a famous South American singer performing in the United States. If her hips remain attractive over the next year, you will not make any payment. If they do not, you will make a payment of $5 million. Her hips will remain attractive with probability 0.89, if she does not hire a bodyguard. If she hires a bodyguard, the probability of hips remaining attractive increases to 0.95. Calculate the highest possible salary you can pay the bodyguard so that the cost of this insurance when the bodyguard is hired, combined with the cost of the bodyguard, is no more than the cost of this insurance when the bodyguard is not hired. A. $200,000 B.$250,000 C. $300,000 D. $350,000 E. $400,000 Solution. First, assume that she does not hire a bodyguard. Then the expected payment for the insurance contract is 0.89 - 0 + (1— 0.89) - $5,000,000 = $550,000. If she hires a bodyguard, expected payment is 0.95 . 0 + (l — 0.95) - $5,000,000 = $250,000. The highest possible salary of the bodyguard is the difference of the two, i.e., $300,000. Answer C. 5. X is a random variable with mean 2 and the coefficient of variation of 2. Find the upper bound of the interval in which X is contained with probability 0.99 based on the Chebyshev’s ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 532 - PRACTICE EXAM 15 Inequality. A. 6 B. 10 C. 14 D. 34 E. 42 Solution. The Chebyshev’s Inequality says that Pr X——E(){.). >r <32._ JVar(X) 7' In this case E(X)=2 and JVar(X) _ 2 E(X) ' so that ,lVar(X) = 4. Therefore, Pr[ X'2 >r]=Pr(|X—2|>4r)<—1;. r This is equivalent to Pr(|X—2| 5 4r) 21-%. r But Pr(|x—2|s4r)=Pr(—4rsX—2s4r)=Pr(2-4rsxs2+4r). Therefore, X lies between 2 - 4r and 2 + 4r with probability of no less than 1— For that probability to equal 0.99, we must have r = 10. The upper bound we are looking for is 2+4r=2+4-10=42. AnswerE. 6. For a random variable X you are given that its third moment about 1 is 7, and its third moment about 2 is 3, and first moment about 1 is zero. Find the coefficient of variation of X. l l A. - B. —- C. 2 D. l E. 2 2 J? J— Solution. The first moment about 1 is E(X -1), and this being 0 implies that the mean of X is 1. We also have 7 = E((X-1)3) = E(X’) — 3E(X2)+ 3E(X)- 1 and 3: E((X—2)3)= E(X3)-6E(X2)+12E(X)— 8. ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 533 - SECTION 19 By subtracting the second equation from the first we get 4 = 3E(X2)- 9E(X)+ 7, and by substituting E (X) = 1, we obtain E (X 2) = 2. This gives Var(X) = E(X2)—(E(X))2 = 2 - 1 = 1. The standard deviation of X is 0'x = JVar(X) =1. The coefficient of variation is (note the use of the common notation [.lx = E(X)) Ea=l=L [ix 1 Answer D 7. You are given a random variable whose probability density function is fx (x) = éxze" for x > 0 and 0 otherwise. Find the moment generating function of X. 1 B. 1 C l D 1 E. 1 ALI—:37 l—t ' (l—t)2 '(l—t)3 (Ft); Solution. By definition +09 12 Mx(t)=E(e‘x)= {ennax e-xdx=%£x2e-x(i-t)dx= _Z=x(1‘t) x=0=>z=0 _ _dz=(1")dx x->°°=>z—->oofort<l_ “"——7aamnaasr--—~ +09 J'zze-zdz = = 2 1 1*” z2 1 1 _ 2(1—1‘)3 2(1—t)3 -(1—t)3' e'z —dz = 2 o (1—1)2 1—: 2(1—:)3 0 Answer D. There are two alternative approaches: 0 Observe that this is the PDF of the gamma distribution with parameters a = 3 and B = 1, and therefore its MGF