{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Practice_Exam_1-Solutions

Practice_Exam_1-Solutions - PRACTICE EXAMINATION NO 1...

Info icon This preview shows pages 1–23. Sign up to view the full content.

View Full Document Right Arrow Icon
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
Image of page 3

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
Image of page 5

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 6
Image of page 7

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 8
Image of page 9

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 10
Image of page 11

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 12
Image of page 13

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 14
Image of page 15

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 16
Image of page 17

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 18
Image of page 19

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 20
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: PRACTICE EXAMINATION NO. 1 PRACTICE EXAMINATION NUMBER 1 SOLUTIONS 1. May 2001 Course 1 Examination, Problem No. 4, also Study Note P-09-08, Problem No. 66 A company agrees to accept the highest of four sealed bids on a property. The four bids are regarded as four independent random variables with common cumulative distribution function F(x)=%(l+sin7rx) for 3 S x S ‘2‘. Which of the following represents the expected value of the accepted bid? . 1—16- (1+sinnx)‘dx C. —1- x(1+sin7tx)4dx xcosztx dx B 16 A.:r N|u0—.u|u N|u'—.N|Ut NIu'-—.M|uu 11: xcos 7rx(1 + sin 7tx)3dx cos 7tx(1 + sin nx)3dx E Din .1 4 4 Nlu'——.Nl'-h Nlu‘——.NIM Solution. Let X1 ,X2 ,X3 and X 4 denote the four independent bids with common distribution function F. Then if we define Y = max(Xl,X2,X3,X4), the distribution function F} of Y is F}(y)=Pr(Y Sy)=Pr(Xl sy)n(X2 Sy)n(X3 Sy)n(X4 Sy)= 4 9 = Pr(Xl S y)Pr(X2 S y)Pr(X3 S y)Pr(X4 S y) = (F(y))4 = %(1+ sin fly) 3 5 for a S y S 5. It then follows that the density function g of Y is given by l f), (y) = 5307) = Z(1+ sin ivy)3 oncosn'y = %cos ny(l + sin my)3 , 3 5 for — S S —. Final] , 2 y 2 y s 5 5 5 fl 3 150’) = 1% Mb = Izycowyfl + sin 70’) dy- 3 3 5 5 Answer E. ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 118 - J ‘ PRACTICE EXAMINATION NO. 1 / ‘ 2. May 2001 Course 1 Examination, Problem No. 5, also Study Note P-09-08, Problem No. W 117 A company is reviewing tornado damage claims under a farm insurance policy. Let X be the portion of a claim representing damage to the house and let Ybe the portion of the same claim representing damage to the rest of the property. The joint density function of X and Yis 61— x+ forx>0, >0,x+ <1, f(x.y)={0( ( y» . y y , otherwrse. Determine the probability that the portion of a claim representing damage to the house is less than 0.2. A. 0.360 B. 0.480 C. 0.488 D. 0.512 E. 0.520 Solution. Always start problems such as this one by drawing a graph of the area where the joint density is positive, and where events considered happen. The area where the joint density of X and Y is positive is shown in the figure below as the triangle bounded by the axes and the line y = l — x. The shaded region is the portion of the domain of the joint density over which X < 0.2. Pr(X <0.2)=TT6(1—(x+y))dydx= 60f[y_w_ly2] x=0 2 ' =-0.83+1=0.488. Answer C. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 119 - PRACTICE EXAMINATION N0. 1 3. May 2001 Course 1 Examination, Problem No. 6, also Study Note P-09-08, Problem No. 20 An insurance company issues life insurance policies in three separate categories: standard, preferred, and ultra-preferred. Of the company’s policyholders, 50% are standard, 40% are preferred, and 10% are ultra-preferred. Each standard policyholder has probability 0.010 of ' dying in the next year, each preferred policyholder has probability 0.005 of dying in the next year, and each ultra-preferred policyholder has probability 0.001 of dying in the next year. A policyholder dies in the next year. What is the probability that the deceased policyholder was ultra-preferred? A. 0.0001 B. 0.0010 C. 0.0071 D. 0.0141 E. 0.2817 Solution. Start by labeling the events: let S be the event of a policyholder having a standard policy, P be the event of a policyholder having a preferred policy, U be the event of a policyholder having an ultra-preferred policy, and D be the event that a policyholder dies. Then by the Bayes’ Theorem: _ Pr(D|U)Pr(U) _ MUID) _ Pr(D|S)Pr(S)+Pr(D|P)Pr(P)+Pr(D|U)P(U) — 0.001.o.1o = —-—————— = 0.0141. 