北美精算师 exam P 2003 真题
2003年美赛题(AB)
PROBLEM A: The Stunt PersonAn exciting action scene in a movie is going to be filmed, and you are the stunt coordinator! A stunt person on a motorcycle will jump over an elephant and land in a pile of cardboard boxes to cushion their fall. You need to protect the stunt person, and also use relatively few cardboard boxes (lower cost, not seen by camera, etc.).Your job is to:∙determine what size boxes to use∙determine how many boxes to use∙determine how the boxes will be stacked∙determine if any modifications to the boxes would help∙generalize to different combined weights (stunt person & motorcycle) and different jump heightsNote that, in "Tomorrow Never Dies", the James Bond character on a motorcycle jumps over a helicopter.PROBLEM B: Gamma Knife Treatment PlanningStereotactic radiosurgery delivers a single high dose of ionizing radiation to a radiographically well-defined, small intracranial 3D brain tumor without delivering any significant fraction of the prescribed dose to the surrounding brain tissue. Three modalities are commonly used in this area; they are the gamma knife unit, heavy charged particle beams, and external high-energy photon beams from linear accelerators.The gamma knife unit delivers a single high dose of ionizing radiation emanating from 201 cobalt-60 unit sources through a heavy helmet. All 201 beams simultaneously intersect at the isocenter, resulting in a spherical (approximately) dose distribution at the effective dose levels. Irradiating the isocenter to deliver dose is termed a “shot.” Shots can be represented as different spheres. Four interchangeable outer collimator helmets with beam channel diameters of 4, 8, 14, and 18 mm are available for irradiating different size volumes. For a target volume larger than one shot, multiple shots can be used to cover the entire target. In practice, most targetvolumes are treated with 1 to 15 shots. The target volume is a bounded, three-dimensional digital image that usually consists of millions of points.The goal of radiosurgery is to deplete tumor cells while preserving normal structures. Since there are physical limitations and biological uncertainties involved in this therapy process, a treatment plan needs to account for all those limitations and uncertainties. In general, an optimal treatment plan is designed to meet the following requirements.1.Minimize the dose gradient across the target volume.2.Match specified isodose contours to the target volumes.3.Match specified dose-volume constraints of the target and critical organ.4.Minimize the integral dose to the entire volume of normal tissues ororgans.5.Constrain dose to specified normal tissue points below tolerance doses.6.Minimize the maximum dose to critical volumes.In gamma unit treatment planning, we have the following constraints:1.Prohibit shots from protruding outside the target.2.Prohibit shots from overlapping (to avoid hot spots).3.Cover the target volume with effective dosage as much as possible. But atleast 90% of the target volume must be covered by shots.e as few shots as possible.Your tasks are to formulate the optimal treatment planning for a gamma knife unit as a sphere-packing problem, and propose an algorithm to find a solution. While designing your algorithm, you must keep in mind that your algorithm must be reasonably efficient.。
2003年6月PRETCO考试B级真题试卷(含答案)
样题2003年6月PRETCO考试(kǎoshì)B级真题试卷（含答案）３级考试(kǎoshì)试卷是全英的，一定要看清题目！Part I Listening Comprehension ( 15 minutes)【听力(tīnglì)部分】Directions: This part is to test your listening ability. It consists of 3 sections. Section ADirections: This section is to test your ability to understand short dialogues. There are 5 recorded dialogues in it. After each dialogue, there is a recorded question. Both the dialogues and questions will be spoken only once. When you hear a question, you should decide on the correct answer from the 4 choices marked A) , B) , C) and D) given in your test paper. Then you should mark the corresponding letter on the Answer Sheet with a single line through the center.Example: You will hear:You will read: A) New York City.B) An evening party.C) An air trip.D) The man’s job.From the dialogue we learn that the man is to take a flight to New York. Therefore, C) An air trip is the correct answer. You should mark C) on the Answer Sheet with a single line through the center.