Practice_Exam_2-Solutions

Practice_Exam_2-Solutions - PRACTICE EXAMINATION NO 2...

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Unformatted text preview: PRACTICE EXAMINATION NO. 2 PRACTICE EXAMINATION NUMBER 2 SOLUTIONS 1. November 2000 Course 1 Examination, Problem No. 27, also Study Note P-09-08, Problem No. 98 Let Xl ,X2 ,X3 be a random sample from a discrete distribution with probability function , forx=0, , forx=1, , otherwise. Determine the moment generating function M (t) of Y = X,X2X3. 19 s 1 2 3 1 8 3, 1 2,, .—+—' B.l+2’ c. —+—' D.—+—e E.—+—e 27 27e e [3 39] 27 27 3 Solution. Let f, be the probability function for Y = X1X2X3. Note that Y = 1 if and only if XI = X2 = X3 =1. Otherwise, Y: 0. We also know that Pr(Y= 1)= Pr({xl =1}n{X2 =1}n{x3 =1})= 2 3 8 = Pr(Xl =1)Pr(X2 =1)1>r(x3 = 1) = [3] = .27, We conclude that %, fory=0, —, fory=1, 27 and is zero otherwise. Y is a Bernoulli Trial random variable and 19 8 Mt =M 1=E ‘Y =—+— ', () r() (e ) 27 273 AnswerA. 2. November 2000 Course 1 Examination, Problem No. 28, also Study Note P-09-08, Problem No. 12 A doctor is studying the relationship between blood pressure and heartbeat abnormalities in her patients. She tests a random sample of her patients and notes their blood pressures (high, low, or (WV normal) and their heartbeats (regular or irregular). She finds that: ASM Study Manual for Course PI] Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 149 - PRACTICE EXAMINATION N0. 2 (i) 14% have high blood pressure. (ii) 22% have low blood pressure. ' (iii) 15% have an irregular heartbeat. (iv) Of those with an irregular heartbeat, one-third have high blood pressure. (v) Of those with normal blood pressure, one-eighth have an irregular heartbeat. What portion of the patients selected have a regular heartbeat and low blood pressure? A. 2% B. 5% C. 8% D. 9% E. 20% Solution. We are directly given that Pr(Irregular heartbeat) = 0.15, Pr(High blood pressure) = 0.14, and Pr(Low blood pressure) = 0.22. We calculate other relevant probabilities and place all probabilities in a table (below): Pr(Normal blood pressure) = l— 0.14 — 0.22 = 0.64, Pr ({High blood pressure} A {Irregular heartbeat}) = = Pr(High blood pressure] Irregular heartbeat) - Pr(lrregular heartbeat) = g - 0.15 = 0.05, Pr({Irregular heartbeat} m {Normal blood pressure}) = = Pr(Irregular heartbeat| Normal blood pressure) o Pr(Normal blood pressure) = = 1-0.64 =0.08, 8 Pr (Regular heartbeat) = 1— 0.15 = 0.85, Pr({High blood pressure} n {Regular heartbeat}) = 0.14 — 0.05 = 0.09, Pr({Normal blood pressure} 0 {Regular heartbeat}) = 0.64 - 0.08 = 0.56, Pr({Low blood pressure} n {Irregular heartbeat}) = 0.15 — 0.05 - 0.08 = 0.02, Pr({Low blood pressure} n {Regular heartbeat}) = 0.22 — 0.02 = 0.20, m We see from this table or from the last calculation that 20% of patients have a regular heartbeat and low blood pressure. Answer E. ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 20044008 by Krzysztof Ostaszewski - 150 - PRACTICE EXAMINATION N0. 2 3. November 2000 Course 1 Examination, Problem No. 30, also Study Note P-09-08, Problem No. 11 An actuary studying the insurance preferences of automobile owners makes the following conclusions: (i) An automobile owner is twice as likely to purchase collision coverage as disability coverage. (ii) The event that an automobile owner purchases collision coverage is independent of the event that he or she purchases disability coverage. (iii) The probability that an automobile owner purchases both collision and disability coverages is 0.15 . What is the probability that an automobile owner purchases neither collision nor disability coverage? A. 0.18 B. 0.33 C. 0.48 D. 0.67 E. 0.82 Solution. We begin by labeling the events. Let C be the event that a