0.01 - 0.50 + 0.005 .0.40 + 0.001 - 0.10 Answer D. 4. May 2001 Course 1 Examination, Problem No. 7, also Study Note P-09-08, Problem No. 104 A joint density function is given by f( ) kx, for0<x<l, O<y<1, x. = . y 0, otherwrse, where k is a constant. What is Cov(X,Y)? A. -— B.0 C .l D 13.3 6 9 3 l . 6 Solution. The joint density is a product of functions of x, fx (x) = k, -x, and y, f, (y) = k2 -l, where kl -k2 = k, on a rectangular region, thus X and Y are independent, and their covariance must be zero. This is an important shortcut to remember: a necessary condition for independence of two random variables is that the region where their joint density is positive is a rectangle with sides parallel to the axes (this could be the whole plane, or a quadrant of it, or in a discrete case a collection of points positioned in a rectangular shape, with points present in each position along each vertical or horizontal line) and if also that joint density is a product of a function of the first ASM Study Manual for Course PI] Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski ~ 120 - \ PRACTICE EXAMINATION N0. 1 variable only and a function of the second variable only, then the two variables are independent. Answer B. 5. May 2001 Course 1 Examination, Problem No. 9, also Study Note P-09-08, Problem No. 13 An actuary is studying the prevalence of three health risk factors, denoted by A, B , and C, within a population of women. For each of the three factors, the probability is 0.1 that a woman in the population has only this risk factor (and no others). For any two of the three factors, the probability is 0.12 that she has exactly these two risk factors (but not the other). The probability that a woman has all three risk factors, given that she has A and B, is 31);. What is the probability that a woman has none of the three risk factors, given that she does not have risk factor A? A. 0.280 B. 0.311 C. 0.467 D. 0.484 E. 0.700 Solutiong"? . Let A, B,‘ and C be the events of having the three health risk factors A, B , and C, respectively. The Venn diagram below summarizes the unconditional probabilities of all mutually exclusive pieces (e.g., Pr(An BC n CC) = 0.10, Pr(A n B nCC) = 0.12, etc.) given in the problem. S 0.10 Furthermore, we are told that l:Pr(AanC|AnB)=——Pr(AanC)= x . 3. Pr(AnB) x+0.12 Itfollowsthat x=l(x+0.12)=lx+0.04. 3 3 Therefore, %x = 0.04, and x = 0.06. The probability we are looking for is ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20042008 by Knysztof Ostaszewski - 121 - PRACTICE EXAMINATION NO. 1 We add u probabilities of mutually exclusive pieces in mated in the Venn diagram above r—’— l—Pr(AUBuC) 1— (3-0.10+3-0.12+0.06) 0.28 =—————-=——=—=0.4667. l—Pr(A) 1— (0.10+2-0.12+0.06) 0.60 w We add u probabilities of mutually exclusive pieces in icated in the Venn diagram above Answer C. 6. May 2001 Course 1 Examination, Problem No. 10, also Study Note P-09-08, Problem No. 113 Two life insurance policies, each with a death benefit of 10,000 and a one-time premium of 500, are sold to a couple, one for each person. The policies will expire at the end of the tenth year. The probability that only the wife will survive at least ten years is 0.025, the probability that only the husband will survive at least ten years is 0.01 , and the probability that both of them will survive at least ten years is 0.96. What is the expected excess of premiums over claims, given that the husband survives at least ten years? A. 350 B. 385 C. 397 D. 870 E. 897 Solution. Start by labeling the events. Let W be the event that wife survives at least 10 years, and H be the event that husband survives at least 10 years. Let us also denote by B the amount of benefit paid, and H be the profit (i.e., excess of premiums over claims) from selling policies. We have: Pr(H)= Pr(Hnw)+ Pr(H nWC) = 0.96+0.01= 0.97. Furthermore: Pr H 0Wc 0.01 Pr(W CIH)=%=6§7=00103. It follows that: Elan)= E((1000-B)|H)=1000-E(B|H)= =1000-(0-Pr(W|H)+10000-PI(W‘ H))=1000—10000-0.0103 = 897. Answer E. 7. May 2001 Course 1 Examination, Problem No. 12, also Study. Note P-09-08, Problem N0. 