is l3 a B“ 1 M x (t ) = [_ = a = 3 ' B-t (fi-O 0-0 - Observe that this is the PDF of the gamma distribution with a = 3 and B =1, and such a gamma distribution is a sum of three IID exponential distributions with hazard rate equal to [3, ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 534 - PRACTICE EXAM 15 so that its MGF is the product of 3 MGF’s of the exponential distribution with hazard rate equal [3 to [3, each of which equals —, so that fi—t MW): = meat)“ 71—505“ Answer D, in all methods. 8. You are given two independent random variables X and Y. These random variables have the same expected value of 5, and the second moment of X equals 30. Coefficient of variation of X is half of that of Y. Find coefficient of variation of ix + Y. A. 0.5000 B. 0.5895 C. 0.6146 C. 0.6666 D. 0.7500 Solution. Let us also write ox for the standard deviation of X, and 0', be the standard deviation of Y. The coefficient of variation of X is E(X) 5 ' and the coefficient of variation of Y is twice that, so that —Vvar(Y) = 3’- : 0.2a . E(Y) 5 ” This implies that 0', = 20x and Var(Y) = 4Var(X). On the other hand Var(X) = E(X’)—(E(X))z = 30 - 25 = 5. This gives Var(Y) = 4Var(X) = 20. 0.40,. = Finally, the coefficient of variation of —l-X + Y is 2 l 1 5 85 Var[—X+Y) ./—Var(X)+Var(Y) "—+20 — 2 = 4 = 4 4 JE=06146363. X+Y) E( 41 I 2 5 _ X __ 2 )+E(Y) 2+5 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 535 - SECTION 19 9. You are given that M x (t) is the moment generating function of a random variable X. Which of the following is also a moment generating function of a random variable? I. Mx(t)-Mx(2t) II. 5Mx(t) III. e' .M, (t) A.Ionly B.IandII C.IandIH D. II and III E.I,IIandIII Solution. We begin with consideration of I. We have M X (t) = E(e’x). Therefore, M, (2:) = E(e2"‘) = E(e"2x) is the MGF of the random variable 2X. If we choose X and Y to be independent, but Y having the same distribution as 2X, then M x (t) - M x (2t) is the MGF of X + Y, therefore I can be an MGF. For II, we have SMX (0) = 5, implying that the function given in II has the value of 5 at zero, and therefore it cannot be an MGF. For III, we have er .Mx(t)= et .E(e‘x)= E(et ,eIX)= E(et(X+-l))’ so that e' .MX (t) is the MGF ofX + 1, a random variable, Answer C. 10. Suppose that the joint distribution of random variables X and Y is such that fxm (xsy) = K(7 - x) for x = l, 2, 3, 4, 5, or 6 and for y assuming only integer values such that 0 S y < x. Find the probability that Y > 3. A. i B. i c. E D. i E. i 14 56 56 28 14 Solution. For each value of X, the only possible values of Y are those meeting the condition 0 S y < x. Therefore, Forx = 1, the only possible value of Y is y = 0, For x = 2, the only possible values of Y are y = 0, or y = 1, Forx = 3, the only possible values of Y are y = 0, 1, or 2, Forx = 4, the only possible values of Yare y = 0, 1, 2, or 3, Forx = 5, the only possible values of Yare y = 0,1, 2, 3, or 4, Forx = 6, the only possible values of Yare y = 0,1,2, 3, 4, or 5. We do not know the value of the parameter K, but we should be able to find it by adding up all values of f“, (x, y) , as those values must add up to 1. Hence, ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004.2008 by Knysztof Ostaszewski - 536 - 1: This