policyholder buys collision coverage, and let D be the event that a policyholder buys disability coverage. We are given that Pr(C) = 2Pr(D) and Pr(C n D) = 0.15. Using independence of C and D, we see that 0.15 = Pr(C n D) = Pr(C)Pr(D) = 2Pr(D)Pr(D) = 2(1’r(D))2 and therefore (Pr(D))2 = g = 0.075. From that Pr(D) = , and Pr(C) = 2%. The problem asks for Pr((C u D)C) = Pr(CC 0 DC). Independence of C and D also implies the independence of CC and DC. As a result, we have Pr(Cc nDc) = Pr(CC) - 19(06): (l—Pr(C))-(1 — Pr(D)) = =(1—2W)-(1-W)=0.3284. We could also calculate the answer as Pr((Cu D)C) =1-Pr(CU D)=1—(Pr(C)+Pr(D)— Pr(CnD)) = = 1- (240.075 + J0.075 — 0.15) = 1.15 - 340.075 = 0.3284. Answer B. 4. November 2000 Course 1 Examination, Problem No. 32, also Study Note P-09-08, Problem No. 75 The monthly profit of Company I can be modeled by a continuous random variable with density function f. Company 11 has a monthly profit that is twice that of Company I. Detemiine the probability density function of the monthly profit of Company 11. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 151 - PRACTICE EXAMINATION N O. 2 A. B.f(§] C. 24%] D. 2f(x) E.2f(2x) Solution. Let X and Y be the monthly profits of Company I and Company H, respectively. Let us write fx for the PDF of X, instead of just f. Then Y = 2X, so that X = éY. Therefore dx 1 l = x - — = - --. my) f.( (a) dy f.[2y] 2 You can, of course, also use the CDF approach. In this case 5(y)= Pr(Y Sy)=Pr(2XSy)=Pr[XS%)= Fx(%], and f.(y)=F;(y)=in[Z)=lF;[l]=§f.[z). dy 2 2 2 2 Answer A. 5. November 2000 Course 1 Examination, Problem No. 34, also Study Note P-09-08, Problem No. 29 The number of days that elapse between the beginning of a calendar year and the moment a high- risk driver is involved in an accident is exponentially distributed. An insurance company expects that 30% of high-risk drivers will be involved in an accident during the first 50 days of a calendar year. What portion of high-risk drivers are expected to be involved in an accident during the first 80 days of a calendar year? A. 0.15 B. 0.34 C. 0.43 D.0.57 E. 0.66 Solution. Let The the number of days that elapse before a high-risk driver is involved in an accident. We know that T is exponentially distributed with unknown parameter A. We are also given that 0.3 = Pr(T s 50)=1—e"°‘. ln0.7 Therefore, e"°" = 0.7 and 2. = — . It follows that Q Pr(T s 80) = 1- e’m =1— e50'1"°'7 =1— 0.7‘-6 = 0.4349. Answer C. 6. November 2000 Course 1 Examination, Problem No. 36, also Study Note P-09-08, Problem No. 91 An insurance company insures a large number of drivers. Let X be the random variable ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 152 - PRACTICE EXAMINATION N0. 2 representing the company’s losses under collision insurance, and let Yrepresent the company’s losses under liability insurance. X and Y have joint density function 2x+2—y f000= , for0<x<land0<y<2, 0. otherwise. What is the probability that the total loss is at least 1? A. 0.33 B. 0.38 C.0.4l D.0.71 E. 0.75 Solution. We want to find Pr(X + Y > 1). We start by drawing a graph of the area where the density of this random vector “plays.” The probability is calculated as an integral of the joint density over the region indicated with dotted lines in the figure below. Therefore ‘2 2x+2— ‘ 1 1 1 ' Pr(X+Y>1)=IJl7lldydx=llliw+5y7y2l 01-: l =J[x+l—l—lx(l—x)—%(l—x)+l(l-x)2] = 0 l x=l =“if+3x+ljdx=(ix3+2x2+lx) =1 2 l=fl o 8 8 24 8 =0 24 8 8 24 Answer D. 7. November 2000 Course 1 Examination, Problem No. 38, also Study Note P-09-08, Problem No. 101 The profit for a new product is given by Z = 3X - Y — 5. X and Y are independent random variables with Var(X) = l and Var(Y) = 2. What is the variance of Z? ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 153 - PRACTICE EXAMINATION NO. 2 A.l B.5 C.7 D.ll E. 16 Solution. Due to independence of X and Y Var(Z) = Var(3X—Y —5) = Var(3X)+Var(Y) = 32Var(X)+Var(Y) = 9-1+ 2 =11. AnswerD. 8. November 2000 Course 1 Examination, Problem No. 40, also Study Note P-09-08, Problem No. 122 A device contains two circuits. The second circuit is a backup for the first, so the second is used only when the first has failed. The device fails when and only when the second circuit fails. Let X and Y be the times at which the first and second circuits fail, respectively. X and Y have joint probability density fimction 6 " '2’ , < < < oo, f(x’y)={ e e forO .x y 0, othermse. What is the expected time at which the device fails? A. 0.33 B. 0.50 C. 0.67 D. 0.83 E. 1.50 Solution. Note that the joint density is positive only when 0 < x < y. Therefore, the marginal density of Y is calculated as y 3’ fr (3’) = Ilse-xe'z’dx = 6e'2’ Ie‘QLx = -6e'2’e" + 6e"2y = 6e'2’ — 6e'” 0 0 for 0 < y < co. Given this, E(Y) = Tyfr (y)dy = T(6ye-2y " 6ye"’)dy = 6 °° 6 °° 6 1 6 1 3 2 5 =_ ,2 -2yd __ .3 -3.vd = _____ ._=___=_=0.83. 2 {y e y 3 {y e y 2 2 3 3 2 3 6 fir—J ‘—w-—’ Mean of an ex nemial Mean of an ex nential random vnriab c with random varinb c with hazard late 2 hazard rate 3 Answer D. 9. May 2000 Course 1 Examination, Problem No. 1, also Study Note P-09-08, Problem No. 2 The probability that a visit to a primary care physician’s (PCP) office results in neither lab work nor referral to a specialist is 35%. Of those coming to a PCP’s office, 30% are referred to specialists and 40% require lab work. Determine the probability that a visit to a PCP’s office results in both lab work and referral to a specialist. ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 154 - PRACTICE EXAMINATION NO. 2 A. 0.05 B. 0.12 C. 0.18 D. 0.25 E. 0.35 Solution. We define L as the event of lab work needed, and R as the event of a referral to a specialist needed. We are given that Pr(R) = 0.3, Pr(L) = 0.4, and Pr((Lu 1a)“) = Pr(LC n R‘) = 0.35. It follows that Pr(Lu R) = 1 -P((LUR)C)= 0.65. Finally, fiom the formula Pr(L u R) = Pr(L)+Pr(R)— Pr(L n R), we get Pr(LnR)= Pr(L)+Pr(R)—Pr(LuR)= 0.3+ 0.4-0.65 =0.05. Answer A. 10. May 2000 Course 1 Examination, Problem No. 2, also Study Note P-09-08, Problem No. 27 A study of automobile accidents roduced the followin ; data: I odel year Proportion of all Probability of involvement vehicles in an accident 1997 m- 1998 m- [EE— m_ An automobile from one of the model years 1997, 1998, and 1999 was involved in an accident. Determine the probability that the model year of this automobile is 1997. A. 0.22 B. 0.30 C. 0.33 D. 0.45 E. 0.50 Solution. Let us start by labeling the events: E97 -- The model year is 1997, 1!?93 -- The model year is 1998, E99 -- The model year is 1999, E0 -- The model year is other than 1997, 1998 or 1999, A -- The car is involved in an accident. We are given Pr(E9.,)=0.l6, Pr(59,)=0.18. Pr(E..,)=o.20, Pr(Eo)=O.46, Pr(A[ 159,) = 0.05, Pr(A| E93) = 0.02, Pr(A|E99) = 0.03, Pr(A|Eo) = 0.04. It might seem that this is a Bayes’ Theorem problem, but note that the accident happened to a ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Krzysztol' Ostaszewski - 155 - PRACTICE EXAMINATION N0. 2 1997, 1998, or a 1999 model, but not one of the other years’ model, so that not all of the pieces of the partition {E97 ,E93 ,E99 ,Eo} are considered, and we can’t use Bayes directly. We are i actually asked to find this conditional probability: Pr(E97lAn(E97 U E93 U E99))' Using the definition of conditional probability M159, n(A n (E,7 u 13,, u E,,))) Pr(E97|Afl(E97UE98UE99))= Pr(An(E UE UE 97 98 99 Pr[A n E97 n [E97 u E93 u E99 ____—J Mutually exclusive Mutually exclusive _ Pr(AnE97) Pr(AnE,,)+Pr(AnE,,)+Pr(AnE,,) Pr(A|E97) ' Pr(E97)+ Pr(A|E9s)'Pr(E9s)+ FAME») ' Pr(E99) = 0.05-0.16 = 0.008 20.4545. 