[A] [B] [C] [D]Now the test will begin1. A) BoatingB) WalkingC) RunningD) Driving2. A) She allows the man to smoke in the room.B) She will open the window for the man.C) She doesn't like the man to smoke in the room.D) She doesn't mind the man's opening the window.3. A) .B) In a bookstore.C) In a library.D) In a hotel.4. A) Write a reporB) Type a reportC) Check a report.D) Read a report.5. A) A waitress.B) A salesgirlC) A housewifeD) A receptionist.Section BDirections: This section is to test your ability to understand short conversations. There are 2 recorded conversations in it. After each conversation, there are some recorded questions. Both the conversations and questions will be spoken only two times. When you hear a question, you should decide on the correct answer from the 4 choices markedA) ,B) ,C) andD) given in your test paper. Then you should mark the corresponding letter on the Answer Sheet with a single line through the center.6. A) OneB) TwoC) ThreeD) Four7. A) GrassB) VegetablesC) RosesD) Trees8. A) In a small yard.B) In a big garden.C) On the roof.D) In the greenhouse.9. A) $70B) $20C) $10D) $1710. A) He ties.B) He didn't have enough money.C) He might get some from his children.D) He had to save money for his birthday.Section CDirections: This section is to test your ability to comprehend short passages. You will hear a recorded passage. After that you will hear five questions. Both the passage and the questions will be read two times. When you hear a question, you should complete the answer to it with a word or a short phrase (in not more than 3 words). The questions and incomplete answers are printed in your test paper. You should write your answer on the Answer Sheet correspondingly. Now the passage will begin.【填空(tiánkòng)，三字以内】11. How was the traveler when he got to the country inn?He was_______________________.12. Why couldn't the traveler get near the fire?Because the room was very____________________.13. Why did the people run out to see the horse?Because they were curious to see a horse eating_______________.14. What did the traveler went outside?He sat down beside the fire and______________.15. Who ate the fish in the end?_____________________________.答案(dá àn)：11. wet and cold12. crowned13. fish14. warmed himself15. The travelerPart II Vocabulary and Structure（15 minutes）【词汇(cíhuì)与语法】Directions: This part is to test your ability to use words and phrases correctly to construct meaningful and grammatically correct sentences. It consists of 2 sections.Section A 【单项选择题】16. —“How about having dinner at Sun Restaurant?” —“ It _________ good.”A) smellsB) looksC) soundsD) appears17. Seldom ______ my boss in such good mood（心情(xīnqíng)）since I came to work in this company .A) I sawB) I have beenC) have I seenD) do I see18. You’d better _______ the whole article at once.A) copyB) copyingC) to copyD) copied19. The machine much noise _______ we have it repaired.A) whenB) becauseC) ifD) unless20. The manager told us never to _______ till tomorrow what we can do today.A) come upB) put offC) turn onD) give out21. The children are getting more and more excited when Christmas is________ near.A) drawingB) joiningC) takingD) operating22. The old man has two daughters, ________ are doctors.A) both of themB) both of whomC) both whoD) they both23. If you travel in a foreign country, a tour ________ may save you a lot of trouble.A) directorB) helperC) guideD) assistant24. Dinner wil l be ready _______. Let’s go and wash our hands.A) at allB) at leastC) just nowD) right away25. If you ________ smoking and drinking, soon.A) gave upB) give upC) had given upD) will give upSection B【变形(biàn xíng)填空】Directions: There are also 10 incomplete statements here. You should fill in each blank with the proper form of the word given in the brackets. Write the word or words in the corresponding space on the Answer Sheet.26. John is the (clever)________ student I have ever taught.27. His suggestions turned out to be very(effect) ________ in the improvement of our production.28. Sixty people(employ) ________in this big factory last year.29. It is difficult for a(foreign) ________ to learn Chinese.30. Both of the twin brothers(be) ________capable of doing technical work at present.31. When Jenny came to Britain, she had to get used to(drive) ________on the left.32. She is well-known for her excellent(achieve) ________in her career.33. The chairman speaker(limit) ________ himself to fifteen minutes.34. He is (confidence) ________ even though he has failed several times.35. No student is supposed(spend) ________so much money in school in a week.