gives PRACTICE EXAM 15 fxm (1’0) + fxm (290) + fxa' (2’1)+ fo (3'0) + fxm (3’1) + fxa’ (3'2) + + fxa' (4'0) + fxx (4’1) + fxy (4’2) + fxx (493) + fxx (5’0) + fx.r(5’1)+ + fx.r (5’2) + fx.r (5’3) + fx.r (5'4) + fx.r (6'0) + fx.r(6»1)+ fx.y (6’2) + +fx.r(6v3)+fx.y (6’4)+fx.r (6'5): =K-6+K-5+K-5+K-4+K-4+Kc4+K-3+K-3+K-3+K-3+ +Kc2+Ko2+Ko2+K-2+K-2+K-1+K-1+K-1+K-1+K-1+K~1+ =K-(6+10+12+12+10+6)=56K. K = —1-. Finally, 56 Pr(Y>3)=Pr(Y=4)+Pr(Y=5):fx.r(5'4)+fx.r(6v4)+fx.r(6!5)= Answer A. W =Pr()’=5) 1 1 l 2 l 1 4 =—- 7—5 +—- 7—6 +—- 7-6 =—+ 56( ) 56( ) 56( ) =Pr(Y=4) —+—=—=—, 56 56 56 56 14 11. You are given the probability density function of a random variable X: fx(x)={0’ 2x, 0 < x <1, otherwise. Find the difference between the third central moment and the second central moment of this random variable. A. i B. E C. —i D. 3&7- E. —30—7 18 135 270 270 270 Solution. We have 1 x=1 E(X)=Ix-2xdx=3x3 =§-. o x= ASM Study Manual for Course PI] Actuarial Examination. (9 Copyright 20044008 by Krzysztof Ostaszewski -537- SECTION 19 l { [2x5 —x4 +§x3 --§-x2] 5 9 27 The value sought is E((x-E(x))3)—E((X—E(X))2)=[__L]_i= 4‘15 = E. 13 270 ‘270° Answer C. 12. November 197 9 Part 2 Examination, Problem No. 5 Suppose X is a random variable with the stande normal distribution and Y is such that Y - N (0.0.01). Suppose also that Z is a discrete random variable with two point masses: Pr(Z=0)=% and m(z=1)=%. Find Pr(X-Z+Y-(l—Z)20.2). A.0.06 B.0.08 C.0.12 D.0.42 B.0.62 Solution. Because Pr(Z = 0) =% and Pr(Z = l) =i— regardless of the values assumed by X and Y, we conclude that Z is independent of X, and also independent of Y. Therefore, we have Pr(X-Z+Y-(l-Z)20.2)= =Pr(X-Z-t-Y-(l—Z)20.2|Z=0)-Pr(Z=0)+ +Pr(X«Z+Y~(l—Z)20.2|Z=1)-Pr(Z=l)= =%.pr(y 20.2|z=o)+%.Pr(X20.2|z=1). But the conditional probabilities in this case are the same as unconditional probabilities, i.e., Pr(Y 2 0.2|z =0) = My 202): Pr[YO‘10 > %]= Pr(Y0_10 > 2]: 1413(2), and Pr(X 2 o.2|z = 1) = Pr(X 2 0.2) = 1 — c1)(o.2). Therefore, Pr(XZ+Y-(l-Z).>.0.2) l! Alb.) Alt» (1—<1>(2))+%(1—<1>(0.2))= ll -(1—0.9772)+%-(1—0.5793)z 0.122275. Answer C. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 538 - PRACTICE EXAM 15 13. You are given three independent random variables X, Y, and Z, all distributed exponentially, such that the hazard rate of X is 1x, the hazard rate of Y is 11,, , and the mean of Z is 4. You are also given that E(Y+Z)=Var(Y—X) and Var(X+Y+Z)= 3E(2Y+Z). Find 11, —Zx. A. -0.05 B. O C. 0.01 D.0.05 E. 0.09 Solution. We have 1 1 E(Y+Z)=E(Y)+E(Z)=—1—+4=Var(Y—X)=Var(Y)+Var(X)=7+7, 1L 1x 1y and Var(X+Y+2)=Var(X)+Var(Y)+Var(Z)=%;+i;+16= X =35(2Y+Z)=6E(Y)+3E(Z)=%+12. From the second equation we get 1 6 l —2=_’4‘—2v 1x 1y 1y and by substituting that in the first equation we obtain 1 6 1 1 6 —+4=——4-—+—=——4. 17 1y 13 I15 11 This results in 1,. = Therefore _1__£_4__1__£_ -fi-@_fl_fi~7_6 a; 2, 2.3 5 25 25 25 25 25' 25 5 This ives A = —=—.Fina11 , g x 76 2J5 y 5 5 —2 =————=0.05146067. 2" X 8 2J19 Answer D. 14. You are given that N is Poisson random variable such that for its cumulative distribution function FN, SFN (2) = 13FN Find the variance of N. A.1 13.2 C.J7 D.3 13.4 Solution. Let E(N) = Var(N) = A. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 539 - SECTION 19 We have 2 l" 2. 