0.05~0.16+0.02-0.18+0.03-0.20 0.0176 ' We could also do this problem by drawing a probability tree, and using only a portion of it, excluding other years’ models: Consider only these three possibilities, N0 acoldent instead of full D Bayes Acoldent Theorem, setup No accident Accident No accident Answer D. “5 ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof OstaszeWSki - 156 - "w PRACTICE EXAMINATION N0. 2 11. May 2000 Course 1 Examination, Problem No. 3, also Study Note P-09-08, Problem No. 37 The lifetime of a printer costing 200 is exponentially distributed with mean 2 years. The manufacturer agrees to pay a full refund to a buyer if the printer fails during the first year following its purchase, and a one-half refund if it fails during the second year. If the manufacturer sells 100 printers, how much should it expect to pay in refunds? A. 6,321 B. 7,358 C. 7,869 D. 10,256 E. 12,642 Solution Z The exponential distribution with a mean of 2 has density fx (x) = 5e 2 , for x > 0, and the l --x cumulative distribution function Fx (1:) = l — e 2 , for x > 0. The probability that a printer will fail in the first year is l Pr(X s 1) = Fx(1)=1— e 2 = 0.39347, so that the expected number of failures in the first year out of 100 printers is 39.347. The probability that a printer will fail in the second year is 2 l __ Pr(l < X: 2)=Fx(2)-Fx(1)= sx(1)-sx(2)= e 2 —e 2 =0.23865, so that the expected number of failures in the second year out of 100 printers is 23.865. The expected amount the manufacturer will pay in refunds is 200 - 39.347 + 100 ~ 23.865 = 10256. Answer D. 12. May 2000 Course 1 Examination, Problem No. 4, also Study Note P-09-08, Problem No. 74 Let T denote the time in minutes for a customer service representative to respond to 10 telephone inquiries. T is uniformly distributed on the interval with endpoints 8 minutes and 12 minutes. Let R denote the average rate, in customers per minute, at which the representative responds to inquiries. Which of the following is the density function of the random variable R on the interval l2 8 12 5 51n(r) 10 5 A.— B. 3-— C.3 - D.- E.— 5 2r r 2 r2 2r2 Solution. Since Thas a uniform distribution on the interval from 8 to 12, its PDF is f, (t) =% for ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 157 - - SSI - msmazsmso 301255291 Kq sooz-vooz lufiufidoa ©'“°!19“!“‘3¥El lam!an [Id 931:10:) 10.1 [mum Rpms wsv 01 [12an pm; uogfiax 91.11 .1er wusuoo s! 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May 2000 Course 1 Examination, Problem No. 6, also Study Note P-09-08, Problem No. 18 Two instruments are used to measure the height, h, of a tower. The error made by the less accurate instrument is normally distributed with mean 0 and standard deviation 0.0056h. The error made by the more accurate instrument is normally distributed with mean 0 and standard deviation 0.0044h. Assuming the two measurements are independent random variables, what is the probability that their average value is within 0.005h of the height of the tower? . A. 0.38 B.0.47 C. 0.68 D. 0.84 E. 0.90 Solution. Let us denote by El and E2 the errors of the first and second instrument, respectively. We are given that 15(5.) = 15(52): 0, Var(E, ) = (0.0056h)2, and Var(E,) = (0.0044h)2. We also know that E, +132 is normal with mean zero, and since El and E2 are independent, Var(E, + 5,) = Var(E,)+ Var(E,) = (0.005%)2 +(o.oo44h)2 = (0.00712h)2. The average of El and E2 is a random variable A = %(E1 + E2), which is also normal with mean 0 and variance Var(A) = ivmzz, + 15,) = (0.00356h)2. The average height is within 0.005h of the height of the tower if the absolute error in the average is less than 