答案(dá àn)：16-26. CAABD BCADB26. illness 27. successful 28. painting 29. drawing 30. have spent 31. crossing 32. highly 33. walk 34. talking 35. have collectedPart III Reading Comprehension ( 40 minutes)【阅读(yuèdú)理解】Directions: This part is to test your reading ability. There are 5 tasks for you to fulfill. You should read the reading materials carefully and do the tasks as you are instructed.Task 1People today are still talking about the generation gap（代沟(dài ɡōu)）. Some parents complain that their children do not show them proper respect, while children complain that their parents do not understand them at all. What has gone wrong? Why has the generation gap appeared?One important cause is that young people want tostyle. In more traditional societies, when children grow up, they are expected to live in the same area as their parents, to many people that their parents like, and often to continue the family occupation.Parents often expect their children to do better than they do, to find better jobs, to make more money, and to do all the things that they were unable to do. Often, however, the high wishes that parents place on their children are another cause of the generation gap.Finally, the high speed of social changes deepens the gap. In a traditionalculture, people are valued for their wisdom, but in our society today the knowledge of lifetime may be out of use overnight（隔夜(gé〃yè)）.36. According to the passage, children today expect their parents to_________.A) give them more independenceB) choose a good job for themC)D) make more money37. Parents often hope that their children will ________.A) make as much money as they doB) be more successful than they areC) choose jobs according to their own willD) avoid doing what their parents can’t do38. The generation gap has become wider than before because of ________.A) the increasing dependence of children on parentsB) the influence of traditional culture on childrenC) the rapid changes of modern societyD) the missing of lifelong occupation39. In today’s society, the knowledge of a lifetime _________.A) is still very much valuedB) becomes out of date quicklyC) is essential for continuing family occupationD) helps the young generation to find a better job40. A proper title for this passage would be ____________.A) Parents’ Viewpoints On Generation GapB) Relationship Between Family MembersC) Generation Gap Between the Young and the OldD) Difference Between Traditional Culture and Modern KnowledgeTask 2Directions: This task is the same as Task 1. The 5 questions or unfinished statements are numbered 41 to 45.For some employers, the policy of lifelong employment is particularly important because it means that they can put money and effort into their staff（职员(zhíyuán)）training and make them loyal to the company. Whatthey do is to select young people who have potential（潜能）and who can be trained. They then give the young people the kinds of skills that will make them suitable employees for the company. In other words, they adjust their training to their particular needs.One recently employed graduate says that she is receiving a great deal ofvaluable training from the company. “This means that I will be a loyal employee,” she says, “And it also means that the company will want to keep me. I am an important investment for them. So the policy is a good onebec ause it benefits both the employer and the employee.”Recently, however, attitudes towards lifelong employment are beginning to change. Employees are slowly beginning to accept the idea that lifelong employment is not always in their best interest and that changing firms can have career advantages.41. The purpose of lifelong employment is to __________.A) adjust the needs of the company to its employeesB) make employees loyal to their companyC) select the best skilled young employeesD) keep the skilled staff satisfied42. By training its employees, a company can make them _________.A) do their work more easilyB) more interested in their workC) willing to invest money into the countryD) posses the necessary qualities for the job43. Talking about the training she has received, a recently employed graduate has the view that _________.A) it is still well-received by all the staff members todayB) it is valuable to the employer and the employeesC) it is helpful for attracting young employeesD) it is both useful and interesting44. Attitudes towards lifelong employment are changing because _________.A) job changes have