2. 1 a. FN(2)=2e"1-—=e' +/'Le’ +—}.2e' , "=0 ’1! 2 and n 1 FN(1) =2e“ -—'= e" +1161. n=0 n ' The condition 5F” (2) = 13FN (1) gives us the following equation: 5(6)‘ + file“ + élze") = 13(e"1 + le"). This results in the following: 5+51+§AZ=I3+13L By moving all terms to the left-hand side we get 2,5,12 — 8/1— 8 = 0, andfromthat 8iJ64+80 8:12 4’ l=—-————=——= 4 5 5 ——. 5 Only the positive solution makes sense, and based on that solution, Var(N) = A = 4. Answer E. 15. The coefficient of variation of a random variable X is 3, while the coefficient of variation of a random variable Yis 4. If E(Y) = 1 .SE (X) > 0 and the correlation coefficient of X and Y is 0.5, what is the coefficient of variation of X + Y? A. 1.5675 B. 2.5000 C. 3.1749 D. 5.0000 E. 7.9373 Solution. We have E(Y)=1.5E(X), ./Var(x) = 3E(X), and ./Var(Y) = 4E(Y). Furthermore, Cov(X,Y) _ 0 5 ./Var(x)./Var(y) ‘ We are looking for the coefficient of variation of X + Y, i.e., om, _ JVar(X+ Y) um E (X + Y) Let us denote that unknown coefficient of variation of X + Y by C. We have ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20044008 by Knysztof Ostaszewski - 540 - PRACTICE EXAM 15 C: .{v‘ ar(X+Y) _ ‘/Var(X)+Var(Y)+2~Cov(X,Y) E(X+ Y) " E(X)+ E(Y) .[Var(x) + Var(Y)+ 2 - 05 -,/Var(X) -,/Var(r) = E(X)+ E(Y) ' Now we use the information that E(Y) = 1.5E(X), JVar(X) = 3E(X), and ,lVar(Y) = 4E(Y) to get C: 9(E(X))2 +16(E(Y))2 +12-E(X)-E(Y) = E(X)+E(Y) 9(15(X))2 +16 155%)2 +12-E(X)-1.5E(X) _ = E(X)+15E(X) ‘ _ ‘/9(E(X))2 +36(E(X))2 +18 .(E(X))2 = fl . M 3 3 17490157 " 2.5E(X) 2.5 E(X) ' ' Answer C. 16. You are given the following joint probability function of random variables X and Y: 0.425, (x,y) = (0,0), 0.275, (x,y)=(l,0), fx-”(x’y)= 0.300, (x,y)=(l,l), 0, othwerwise. Find the correlation coefficient of X and Y. A. -0.2444 B. 0.2100 C. 0.3000 D. 0.4575 E. 0.5628 Solution. We have 0.425, x = 0, fx (x) = 0.575, x = l, 0, otherwise, and 0.7, y = 0, f,,(y)= 0.3, y=1, 0, otherwise. Therefore, X is a Bernoulli Trial with probability of success px = 0.575, and Y is a Bernoulli kw Trial with probability of success p, = 0.3. This gives ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 541 - SECTION 19 E(X) = px =0.575, Var(X) = px .(1—p,)= 0.575 - 0.425 = 0.244375, E(Y)= p, = 0.3, and Var(Y)=p,, -(1-p,.)=0.3.0.7 =0.21. Furthermore, E(XY)=0-0-0.425+1-0-0.275+1-1-0.3=0.3. Finally, p = E(XY)-E(X)oE(Y) = 03—0575 .03 z 0 56282341 x” ,IVar(X) -,/Var(r) J0.244375 4m ' ' Answer E. 17. You are given the joint probability density function of random variables X and Y: 0.32 '°""‘°3”, > O, > 0, fx Y (3323’) ={ e x ' 0, otherw15e. Find the variance of X + Y. A. 1.5625 B. 3.1250 C. 4.6875 D. 5.8594 E. 7.8125 Solution. Note that the region where the joint density is positive is a rectangle (actually, a square: (0,+°°) x (0,+oo)) with sides parallel to the axes, and forx> 0, y > 0, f” (x, y) = 0.32e*’“'°-” = 0.4e'°'4" -0.8e‘°‘“’ , 7.20 40) which implies that X and Y are independent, andX is exponential with hazard rate 0.4, mean 0L4 = 2.5, and variance 2.52 = 6.25, while Y is exponential with hazard rate 0.8, mean —1- = 1.25, and variance 1.252 = 1.5625. For independent random variables the variance of their 0.8 sum is the sum of their variances, so that