3 You are given Pr(AUB)=0.7 and Pr(AUBC)=0.9. Determine Pr(A). ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 122 - PRACTICE EXAMINATION No.1 A. 0.2 B. 0.3 C. 0.4 D. 0.6 D. 0.8 Solution. Since Pr(A U B) = 0.7, we see that Pr((A U B)C) = 0.3. From Pr(A U BC) = 0.9 we get Pr((AUBC)C)=0.l. But (AUB)c =AC nBC and (AUBC)C =Ac OB, so that these two events, (A U B)c and (A U BC )c are mutually exclusive and their union is Ac. This gives Pr(A‘) =0.3+o.1 = 0.4 and Pr(A)= 0.6. Answer D. 8. May 2001 Course 1 Examination, Problem No. 13, also Study Note P-09-08, Problem No. 41 A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year period of time. Individual participants in the study drop out before the end of the study with probability 0.2 (independently of the other participants). What is the probability that at least 9 participants complete the study in one of the two groups, but not in both groups? A. 0.096 B. 0.192 C. 0.235 D. 0.376 E. 0.469 Solution. Define the following random variables: X = number of group 1 participants that complete the study. Y = number of group 2 participants that complete the study. We are given that X and Yare independent. Therefore, Pr(({X 2 9} n {Y < 9}) U ({X < 9} n{Y .>. 9})) = =Pr({X 29}n{Y < 9})+Pr({X<9}n{Y 29}). Because of symmetry and independence between X and Y the above equals: 2Pr({X 2 9} n {Y < 9}) = 2Pr(X 2 9)Pr(Y < 9). Note that probabilities concerning X are same as those concerning Y. Observe that X counts the number of successes in ten Bernoulli trials, where success is defined as completion of the study by an individual participant, so that p = 0.8, and n = 10. Based on this, the above is equal to: 2m(x 29)Pr(X< 9)=2Pr(X 29)(1-Pr(X 29)): =2[[ 1: W£ :3)-03'°1[1-[( ‘9" Wi 10°31} = 2 - 0.3758 - (1 — 0.3758) = 0.4692. Answer E. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 123 - ~ PRACTICE EXAMINATION NO. 1 9. May 2001 Course 1 Examination, Problem No. 14, also Study Note P-09-08, Problem No. 115 The stock prices of two companies at the end of any given year are modeled with random variables X and Y that follow a distribution with joint density function 2x, for0<x<1,x<y<x+1, f(x,y) = . 0, otherW1se. What is the conditional variance of Y given that X = x? A.— B.1 C.x+l D. xz—l E.x2+x+-1- 12 6 2 6 3 Solution. - Let fx (x) be the marginal density function of X. Then x+l fx (x) = I 2xdy = 23:34::+1 = 2x(x + 1) — 2x2 = 2x for 0 < x < l, and‘ fx (x) = 0 otherwise. Consequently, fx Y (x’y) { f y X = x = —-‘——— = A I ) fx (x) We see that the random variable (Y! X = x) is uniform on the interval (x,x +1), and therefore its . . . . l . mean 18 x + 2’ and its variance is 1—2. You could calculate those parameters usmg calculus, but 1, ifx<y<x+1, 0, otherwise. you should not. You should know key properties of the uniform distribution. Answer A. 10. May 2001 Course 1 Examination, Problem No. 17, also Study Note P-09-08, Problem An auto insurance company insures an automobile worth 15000 for one year under a policy 'with a 1000 deductible. During the policy year there is a 0.04 chance of partial damage to the car and a 0.02 chance of a total loss of the car. If there is partial damage to the car, the amount X of damage (in thousands) follows a distribution with a density function a f(x)={0.5003e , for0<x<15, 0, otherwise. What is the expected claim payment? A. 320 B. 328 C. 352 D. 380 E. 540 Solution. ‘. ' ., -. “ Let Ydenote the claim payment made by the insurance company, in thousands. As'the‘fiiilicy ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof OstaszeWSki - 124 . PRACTICE EXAMINATION NO. 1 has a deductible of 1 (thousand), the claim payment is 0, when X <1,with probability 0.94, . . Y = X—l, when l S X <15,note that Pr(l S X < 15) =0.04, WX. 14, when X 2 15,with probability 0.02. Therefore, the expected value ofs this random variable is calculated as follows: E(Y)= 0.94 o+o.04 “iIOHm x-1)e 2ax+o.0214= — __ '3‘ =0-020012'[:1f I-xegdx_ T-edeJ‘I'OZS: u—x v—~ 28x M—J -_ du=dx dv=e 2die =0.04 05003 fi—w———l Integration by pans in the first integral ”0020012 —2xe ‘2‘ :s+2je Zarx— J'e ;dx]+0.28= —30e 75 + 2605+ +[-2e-2] % =0 “020012 = 0.020012 (—30e'75 + 2e'05 — 2e 75 + 2e’05 ) + o .28~ 0 32819738. Since this is E (Y) expressed m thousands, the expected claim payment is approximately 328. Answer B. = 0.020012 [—30e'” + 2e'05 + je" 2 dx] +0 28— _ x x=l5 ' J+028= x=1 11. May 2001 Course 1 Examination, Problem No. 19, also Study Note P-09-08, Problem No. 83 A company manufactures a brand of light bulb with a lifetime in months that is normally distributed with mean 3 and variance 1. A consumer buys a number of these bulbs with the intention of replacing them successively as they burn out. The light bulbs have independent lifetimes. What is the smallest number of bulbs to be purchased so that the succession of light bulbs produces light for at least 40 months with probability at least 0.9772? A. 14 B. 16 C. 20 D. 40 D. 55 Solution. Let X,,...,X,, denote the life‘ spans of the n light bulbs purchased. Since these random variables are independent and normally disuibuted with mean 3 and variance 1, the random variable (-3,. ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 125 - PRACTICE EXAMINATION NO. 1 . S = XI + ...+ X” is also normally distributed with mean it = 3n and standard deviation J 0' = Jn- = J5. We want to choose the smallest value for n such that 0.9772 s Pr(S > 40) =Pr[s_ 3" 40" 3"]. T>Jz We know that 53—3" is standard normal, because S —- N (3n,n). Let Z denote a standard normal n random variable. Since Pr(Z > —2) = 0.9772, n should satisfy the following inequality: 40 — 3n < _2 ‘ J; 0 To find such an n, we solve the corresponding equation for n: 40f“ = —2,resulting in n = 16. n Answer B. ~ 12. May 2001 Course 1 Examination, Problem No. 20, also Study Note P-09-08, Problem A No. 46 ‘ A device that continuously measures and records seismic activity is placed in a remote region. The time, T, to failure of this device is exponentially distributed with mean 3 years. Since the device will not be monitored during its first two years of service, the time to discovery of its failure is X = max(T,2). Determine E(X). 2 4 g A. 2+:e'6 B.2—2e3+5e3 C.3 D.2+3e3 E.5 Solution. _1 I The survival function of Tis s, (t): e 3 for 0 < t < co, while its PDF IS s, (t)- - 3 —e 3 for 0 < t < co, and T is nonnegative with probability one. Using the Darth Vader Rule for E(max(T,2)), we obtain +oo w : =- I 2 E(max(r,2))= 2+ js,(:)dt = 2+ je‘idt = 2+ 3I§e_3dt = 2+ 3e—3. 2 2 2 =Pf(T >2) AnswerD. E(7MM7'Z)>; E(T1V’)+Ei°1/7<>> :Ltw 362751?” 5? PM”) 13. May 2001 Course 1 Examination, Problem No. 22, also Study Note P-09- 08, Problem No.88 The waiting time for the first claim from a good driver and the waiting time for the first claim from a bad driver are independent and follow exponential distributions with means 6 years and 3 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 126 - 3") PRACTICE EXAMINATION NO. 1 years, respectively. What is the probability that the first claim from a good driver will be filed within 3 years and the first claim from a bad driver will be filed within 2 years? 1 -1 1 -1 1 -1 -2 -1 -l A.—l—e3—e2+e‘ B.—e° C.1-e3—e2+e6 18 18 -2 -1 -1 _1 -1 1 -1 D.l-e3—e3+e3 E.l——e3--e2+—e6 6 18 Solution. Let X denote the waiting time for a first claim from a good driver, and let Y denote the waiting time for a first claim from a bad driver. The respective cumulative distribution functions for X x y and Yare Fx(x)=1—e'3 for x>0, and my)=1-e'3 for y>0. Therefore, Pr({X s 3}n{Y s 2}) = Pr(X s 3)-Pr(Y s 2) = Fx(3)F,,(2) = -1 -2 -2 -1 -1 =[l-e 2][l—e 3)=l-e 3-e 2+e 5. 