career advantages.B) it’s boring to work in only one company.C) only the employer benefits from such employment.D) stable employment opportunities45. The passage is mainly about ___________.A) lifelong training of employeesB) policies of lifelong employmentC) attitudes towards lifelong employmentD) employers’ interest in lifelong employment答案(dáàn)：ＤＢＣＡＡＡＢＣＤＡTask 3【根据(gēnjù)短文填词，三字以内】Directions: The following is an advertisement. After reading it, you should complete the information by filling in the blanks marker 46 to 50 in not more than 3 words in the table below.Over a million people visit Hawaii（夏威夷）each year because of its beautiful weather and wonderful scenery（景色(jǐngsè)）. The Hawaiian islands have very mild temperatures. For example, August, the hottest month, average 78.4oF, while February, the coldest month, averages 71.9oF. In addition, the rainfall in Hawaii is not very heavy because mountains on the north of each island stop incoming storms; for instance, Honolulu averages only 23 inches of rain per year. This beautiful weather helps tourists to enjoy Hawaii’s wonderful natural scenery, from mountain waterfa lls to fields of flowers and fruits. And Hawaii’s beautiful beaches are everywhere — from the lovely Kona coast beaches on the large island of Hawaii to Waikiki Beach on Oahu. Warm sunshine and beautiful beaches — itmany people visit Hawaii each year. Are you going to join us? Don’t miss the chance!HawaiiFamous for its: 1) (46)______and2)(47) ______Average Temperature: ranging from (48) ______to 78.4oFAnnual rainfall in Honolulu:(49) ______Attractions for tourists:(50) ______and beautiful beaches答案(dá àn)：46. Brain store 47. two business days 48. phone 49.fax 50. 90 daysTask 4【单词(dāncí)搭配】Directions: The following is a list of terms frequently used in medical services. After reading it, you are required to find the items equivalent to (与……等同)those given in Chinese in the list below. Then you should put the corresponding letters in brackets on the Answer Sheet, numbered 51 through 55.A— answer phoneB— burglar alarmC— date-stampD— electronic display materialE— headed paperF— office information systemG— shorthandH— annual reportI— registered deliveryJ— office automationK— time sheetL— computer packageM— handbookN— waste basketO— card-indexP— blueprintExample: (B) 防盗报警器(G) 速记51. （）年度报告（）电子显示材料52. （）计算机程序包（）邮戳日期53. （）办公自动化（）废纸篓54. （）印有信头的信纸（）录音电话55. （）挂号邮件（）办公室信息系统答案：51. S N 52. G P 53. F A 54. U C 55. M DTask 5【根据短文填词，三字以内】APPOINTMENTSYOUNG Italian girl, student, speaks English and French, seeks post in a school or family, giving lessons or looking after children. — Write Box L. 1367, The Daily—, London, E.C. 4.YOUNG man, once an officer, office work, is willing to go to any part of the world and to do anything legal; speaks several languages; drives all makes of cars; exciting work more important than salary. —Write Box F. 238, The Daily —, London, E. C.MARRIED couple wanted Gardener; country house 2 miles from Oxford, good bus service; family three adults, five children; wages ￡ 9; comfortable rooms with central heating. — Write Box S, 754. The Daily —, London, E. C.56. What kind of work is suitable for the Italian girl?Teach classes or___________________.57. What foreign languages does the Italian girl know?She knows___________________.58. Why is the young man tired of his office work?Because it is ___________________.59. What does the young man think of salary?He thinks that salary is ___________________than exciting work.60. What kind of helper are the find?They are trying to find___________________.答案：56. select your car 57. Contact Information Form 58. confirm your selection 59. confirmed 60. Customer Care RepresentivePart IV Translation —— English into Chinese ( 25 minutes)【翻译、英译汉】61. This is rather for your for you.A) 这是你父亲的决定而不是你的决定。
2003年美赛题(B)