Var(X+ Y) = Var(X)+Var(Y) = 6.25 + 1.5625 = 7.8125. Answer E. 18. You are given that a random variable X is normally distributed, with mean 5 and variance 1. Another random variable, Y, is defined as Y = max(ex —100,0). Find Pr(Y > 500). A.0.08 B.0.06 C.0.04 D.0.02 E. 0.01 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Kmysztof Ostaszewski - 542 - PRACTICE EXAM 15 Solution. We are looking for the probability Pr(Y > 500) = 1>r(max(ex — 100,0) > 500). But because 0 is clearly less than 500, this probability is Pr(e" — 100 > 500) = 1>r(ex > 600) = Pr(X > 111600) = Pr( X—S>ln600-5) 1 1 ' X ‘ 5 is standard normal, and this implies that if we write <1> for the The random variable Z = cumulative distribution function of the standard normal distribution, the probability sought is 1— ¢(13§'#)= 1- <1>(1.39692966). From the standard normal distribution table, <D(1.39)= 0.9177 and @(1 .40) = 0.9192. Using linear approximation, we obtain 1 39692966 — 1.39 CD(1.39692966) = (13(1 .39)+ 1.4 -1.39 = 0.9177 +W109192—09177) = 091873945. 1.4 — 1.39 Therefore, the probability we are seeking is approximately 1 — 0.91873945 = 008126055. You would actually obtain the same answer choice without using linear interpolation. Answer A. ~(<1>(1.4)— <1>(1.39)) = 19. The joint distribution of random variables X and Y is uniform on the region defined by the 2 conditions: 0 <x<5, 0 <|y| <:—5. Find Var(Y|X=4). A. 0.1365 B. 0.2048 C. 0.4096 D. 0.5461 E. 0.6400 Solution. 2 x2 The conditions 0 < x < 5 and 0 < [y| < :—5 defines the regions between the graphs of y = —E x2 and y = E for 0 < x < 5, whose area equals "2 5 25 5 2 «i=5 I Idy dx: 2.x_dx=i.lx3 =1..1_2_5=1_9.. ox, 025 253,=0253 3 -3 Since the joint distribution is uniform over that region, we know that ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Kizysztof Ostaszewski - 543 - SECTION 19 i 0<x<5—£< <x—2 fx.y(x.y)= 10’ ’ 25 y 25’ 0, otherwise. We know from the definition of the conditional distribution that fi(y|X=4)=fo.c_:((%Zl=£€T) and this expression is constant with respect to y, thus implying that (Y I X = 4) is uniform on the interval This means that 25 25 (Li-19D: [3sz — 2 Var(Y|X=4)- 25 1225 —1232252z0.1365333. Answer A. 20. November 1979 Part 2 Examination, Problem No. 14 Suppose X has the uniform distribution on the unit interval, while the density of Y is 2y, 0 S y S 1, ff (y) _ {0, otherwise. You are also given that X and Y are independent. Find Pr (X < Y A. l B. l c. 3 D. 3 E. i 3 4 5 3 5 Solution We have I l l l Pr(x<r)= J[Jf..(x.y)dy]dx= I[Ify(Y)dy]fx(x)dx= 0 x 0 x l l l _l l = [[jzydy] l~dx= flyz :=x)dx= [(1—x2)dx= 0 x 0 0 Answer D. 21. One thousand independent throws of a fair coin will be made. Let X be the number of heads in these thousand independent throws. Using the normal approximation, find the smallest value of n, and integer, such that Pr(500 — n < X < 500 + n) > 0.70. A. 15 B. 16 C. 17 D. 18 E. 19 -544- ASM Study Manual for Course PI] Actuarial Examination. © Copyright 2004-2008 by Krzysztof 0staszewski PRACTICE EXAM 15 Solution. Each throw of a coin is a Bernoulli trial with probability of success p = 0.5. X is a sum of 1000 such Bernoulli trials, independent and identically distributed. Therefore E(X)=1000op =1000~%= 500, and Var(X) = 