0.005h. Thus, we want to find Pr(|A| < 0005):) = Pr(-0.005h < A < 0.005h). We standardize A to get Pr(—0.005h < A < 0.005h) = r M < Varm‘ varm) =Pr(—0.005h—0 Z 0.005h-0 P[—0.005h-E(A) A-E(A) 0.005h—E(A)] < < — 0.00356h 0.0035611 ASM Study Manual for Course Pll Actuarial Examination. © Copyright 2004-2008 by Krzysztof Ostaszewski - 159 - )=Pr(—1.4<Z<l.4), PRACTICE EXAMINATION NO. 2 , . where Z is standard normal. From the standard normal table, Pr(Z < 1.4)= 0.9192, so that Pr(Z < —1.4) = Pr(Z > 1.4) = 1— Pr(Z $1.4): 0.0808 and finally Pr(-1.4 < Z <1.4)= 0.9192 — 0.0808 = 0.8384. Answer D. 15. May 2000 Course 1 Examination, Problem No. 7, also Study Note P-09-08, Problem No. 55 An insurance company’s monthly claims are modeled by a continuous, positive random variable X , whose probability density function is proportional to (1+ x)" , where 0 < x < oo. Determine the company’s expected monthly claims. 1 l 1 A. - B. - C. — D. 1 E. 3 6 3 2 Solution. The PDF is of the form fx (x) = c(1+ x)'4 where c is an unknown positive constant, for x > 0, and fx (x) = 0 for x S 0. We must have .. x-w: 1=£e(1+x)“¢x=[_is-(1+x)"] This gives C: 3 and fx (x): 3(1-1-x)’4 for x >0, fx (x): 0 for x 50. Hence, SAX): Tf(t)dt = +f3(l+t)"a1t =(—(1+t)'3) X is a random variable, which is non-negative almost surely, and therefore x-)°° l -2, x=0 r—m = (1+x)—3. (=3 +00 E(X) = js, (x)dx = j (1 +x)"¢x = Ema-1] o o - You could also see that this a Pareto distribution, of the form a0“ fx 0‘) - W: with 05:3 and 9:1, sothat x=0 16. May 2000 Course 1 Examination, Problem No. 8, also Study Note P-09-08, Problem No. 64 , A probability distribution of the claim sizes for an auto insurance policy is given in the table ASM Study Manual for Course P/l Actuarial Examination. (9 Copyright 2004-2008 by Krzysztof Ostaszewski - 160 - PRACTICE EXAMINATION NO. 2 below: Claim Probability Size 20 0.15 0 .05 What percentage of the claims are within one standard deviation of the mean claim size? A. 45% B. 55% C. 68% D. 85% E. 100% Solution. The mean claim size is: E(X)= 20-0.15+30~0.1+40-0.05+50-0.20+60-0.10+70-0.10+80-0.30 =55. The second moment of the claim size is: E(X2)= 202 -0.15 + 302 -0.1+4o’ -o.05 +502 -0.20+ + 602 -0.10 + 702 -0.10 + 802 030 = 3500. Therefore, the variance of the claim size is: Var(X) = E(X2)— (E(X))2 = 3500 — 552 = 475. The standard deviation is .[Var(x) = Jfi = 21.79. The claim sizes within one standard deviation of the mean claim size of 55 are those claim sizes between 55 — 21.79 = 33.21 and 55 + 21.79 = 76.79. Those claim sizes are 40, 50, 60 and 70. The total probability of those claim sizes is 0.05 + 0.2 + 0.1+ 0.1: 0.45. Answer A. 17. May 2000 Course 1 Examination, Problem No. 9, also Study Note P-09-08, Problem No. 85 The total claim amount for a health insurance policy follows a distribution with density function 1 -L f (x) = We '°°° forx > 0. The premium for the policy is set at 100 over the expected total claim amount. If 100 policies are sold, what is the approximate probability that the insurance company will have claims exceeding the premiums collected? A. 0.001 B. 0.159 C. 0.333 D. 0.407 E. 0.460 Solution. ASM Study Manual for Course P/l Actuarial Examination. © Copyright 2004-2008 by Knysztof Ostaszewski - 161 - PRACTICE EXAMINATION N0. 2 While the problem does not say so directly, you are expected to understand that the 100 policies are independent. This is the PDF of an exponential distribution with a mean of 1000. The expected claim per policy is 1000, and the variance is 10002. The premium collected is 1100 per policy. For 100 policies, a total of 110,000 is collected in premium. The total claim is W=x,+x,+~-X,om and E(W)=100E(X)=100,000, while, due to independe...
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