PROBLEM B: Gamma Knife Treatment PlanningStereotactic radiosurgery (立体定向放射外科)delivers a single high dose of （大剂量）ionizing radiation（电离辐射）to a radiographically （放射照相的）well-defined（边界清晰的）, small intracranial （颅内）3D brain tumor（脑肿瘤）without delivering any significant fraction of the prescribed dose to the surrounding brain tissue（周围的脑组织）. Three modalities（形式）are commonly used in this area; they are the gamma knife unit（伽马刀单元）, heavy charged particle beams（高能粒子束）, and external high-energy photon beams（外部高能光子束）from linear accelerators（直线加速器）.The gamma knife unit delivers a single high dose of ionizing radiation emanating （放射）from 201 cobalt-60 unit sources through a heavy helmet（沉重的钢盔）. All 201 beams simultaneously同时intersect 交叉at the isocenter等深点, resulting in a spherical球形的(approximately近似) dose剂量distribution at分布在the effective dose levels有效剂量领域. Irradiating照射the isocenter 等中心点to deliver dose is termed a “shot.” Shots can be represented as different spheres球体. Four interchangeable outer collimator helmets四个可交替的外部瞄准仪头盔with beam channel diameters射线直径of 4, 8, 14, and 18 mm are available for irradiating 照射different size volumes体积. For a target volume目标体积larger than one shot, multiple shots can be used to cover the entire target. In practice, most target volumes are treated with 1 to 15 shots. The target volume is a bounded有界限的, three-dimensional digital image三维立体的数字图像that usually consists of millions of points.The goal of radiosurgery放射外科is to deplete tumor cells消除肿瘤细胞while preserving normal structures保留正常的构造. Since there are physical limitations体力限制and biological uncertainties生物学的不确定性involved in this therapy process治疗方法, a treatment plan needs to account for all those limitations and uncertainties. In general, an optimal treatment plan is designed to meet the following requirements.一项手术计划需要满足以下要求1.Minimize the dose gradient across the target volume. 使穿透目标体积的辐射量最小2.Match specified isodose contours to the target volumes. 目标体积应满足规定的等剂量线3.Match specified dose-volume constraints of the target and critical organ.考虑对目标肿瘤和关键器官的规定辐射量4.Minimize the integral dose to the entire volume of normal tissues ororgans. 使正常组织和器官受到的辐射量最小5.Constrain dose to specified normal tissue points below tolerance doses.使正常组织所受辐射量低于耐受剂量6.Minimize the maximum dose to critical volumes. 使临界体积的最大辐射量最小化In gamma unit treatment planning, we have the following constraints: 伽马刀治疗包括以下过程1.Prohibit shots from protruding outside the target. 不能使辐射照射到目标外边2.Prohibit shots from overlapping (to avoid hot spots). 避免辐射重叠3.Cover the target volume with effective dosage as much as possible. But atleast 90% of the target volume must be covered by shots. 至少要杀死90% 的肿瘤e as few shots as possible. 尽可能少的照射Your tasks are to formulate起草the optimal treatment planning for a gamma knife unit as a sphere-packing球状填充problem, and propose an algorithm 提出一种算法to find a solution. While designing your algorithm, you must keep in mind that your algorithm must be reasonably efficient.立体定位放射外科, 用单一高剂量离子化射束在X光机精确界定下照射颅内的一个小的3D 脑瘤, 与此同时, 并没有处方剂量的任何显著份额伤及周边的脑组织. 在这个领域中,一般有三种形式的射束可以采用，分别是Gamma刀单元, 带电重粒子射束, 以及来自直线加速器的外用高能光子束. Gamma刀单元具备的单一高剂量离子化射束, 是201个钴-60单位源通过厚重的盔状物发射出来的。
北美精算考试P练习题
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for −2 ≤ x ≤ 4 otherwise
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6. The time of failure of a component in an electronic device has an exponential distribution with a median of four hours. Calculate the probability that the component will work without failing for at least five hours. (a) 0.07 (b) 0.38 (c) 0.42 (d) 0.29 (e) 0.57
where c is a constant. Calculate the moment generating function of X .
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2. An insurance policy pays for a random loss X subject to a deductible of C , where 0 < C < 1. The loss amount is modeled as a continuous random variable with density function f (x) = 2x for 0 < x < 1 0 otherwise
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Quiz 3 Duration: 50 min Total: 10 Marks 1. A random variable X has thction: for x < 1 0 x2 −2x+2 F (x) = for 1 ≤ x < 2 2 1 for x ≥ 2 Find the variance of X. (a) (b) (c) (d) (e)
AMC 美国数学竞赛 2003 AMC 10B 试题及答案解析
2003 AMC 10B1、Which of the following is the same asSolution2、Al gets the disease algebritis and must take one green pill and one pink pill each day for two weeks. A green pill costs more than a pink pill, and Al’s pills cost a total of for the two weeks. How much does one green pill cost?Solution3、The sum of 5 consecutive even integers is less than the sum of the first consecutive odd counting numbers. What is the smallest of the even integers?Solution4、Rose fills each of the rectangular regions of her rectangular flower bed with a different type of flower. The lengths, in feet, of the rectangular regions in her flower bed are as shown in the figure. She plants one flower per square foot in each region. Asters cost 1 each, begonias 1.50 each, cannas 2 each, dahlias 2.50 each, and Easter lilies 3 each. What is the least possible cost, in dollars, for her garden? Solution5、Moe uses a mower to cut his rectangular -foot by -foot lawn. The swath he cuts is inches wide, but he overlaps each cut by inches to make sure that no grass is missed. He walks at the rate offeet per hour while pushing the mower. Which of the following is closest to the number of hours it will take Moe to mow his lawn?Solution.6、Many television screens are rectangles that are measured by the length of their diagonals. The ratio of the horizontal length to the height in a standard television screen is . The horizontal length of a “-inch” television screen is closest, in inches, to which of the following?Solution7、The symbolism denotes the largest integer not exceeding . For example. , and . ComputeSolution.8、The second and fourth terms of a geometric sequence are and . Which of the following is a possible first term?Solution9、Find the value of that satisfies the equationSolution10、Nebraska, the home of the AMC, changed its license plate scheme. Each old license plate consisted of a letter followed by four digits. Each new license plate consists of three letters followed by three digits. By how many times is the number of possible license plates increased? Solution11、A line with slope intersects a line with slope at the point . What is the distance between the -intercepts of these two lines?Solution12、Al, Betty, and Clare split among them to be invested in different ways. Each begins with a different amount. At the end of one year they have a total of . Betty and Clare have both doubled their money, whereas Al has managed to lose . What was Al’s original portion?Solution.13、Let denote the sum of the digits of the positive integer . For example, and . For how many two-digit values of is ?Solution14、Given that , where both and are positive integers, find the smallest possible value for .Solution15、There are players in a singles tennis tournament. The tournament is single elimination, meaning that a player who loses a match is eliminated. In the first round, the strongest players are given a bye, and the remaining players are paired off to play. After each round, the remaining players play in the next round. The match continues until only one player remains unbeaten. The total number of matches played isSolution16、A restaurant offers three desserts, and exactly twice as many appetizers as main courses. A dinner consists of an appetizer, a main course, and a dessert. What is the least number of main courses that the restaurant should offer so that a customer could have a different dinner each night in the year ?Solution.17、An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies of the volume of the frozen ice cream. What is the ratio of the cone’s height to its radius?Solution18、What is the largest integer that is a divisor offor all positive even integers ?Solution19、Three semicircles of radius are constructed on diameter of a semicircle of radius . The centers of the small semicircles divide into four line segments of equal length, as shown. What is the area of the shaded region that lies within the large semicircle but outside the smaller semicircles?Solution20、In rectangle , and . Points and are onso that and . Lines and intersect at . Find the area of .Solution21、A bag contains two red beads and two green beads. You reach into the bag and pull out a bead, replacing it with a red bead regardless of the color you pulled out. What is the probability that all beads in the bag are red after three such replacements?Solution22、A clock chimes once at minutes past each hour and chimes on the hour according to the hour. For example, at 1 PM there is one chime and at noon and midnight there are twelve chimes. Starting at 11:15 AM on February , , on what date will the chime occur?Solution23、A regular octagon has an area of one square unit. What is the area of the rectangle ?Solution24、The first four terms in an arithmetic sequence are , , , and , in that order. What is the fifth term?Solution25、How many distinct four-digit numbers are divisible by and have as their last two digits?Solution。
2003年5月北美精算第四门考试试题