14. May 2001 Course 1 Examination, Problem No. 23‘, also Study Note P-09-08, Problem No. 28 A hospital receives 1/5 of its flu vaccine shipments from Company X and the remainder of its shipments from other companies. Each shipment contains a very large number of vaccine vials. For Company X’s shipments, 10% of the vials are ineffective. For every other company, 2% of Answer C. the vials are ineffective. The hospital tests 30 randomly selected vials from a shipment and finds ; . that one vial is ineffective. What is the probability that this shipment came from Company X? A. 0.10 B.O.14 C.0.37 D.0.63 E.0.86 Solution. As always, we start by labeling the events. Let X be the event that shipment came from Company X, and F be the event that exactly one of the vaccine vials tested is ineffective. By applying the Bayes Theorem we get: Pr(F|X)Pr(X) MXIF): Pr(F|X)Pr(X)+Pr(F|XC)Pr(XC)' Now we use the data given in the problem. We have Pr(X) = g, and therefore Pr(Xc) = %. Furthermore Pr(F|X)=[ 310 J-0.10-0.9029=0.141 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 127 - PRACTICE EXAMINATION NO. 1 and Pr(FIXC)=[ 31° ].o.02-0.98”=o.334. Substituting these values into the Bayes formula, we get 0.141-l Pr(X|F) = —15—4 = 0.096. 0'141‘§+0'334'§ Answer A. 15. May 2001 Course 1 Examination, Problem No. 24, also Study Note P-09-08, Problem No. 79 A device contains two components. The device fails if either component fails. The joint density function of the lifetimes of the components, measured in hours, is f (s,t) , where 0 < s < 1 and 0 < t < 1. What is the probability that the device fails during the first half hour of operation? 0505 10.5 I l A. I Jf(s,t)ds dt B. ij(s,t)ds dt C. I Jf(s,t)ds dt 0 0 0 0 0.50.5 o. fIf(s,t)ds dt+ Iff(s,t)ds dt E. ff f(s,t)ds dt+j°ff(s,t)ds a: 0 D 0 0 0 0.5 0 0 Solution. Always begin by drawing a graph indicating the areas where the density is positive. In this case that region where the density is positive is the unit square. The shaded region is the portion of the domain of the density over which the device fails sometime during the first half hour. We divided it into two subregions: A and B, as indicated in the figure. l S ASM Study Manual for Course P/l Actuarial Examination. © Copyright 20044008 by Krzysztof Ostaszewski - 128 - J PRACTICE EXAMINATION NO. 1 The probability sought is l l l '2' E l Pr[{SS flu {T < 3): Hm.) t )(dsdHHf s, r )dsdt, o o o 1 2 where the first integral is over the sub-region indicated 1n the figure as A and the second integral is over the sub-region indicated in the figure as B. Answer E. 16. May 2001 Course 1 Examination, Problem No. 26, also Study Note P-09-08, Problem No. 109 A company offers earthquake insurance. Annual premiums are modeled by an exponential random variable with mean 2. Annual claims are modeled by an exponential random variable with mean 1. Premiums and claims are independent. Let X denote the ratio of claims to premiums. What is the density function of X? A. l B. #2 c. e" D. 29'” E. xe" 211: +1 (2x + 1) Solution. Let U be the annual claims and let Vbe the annual premiums (also random), and let fay (u,v) be the joint density of them. Furthermore, let fx be the density of X and let Fx be its cumulative distribution function. We are given that U and V are independent, and hence -1 1 -1 fu'v(u,v)=e'" 3e 2 =Ee'“e 2 ‘ for 0 < u < no.0 < v < co. Also, noting the graph below, we have: ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 129 - on V1 Fx(x)=Pr(XSx)=Pr[%—Sx)= "EV! ‘5‘ 1e "e 2dudv=I 2 1w 0 2 ll °'—.3 14:0 0 ll °'—»3 Finally, x(x)= F (10’ (2x+ 1) Answer B. This problem can also be done with the use of bivariate transformation. We will now give an alternative solution using that approach. Consider the following transformation 22:9, y=v. V Then the inverse transformation is U = XY, V = Y. We know that f,,.v(u.v)=§e-ue‘E for u > 0, v > 0. It follows that 3(u, v) fx.y (LY) = fay ("(x’Y)’v(xvY)) a(x’y) for xy > 0 and y > 0, o...
View Full Document

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern