COURSE 1 REVISED SAMPLE EXAM

Transcription

1 COURSE REVISED SAMPLE EXAM A table of values for the normal distribution will be provided with the Course Exam.

2 Revised August 999

3 Problem # A marketing survey indicates that 60% of the population owns an automobile, 30% owns a house, and 0% owns both an automobile and a house. Calculate the probability that a person chosen at random owns an automobile or a house, but not both. A. 0.4 B. 0.5 C. 0.6 D. 0.7 E

4 Problem # In a country with a large population, the number of people, N, who are HIV positive at time t is modeled by N = 000 ln( t + ), t 0. Using this model, determine the number of people who are HIV positive at the time when that number is changing most rapidly. A. 0 B. 50 C. 500 D. 693 E

5 Problem # 3 Ten percent of a company=s life insurance policyholders are smokers. The rest are nonsmokers. For each nonsmoker, the probability of dying during the year is 0.0. For each smoker, the probability of dying during the year is Given that a policyholder has died, what is the probability that the policyholder was a smoker? A B. 0.0 C D E

6 Problem # 4 Let X and Y be random losses with joint density function -( x+ y) f ( x, y) = e for x > 0 and y > 0. An insurance policy is written to reimburse X+Y. Calculate the probability that the reimbursement is less than. A. e - B. e - C. - e - D. - e - E. - e - -4-

7 Problem # 5 The rate of change of the population of a town in Pennsylvania at any time t is proportional to the population at time t. Four years ago, the population was 5,000. Now, the population is 36,000. Calculate what the population will be six years from now. A. 43,00 B. 5,500 C. 6,08 D. 77,760 E. 89,580-5-

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9 Problem # 7 As part of the underwriting process for insurance, each prospective policyholder is tested for high blood pressure. Let X represent the number of tests completed when the first person with high blood pressure is found. The expected value of X is.5. Calculate the probability that the sixth person tested is the first one with high blood pressure. A B C D E

10 Problem # 8 At time t = 0, car A is five miles ahead of car B on a stretch of road. Both cars are traveling in the same direction. In the graph below, the velocity of A is represented by the solid curve and the velocity of B is represented by the dotted curve. miles/hr area= area=0 area= hours Determine the time(s), t, on the time interval (0, 6], at which car A is exactly five miles ahead of car B. A. at t =. B. at t = 3. C. at some t, 3 < t < 5, which cannot be determined precisely from the information given. D. at t = 3 and at t = 5. E. Car A is never exactly five miles ahead of Car B on (0,6]. -8-

11 Problem # 9 The distribution of loss due to fire damage to a warehouse is: Amount of Loss Probability , , , , Given that a loss is greater than zero, calculate the expected amount of the loss. A. 90 B. 3 C.,704 D.,900 E. 3, -9-

12 Problem # 0 An insurance policy covers the two employees of ABC Company. The policy will reimburse ABC for no more than one loss per employee in a year. It reimburses the full amount of the loss up to an annual company-wide maximum of The probability of an employee incurring a loss in a year is 40%. The probability that an employee incurs a loss is independent of the other employee s losses. The amount of each loss is uniformly distributed on [000, 5000]. Given that one of the employees has incurred a loss in excess of 000, determine the probability that losses will exceed reimbursements. A. B. C. D. E

13 Problem # The risk manager at an amusement park has determined that the expected cost of accidents is a function of the number of tickets sold. The cost, C, is represented by the function C(x) = x 3-6x + 5x, where x is the number of tickets sold (in thousands). The park self-insures this cost by including a charge of 0.0 in the price of every ticket to cover the cost of accidents. Calculate the number of tickets sold (in thousands) that provides the greatest margin in the insurance charges collected over the expected cost of accidents. A B C..00 D E

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16 Problem # 4 Workplace accidents are categorized into three groups: minor, moderate and severe. The probability that a given accident is minor is 0.5, that it is moderate is 0.4, and that it is severe is 0.. Two accidents occur independently in one month. Calculate the probability that neither accident is severe and at most one is moderate. A. 0.5 B C D E

17 Problem # 5 An insurance company issues insurance contracts to two classes of independent lives, as shown below. Class Probability of Death Benefit Amount Number in Class A B The company wants to collect an amount, in total, equal to the 95 th percentile of the distribution of total claims. The company will collect an amount from each life insured that is proportional to that life=s expected claim. That is, the amount for life j with expected claim E[X j ] would be ke[x j ]. Calculate k. A..30 B..3 C..34 D..36 E

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19 Problem # 7 An actuary is reviewing a study she performed on the size of claims made ten years ago under homeowners insurance policies. In her study, she concluded that the size of claims followed an exponential distribution and that the probability that a claim would be less than $,000 was The actuary feels that the conclusions she reached in her study are still valid today with one exception: every claim made today would be twice the size of a similar claim made ten years ago as a result of inflation. Calculate the probability that the size of a claim made today is less than $,000. A B. 0.5 C D E

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24 Problem # A dental insurance policy covers three procedures: orthodontics, fillings and extractions. During the life of the policy, the probability that the policyholder needs: orthodontic work is / orthodontic work or a filling is /3 orthodontic work or an extraction is 3/4 a filling and an extraction is /8 The need for orthodontic work is independent of the need for a filling and is independent of the need for an extraction. Calculate the probability that the policyholder will need a filling or an extraction during the life of the policy. A. 7/4 B. 3/8 C. /3 D. 7/4 E. 5/6 --

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26 Problem # 4 An automobile insurance company divides its policyholders into two groups: good drivers and bad drivers. For the good drivers, the amount of an average claim is 400, with a variance of 40,000. For the bad drivers, the amount of an average claim is 000, with a variance of 50,000. Sixty percent of the policyholders are classified as good drivers. Calculate the variance of the amount of a claim for a policyholder. A. 4,000 B. 45,000 C. 66,000 D. 0,400 E. 35,000-4-

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30 Problem # 8 The graphs of the first and second derivatives of a function are shown below, but are not identified from one another. y x Which of the following could represent a graph of the function? A. D. y y x x B. E. y y x x C. y x -8-

31 Problem # 9 An insurance company designates 0% of its customers as high risk and 90% as low risk. The number of claims made by a customer in a calendar year is Poisson distributed with meanθ and is independent of the number of claims made by a customer in the previous calendar year. For high risk customers θ = 0.6, while for low risk customers θ = 0.. Calculate the expected number of claims made in calendar year 998 by a customer who made one claim in calendar year 997. A. 0.5 B. 0.8 C. 0.4 D E

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34 Problem # 3 Curve C is represented parametrically by x = t +, y = t.curve C is represented parametrically by x = t +, y = t + 7. Determine all the points at which the curves intersect. A. (3,8) only B. (,0) only C. (3,8) and (-,8) only D. (,0) and (,7) only E. The curves do not intersect anywhere. -3-

35 Problem # 33 The number of clients a stockbroker has at the end of the year is equal to the number of new clients she is able to attract during the year plus the number of last year=s clients she is able to retain. Because working with existing clients takes away from the time she can devote to attracting new ones, the stockbroker acquires fewer new clients when she has many existing clients. The number of clients the stockbroker has at the end of year n is modeled by C n 9000 = Cn + 3 C. n The stockbroker has five clients when she starts her business. Determine the number of clients she will have in the long run. A. 3 B. 5 C. 0 D. 30 E

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37 Problem # 35 Suppose the remaining lifetimes of a husband and wife are independent and uniformly distributed on the interval [0,40]. An insurance company offers two products to married couples: One which pays when the husband dies; and One which pays when both the husband and wife have died. Calculate the covariance of the two payment times. A. 0.0 B C D E

38 Problem # 36 An index of consumer confidence fluctuates between - and. Over a two-year period, beginning at time t = 0, the level of this index, c, is closely approximated by c( t) t cos( t ), = where t is measured in years. Calculate the average value of the index over the two-year period. A. 4 sin( 4) B. 0 C. 8 sin( 4) D. 4 sin( 4) E. sin( 4) -36-

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40 Problem # 38 An investor bought one share of a stock. The stock paid annual dividends. The first dividend paid one dollar. Each subsequent dividend was five percent less than the previous one. After receiving 40 dividend payments, the investor sold the stock. Calculate the total amount of dividends the investor received. A. $ 8.03 B. $7.43 C. $0.00 D. $3.0 E. $

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42 Problem # 40 A small commuter plane has 30 seats. The probability that any particular passenger will not show up for a flight is 0.0, independent of other passengers. The airline sells 3 tickets for the flight. Calculate the probability that more passengers show up for the flight than there are seats available. A B C D. 0. E

43 Problem Key B D 3 C 4 D 5 C 6 C 7 B 8 C 9 D 0 B E C 3 B 4 E 5 E 6 A 7 C 8 C 9 B 0 E D D 3 D 4 D 5 D 6 A 7 C 8 A 9 C 30 E 3 A 3 C 33 D 34 C 35 C 36 C 37 E 38 B 39 D 40 E

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47 COURSE SOLUTIONS Solution #: B Let A be the event that a person owns an automobile. Let H be the event that a person owns a house. We want Pb A H g Pb A H g. PbA Hg = Pb Ag + PbHg Pb A Hg P A H = P b g ba Hg = 07. b g b g P A H P A H = = 05. Solution #: D dn dt = change in number of people who are HIV positive. We want to maximize dn dt. dn 000 = 000 = dt t + t t + is strictly decreasing for t 0. Therefore, N is maximized when t = 0. b g 000 So, N = ln t + = = = 693.

48 Solution #3: C P S D PcS Dh b g = PbDg PbS Dg = b0. gb005. g = PcS Dh = b0. 9gb00. g = b g b g c h c h = = P D = P S D + P S D = P S D Solution #4: D zz x b 0 0 g x+ y e dy dx = e Solution #5: C dp dt = kpbtg Separation of variables = d P = kdt ln P = kt + c P =, Pb4g = 36, 000 We are given: P b g b g b g b g b g b g ln 5, 000 = k 0+ c c = ln 5, 000 = ln 36, 000 = k 4 + ln 5, 000 ln 36, 000 ln 5, 000 k = = ln P 0 = b g b ln P = gb g P = e = 6, 08

49 Solution #6: C r x + y y = r sinθ x = r cosθ z z π / 4 π Change to polar coordinates c + r h rdrdθ = Solution #7: B b g = 5. E X b g = = P (person has high blood pressure) = E X 5. P (sixth person has high blood pressure) 008. = P (first five don t have high blood pressure) P (sixth has high blood pressure) 5 b g b g = = Solution #8: C A starts out 5 miles ahead of B. The distances traveled are given by the areas under the respective graph. Therefore, the distance between A and B increases by the area between the solid graph and the dotted graph, when the solid graph is above the dotted and decreases by the area when the solid graph is below the dotted. Therefore, at t = 3, the cars are miles apart. At t = 5, the cars are mile apart. So the cars must be 5 miles apart at some time between t=3 and t = 5. Solution #9: D b500gb0. 06g + b000gb0. 03g + b0, 000gb0. 008g + b50, 000gb0. 00g + b00, 000gb0. 00g = 0. = 900 b g 3

50 Solution #0: B Let C = the claim for employee j, for j =, j P( C > 0) = 040. j zx 5000 x P( Cj > x Cj > 0) = dx = 5000, 000 x Note: if C 3000, then total losses cannot exceed j z y P( C + C > 8000 C > 000) = dxdy = L NM z O = QP 3000 y 3000y = L NM z y 3000 dy 6 0 O QP = 5 Solution #: E b g = the margin as a function of the number of tickets sold. b g = 000 b00. g b g 3 = 0x cx 6x + 5xh Let M x M x x C x 3 = x + 6x 5x b g 0 = M x = 3x + x 5 0 = 3x x + 5 ± x = 6 6 = ± 3 = ± 3 = 353. or 047. x b g M x

51 Solution #: C I c = lim 00 + n n c = 00lim + n n c = 00 lim + n n = 00 e c F HG F HG L F NM HG I K J I K J n n I K J n c O QP c Solution #3: B b g b g b g P X = 0 = P 0, + P 0, = + = 4 P X = = P + P 3 = b g b, g b, g = 4 Solution #4: E Type Probability Minor 0.5 Moderate 0.4 Severe 0. Probability that both are minor = b05. gb 05. g = 0. 5 Probability that is minor and is moderate = = = 0.65 b gb g 5

52 Solution #5: E X j claim distribution d i R S T 0 780/ / 800 5/ µ = E X j = = = = = EdX ji 3 = 00 + = d ji d ji b g Var X = E X E X = σ d ji = Var X = Let n = 800 n T = X j E T = 800 E X j = 500 j= b g d i Find t so that: b L N = P T < t = P g b g F HG I KJ b g T nµ t < nµ nσ n σ O Q t 0, < nµ PMN P P N 0, < 645. n σ 645. = t n µ n σ b b g g t = 645. n σ + nµ = k E T = t t k = = = E T 500 b g 6

53 Solution #6: A For each policy k: B k is the random variable representing the benefit amount, given that there is a claim. I k is the claim indicator random variable. Ck = Ik Bk is the claim random variable. T = 3 C k k = is the total claims. b g = 6 b kg z b g 0 3 btg EbC k g E I k E B = y y dy = 3 E = = 3 = 8 k = 6 9 7

54 Solution #7: C Exponential Distribution: b g = < x/ λ Density f x e for 0 x. λ x Distribution Fbxg z = fbtgdt = e x 0 Today 05. = F 000 = Problem: b b 000 λ = ln Find F 000 = e g b = e g g x/ λ ln = 4 3 = 73. = = 034. b g / e x λ /λ. 8

55 Solution #8: C x b g b g z b g F x = P X x = f t dt z 0 x = 3t dt 0 3 x = t = x z 3 G = x x dx F HG F HG x x = 4 I K J = 4 = I KJ 0 Solution #9: B Find points where slope is zero, changing from positive to negative. Dec. 96 is the only one. Solution #0: E u Set gbug z = sin t dt g u = sin u Fundamental theorem of calculus. b g b g b g b g b g b g b g f x = g 3x f x = g 3x 3 = 3 g 3x = 3 sin b3xg f x = 8sin 3x cos 3x b g b g b g 9

56 Solution #: D b g b g b g L NM b g D t = 00 E t I t dd dt If t E de di = 00 I + E dt dt F HG I K JO QP = t 0 on June 30, 996, we are given: Ebt 0 g = Ibt 0 g = de t dt di t dt b 0g =. b 0g = dd t dt b 0g b gb gb g b g b g = =

57 Solution #: D Let: PbOg= probability of orthodontic work P F = probability of a filling b g = probability of an extraction P E We are given: P O g = / F = / 3 E = 3/ 4 b Eg = / 8 P O P O P F b g b g b g bo Eg = bog beg. Since O and F are independent, P O F = P O P F and since O and E are independent, P P P b g b g b g b g. bo Eg bog beg bo Eg beg beg beg = 4, PbEg =. bo Fg bog bfg bo Fg bfg bfg bfg = 6, PbFg = 3. F E = F + E F E = + b g b g b g b g 3 8 = 7 4 We are asked to find P F E = P F + P E P F E = P = P + P P = + P P 3 4 So P = P = P + P P = + P P 3 So P Then P P P P

58 Solution #3: D If f t t b g is the time to failure, we are given f btg = e µ = µ µ where 0. The expected payment =z 7 vbtg fbtgdt 0 z7 t 7 0. t 0 = e e dt 0 0 = 0 = 0 z t 0. t e e dt 0 z t e dt 0 F = I t 7 HG K Je = 7. 7 ce e h 3 = b = g

59 Solution #4: D Let: X = amount of a claim. Y = 0 or according to whether the policyholder is a good or bad driver. b g c h c h Apply the relationship Var X = E Var X Y + Var E X Y. We are given: b g c h c h b g c h c h P Y = 0 = 0. 6 E X Y = 0 = 400 Var X Y = 0 = 40, 000 P Y = = 0. 4 E X Y = = 000 Var X Y = = 50, 000 Then: E VarcX Yh = 0. 6b40, 000g b50, 000g = 4, 000 E Ec X Yh = 06. b400g b000g = 640 Var EcX Yh = 0. 6b g b g = 86, 400 Varb Xg = 4, , 400 = 0, 400 3

60 Solution #5: D y S 4 x From y = x y = Substitute x y = zzs yda = = x zy z + 0 y z z z yx y dxdy x= y+ x= y dy = y y + y dy 3 = y + y y dy 0 F HG F HG = y + y y = = c 3 4 y y = 0 by gby + g = 0 y = h I K J b g I K J O Q P y= y= 0 4

61 Solution #6: A b g = M t X t X t X t 3 3 E E E 3! dm dt E X E X dm dt O = QP t= 0 d M t dt d M t dt b g dmbtgo = dt Q P t= 0 9 t d F + e I F = dthg 3 HG 8 F + I 9HG K J = d M t O = dt Q P = b go Q d dt t= 0 b g 3e t F HG t= P = 3 c h e KJ = I t F HG I KJ t e 3 8 t 8 t + e t e 3e 3 KJ = + 3 F HG b g I K J +... t I tf e 3 8e KJ + + I HG 3 KJ F I HG K J = + = 8 7 Var X = E X E X = 3 = t e e

62 Solution #7: C Graph = cos t π t t k cos π 6 t t k cos π 6 + c t 6

63 Solution #8: A fbxg b g b g b g b g b g Let g x = f x be the function h x = g x = f x. Then in the given graph, looking for example, at x =, the function with y = 0 must be the derivative of the function with a relative maximum. Therefore, the graph of g x b g is y x σ Let a be the value on the x axis so that gbag = 0, 0 < a <.. Then: b g b g < < < g x 0 for 0 x a g x > 0 for a < x < 4 So: b g b g f x is decreasing for 0 = x < a. f x is increasing for a < x < 4. Only alternative A satisfies these conditions. 7

64 Solution #9: C Let: S = the sample space of customers. H = the subset of S consisting of high-risk customers. L = the subset of S consisting of low-risk customers. N y = the number of claims in year y for customer C. Then the problem is to find E N 98. c hb g c hb g E N = P c H N = P c L N = k e We have P N = 97 k = θ b g k! P P c c h e N97 = c H = = e N97 = c Lh = = H and L partition S, applying Bayes Theorem c 97 h b gb g P c H N = = P c L N = = P c H N = = 0. 7 θ where k = 0,,, 3 and θ is the mean of N b gb g b g c 97 h c 97 h And: b gb g b gb g E N 98 = = =

65 Solution #30: E Consider ln e x x x c h c h + 3x = ln e + 3x x c h c h c h + x as x 0 ln e + 3x 0 lim ln d x x e + 3x ln e + 3x so = lim dx + x x 0 d dx x Therefore: lim = lim+ x 0 e x = = 4 c + 3x e x + 3 x x lim ln x e x x e + 3 j 4 ce + 3xh = e = e + + x 0 x 0 h 9

66 Solution #3: A y x Expected payment is: zz 0 y zz 0 y x x + y dx dy = x + xy x dx dy 3 x = + x y x 3 3 y 3 y y y y y QP = = + y + y y + y 3 y + y y + y + y dy = z z 0 0 L NM e b gb g L NM O O y y y + y dy = QP = 3 = = 4 0 b g b g b g j 0

67 Solution #3: C C : x = t +, y = t, thus t = x + t > ± y x = ±, x x + = y or y = x 4x + y x C: x = t +, y = t + 7 t = + t = ± y 7 x x x + = ± 4 7, or = y 7 4 x x x 9 y = x + + 7, y = y = x x c x 4x + = x x + 9, 8x 6x + 8 = x x + 9 7x 4x = 0 x x 3 = 0 b gb g x 3 x + = 0 or x = 3, x = x = 3, y = 8 x =, y 8 4 c h h Solution #33: D Long run number of clients stabilizes at C 9000 C = C + 3 C C = C C C 3 n = 7, =, = = n lim C. n

68 Solution #34: C b g b g b g = E X = x 4 x 3 x 4 x dx + dx 9 9 x x x x dx z dx x x O x x P L L = N M b gb g b gb gq 4 36 = = = = z z 0 0 z 0 7 NM O QP 3

69 Solution #35: C Let H = time to death of the husband E E W = time to death of the wife J = time to payment after both husband and wife have died = max ( H, W) f ( h) = f w = H = E W = P max( H, W) < j j f Jb jg = 800 = F j H G I K J F dhdw H G I K J = E J z 40 3 j = dj = = f H b 0 W b j, hg = b g R S T 40 0 J, H W H J, H z 0 h g zj zj 0 j < h f b jg f bhg = j > h 40 h FW bhg f Hbhg = j = h 40 z z z jh h h h h = djdh + h 40 dh = + dh = = cov J, H = E J, H E J E H = =

70 Solution #36: C Average value = value of integral width of interval z t cosct h F = = H G I K JF H G I K J 0 F = H G I K JF H G I K J sin = 4 8 e z0 dt t cos t c h sin t j 0 b g c h Solution #37: E Distribution is: P (0,0,0) = 0 P (0,0,) = k P (0,0,) = k P (0,,0) = k P (0,,) = 3k P (0,,) = 4k P (0,,0) = 4k P (0,,) = 5k P (0,,) = 6k P (,0,0) = k P (,0,) = k P (,0,) = 3k P (,,0) = 3k P (,,) = 4k P (,,) = 5k P (,,0) = 5k P (,,) = 6k P (,,) = 7k K = /63 P (0,0/X = 0) P (,0/X = 0) = 4/7 P (0,/X =0) = /7 P (,/X = 0) = 5/7 P (0,/X = 0) = /7 P (,/X =0) = 6/7 P (,0/X = 0) = /7 P (,/X = 0) = 3/7 P (,/x = 0) = 4/7 Expected number of unreimbursed equals ( accident) (9/7) + ( accidents) (6/7) = 9/7 + /7 = /7 = 7/9. 4

71 Solution #38: B First dividend paid = $.00 Second dividend paid = (0.95) (.00) Third dividend paid = (0.95) (.00) th dividend paid = (0.95) 39 (.00) Total dividends = (0.95) + + (0.95) 39 Geometric series; S 40 b g 0 95 = =.. = = $ Solution #39: D Mode is max of probable distribution f x b g = 4x x 9 9. Maximize f bxg, so take derivative: 4 x f bxg = = x = 4 x = 5

72 Solution #40: E Let X = number of passengers that show for a flight. b g b g You want to know P X = 3 and P X = 3. All passengers are independent. Binomial distribution: b g b g F n P X = x = P P H G I xk J x n x P = 90 P = 00. n = 3 F P X = = H G 3I 3 3 b 3g K J b0. 90g b00. g = b3gb0. 90g 00. = 0. F 3 PbX = g = H G 3I 3 3 K J b090. g b00. g = P ( more show up than seats) = =