1000-p(1- p)=1000--1--— = 250. l 2 2 Note that n X—500 n PrSOO— <X<500+ =Pr- <X—500< =Pr - < < . ( n n) ( n n) ( J250 J250 250] X—SOO . But 250 15 standard normal, and therefore Pr(500-n<X<500+n)=d>[ " )—q>[— " ]= J250 J250 anal—(I-cbtlafimt—e—oJ—L If we set that value equal to 0.7, we obtain 12 2(1) -1 =0.7 (4250] ¢[J2”5_0] = 0.85. From the table, <I>(1.03)= 0.8485 and <D(l .04) = 0.8508. Using linear interpolation, we obtain and n 1.04 — 1.03 = 1 .03+—— - 0.85 — 0.8485 =1.03652174. JZSO 0.8508 - 0.8485 ( ) Therefore, n ==1036521744250 =16.3888477. But n must be an integer, and we want the probability Pr(500—n<X<500+n) to be more than 0.7, so that we must go to the next (larger) integer value, and n = 17. Answer C. 22. Two fair six-faced die are rolled simultaneously. Find the probability that the sum of the numbers obtained on both of them is divisible by 3. A. l B. l C. l D. — E. l 6 3 2 12 9 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 545 - SECTION 19 Solution. The total number of possible outcomes is 36 (six possible numbers on the first die and for each of those, six possible numbers on the second die). The outcomes resulting in the total being divisible by 3 are: (1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,l),(5,4),(6,3),(6,6), for a total of 12. This means that the probability sought is g = Answer B. 23. May 1985 Course 110 Examination, Problem No. 27 Let X and Y have the joint density function x+y, for0<x<l and0<y<1, fx.r(xv}’)= 0, elsewhere. What is the conditional mean E (Y |X = g)? Ag 13,—5- c.1 13.1 13.3 8 12 2 12 5 Solution. Wehave f 1y 1 1 1 ‘ 1 X.Y 3’ y+§ y‘l‘g‘ y+— 6 2 f}, =—=l—= y__l= 1=—y+— 3 in Illwldy 1 5 5 X 3 o 3 y=o for 0 < y <1. Therefore 1 ' 6 2 2 1 EYX=—= —+—d=—3+—2] [I 3) £45)? Sjy (5y 5y Answer E. y=l 3 i=0 24. May 1985 Course 110 Examination, Problem No. 28 Let X] and X2 be independent continuous random variables, each with density function —h were i 0, elsewhere, where A > 0. Let Y1 = X1+ 2X2 and Y2 = 2Xl + X2. What is the joint density function g(y,,y2) for yl >0 and y2 >0? ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 546 - 1 _’1()'i +)'2) _1()'| 'U'z) _}-()‘1 H2) A glze 3 B. Aze 3 C. 3/12e 3 “n+z) D. 3A2e-31-(ywz) E £7129" ysy Solution. Because X, and X2 are independent, the joint PDF of XI and X2 is 2 WP“: > o, > o, fx x (xl,x2)= A e ’ xl {‘2 " ’ 0, otherwxse. The transformation we are considering is Y1=Xl+2X2, Y2=2X,+X2. Its inverse is 2 1 2 1 Xl=§Y2-§Yl! X2=§Yl-§Y1' Therefore, % £ 1 2 3(JCIJC2) = det ayl ayz = 3 3 901.35) 33 Bi 3 -1 9 9 3 ayl By: 3 3 Finally,for yl >0 and )72 >0, 3 x ,x fy,,y, (yi:)’2)=fx,.x,(x1(yl’y2)'x2 (M9)”). aEyI’yg = 2 2 l ’1 +| = Aze-{gn-ngJ-ign-gh) .1: 1/12; 0'13 y) 3 3 Answer A. 25. May 1985 Course 110 Examination, Problem No. 29 Let Eand Fbe events such that Pr(E) = %, Pr(F) = g, and Pr(EC nFC) = Pr(EUFC) = A. l B. 3 c. 3 D. 3 E. 1 4 3 4 6 Solution. Note that ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski PRACTICE EXAM 15 1. Then 3 ~547- SECTION 19 Pr(EuF°)=Pr EU(ECnFC) =Pr(E)+Pr(E‘-‘ nF‘): Mutunlly exclusive 1 l 5 —+—=—. 2 3 6 Answer D. 26. May 1985 Course 110 Examination, Problem No. 34 In rolling a pair of fair dice, what is the probability that a sum of 7 is rolled before a sum of 8 is rolled? 6 ll 5 6 25 A. -— B. — C. — D. — E. — 36 36 ll 11 36 Solution. There are 36 possible outcomes for a single roll of two dice, and 6 of them result in a sum of 7: 1+6,2+5,3+4,4+3,5+2,6+1. Thus the probability of getting a sum of 7 in a single roll of two dice is 3—66 = There are 5 ways to get a sum of 8 in a single roll of dice: 2+6,3+5,4+4,5+3,6+2. Thus the probability of getting a sum of 8 in a single roll of two dice is Combining these two pieces of information we conclude that the probability of rolling either a 7 or an 8 as a sum in a single roll of two dice is :17;- , and the probability of getting any other sum is The probability that a sum of 7 is rolled before a sum of 8 is rolled is the sum of the probabilities of the following events: - Sum of 7 is rolled in the first roll, 0 First roll produces neither 7 nor 8, and the second one produces the sum of 7, 0 First two roll produce neither 7 nor 8, and the second one produces the sum of 7, etc. By adding probabilities of those events we get: 1+§.1+(§]’.i,(2§)’;+ -1.“[§)"_1._1_-3 6 366 36 6 36 6 6,30 36 61_2511' 36 Answer D. 27. May 1985 Course 110 Examination, Problem No. 35 A bin of 10 light bulbs contains 4 that are defective. If 3 bulbs are chosen without replacement from the bin, what is the probability that exactly k of the bulbs in the sample are defective? ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20044008 by Krzysztof Ostaszewski - 548 - PRACTICE EXAM 15 Solution. 4 6 There are [ k] ways of choosing k defective bulbs, and [ 3 k) ways of choosing 3— k non- 10 defective bulbs. There are [ 3 J ways of choosing 3 bulbs. The probability we are seeking is my 28. May 1985 Course 110 Examination, Problem No. 36 Let X, Y, and Z be independent random variables, each uniformly distributed on the interval [0.2]. What is the number c for which the probability is 0.75 that at least one of X, Y, or Z will exceed c? l l _ .l. l l A. 0.753 B. 63 C. 2—63 D. 23 B. 2-23 Solution. We have 0.75 =Pr({X>c}u{Y>c}u{Z>c})=Pr({XSc}cu{YSc}CU{ZSc}C)= =Pr(({XSc}n{YSc}n{ZSc})c)=l—Pr({XSc}n{YSc}n{ZSc})= c—O 3 c3 =l-Pr(XSc)'Pr(YSc)-Pr(ZSc)=l—(2 0) =1—E. I This gives c = 23. Answer D. ASM Study Manual for Course Pll Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Osraszewski - 549 - SECTION 19 29. May 1985 Course 110 Examination, Problem No. 37 Let E (Y|X = x) = 2x, Var(Y|X = x) = 4x”, and let X have a uniform distribution on the interval (0,1). What is Var(Y)? A. 31,,- B. E- C. 3- D. % E. Cannot be determined from the information given Solution. We have Var(Y) = E(Var(Y|X)) + Var(E(Y|X)) E(4X2) +Var(2X) = =4E(X2)+4Var(X)=4(Var(X)+(E(X))2)+4VaI'(X)= 2 =4. +4.i=4.(i+l]+ 12 2 12 12 4 30. May 1985 Course 110 Examination, Problem No. 38 Let X be a continuous random variable with density function fx (x) = {Ze'z‘ forx > 0, 0 otherwise. Let Y = e"‘ . What is the density function g(y) of Y where non-zero? Answer C. A. y B. Zyz C. y: D. F E. 2y y Solution. We will write f, (y) for g(y). The transformation is Y = e"x and its inverse is X = —1nY so that g- : —%. Therefore for y > O dY = 2e2lny . l = 2y. 3’ y fy (y) = fx (x(y)) . |%:.| = 2e-Zl-lny) , Answer E. ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 550 - ...
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