Course 4Fall 2003 Society of Actuaries**BEGINNING OF EXAMINATION** 1. You are given the following information about a stationary AR(2) model:=.(i) ρ105(ii) ρ201=.Determine φ2.(A) –0.2(B) 0.1(C) 0.4(D) 0.7(E) 1.0(i) Losses follow a loglogistic distribution with cumulative distribution function:F x x x b g b g b g =+//θθγγ1(ii)The sample of losses is:10 35 80 86 90 120 158 180 200 210 1500Calculate the estimate of θ by percentile matching, using the 40th and 80th empirically smoothed percentile estimates.(A) Less than 77(B) At least 77, but less than 87(C) At least 87, but less than 97(D) At least 97, but less than 107(E) At least 107(i) The number of claims has a Poisson distribution.(ii) Claim sizes have a Pareto distribution with parameters θ=0.5 and α=6.(iii) The number of claims and claim sizes are independent.(iv) The observed pure premium should be within 2% of the expected pure premium 90% of the time.Determine the expected number of claims needed for full credibility.(A) Less than 7,000(B) At least 7,000, but less than 10,000(C) At least 10,000, but less than 13,000(D) At least 13,000, but less than 16,000(E) At least 16,0004. You study five lives to estimate the time from the onset of a disease to death. The times todeath are:2 3 3 3 7Using a triangular kernel with bandwidth 2, estimate the density function at 2.5.(A) 8/40(B) 12/40(C) 14/40(D) 16/40(E) 17/405. For the model i i i Y X αβε=++, where 1,2,...,10i =, you are given:(i) X i i =R S T1, if the th individual belongs to a specified group 0, otherwise(ii) 40 percent of the individuals belong to the specified group.(iii) The least squares estimate of β is β=4.(iv) ()2ˆˆ92i i Y X αβ−−=∑Calculate the t statistic for testing H 00:β=.(A) 0.9(B) 1.2(C) 1.5(D) 1.8(E) 2.1(i) Losses follow a Single-parameter Pareto distribution with density function:()()1,1f x x xαα+=>, 0 < α < ∞ (ii) A random sample of size five produced three losses with values 3, 6 and 14, and twolosses exceeding 25.Determine the maximum likelihood estimate of α.(A) 0.25(B) 0.30(C) 0.34(D) 0.38(E) 0.42(i) The annual number of claims for a policyholder has a binomial distribution withprobability function:()()221x x p x q q q x −⎛⎞=−⎜⎟⎝⎠, x = 0, 1, 2(ii) The prior distribution is:()34,01q q q π=<<This policyholder had one claim in each of Years 1 and 2.Determine the Bayesian estimate of the number of claims in Year 3.(A) Less than 1.1(B) At least 1.1, but less than 1.3(C) At least 1.3, but less than 1.5(D) At least 1.5, but less than 1.7(E) At least 1.78. For a sample of dental claims 1210,,...,x x x , you are given:(i) 23860 and 4,574,802i i x x ==∑∑(ii) Claims are assumed to follow a lognormal distribution with parameters µ and σ.(iii)µ and σ are estimated using the method of moments.Calculate ∧ for the fitted distribution.(A) Less than 125(B) At least 125, but less than 175(C) At least 175, but less than 225(D) At least 225, but less than 275(E) At least 2759. You are given:(i)Y tij is the loss for the j th insured in the i th group in Year t . (ii)ti Y is the mean loss in the i th group in Year t . (iii)X j i j i ij =R S T0, if the th insured is in the first group (=1)1, if the th insured is in the second group (=2) (iv)21ij ij ij ij Y Y X δφθε=+++, where 1,2i = and 1,2,...,j n = (v)Y Y Y Y 2122111230374041====,,, (vi) ˆ0.75φ=Determine the least-squares estimate of θ.(A) 5.25(B) 5.50(C) 5.75(D) 6.00(E) 6.2510. Two independent samples are combined yielding the following ranks:Sample I: 1, 2, 3, 4, 7, 9, 13, 19, 20Sample II: 5, 6, 8, 10, 11, 12, 14, 15, 16, 17, 18You test the null hypothesis that the two samples are from the same continuous distribution.The variance of the rank sum statistic is:()112n m n m ++Using the classical approximation for the two-tailed rank sum test, determine the p -value.(A) 0.015(B) 0.021(C) 0.105(D) 0.210(E) 0.420(i) Claim counts follow a Poisson distribution with mean θ. (ii) Claim sizes follow an exponential distribution with mean 10θ. (iii) Claim counts and claim sizes are independent, given θ. (iv) The prior distribution has probability density function:b g=5, θ>1πθθCalculate Bühlmann’s k for aggregate losses.(A) Less than 1(B) At least 1, but less than 2(C) At least 2, but less than 3(D) At least 3, but less than 4(E) At least 4(i) A survival study uses a Cox proportional hazards model with covariates Z 1 and Z 2,each taking the value 0 or 1.(ii) The maximum partial likelihood estimate of the coefficient vector is:, .,.ββ12071020e j b g=(iii) The baseline survival function at time t 0 is estimated as .S t 0065b g =.Estimate S t 0b gfor a subject with covariate values 121Z Z ==.(A) 0.34(B) 0.49(C) 0.65(D) 0.74(E) 0.84(i) Z 1 and Z 2 are independent N(0,1) random variables.(ii) a , b , c , d , e , f are constants.(iii) Y a bZ cZ X d eZ f Z =++=++1212 andDetermine ()E Y X .(A) a(B) ()()a b c X d ++−(C) a be cf X d ++−b gb g(D) a be cf e f +++g d /22(E) a be cf e f X d +++−g d g/22(i) Losses on a company’s insurance policies follow a Pareto distribution with probabilitydensity function:()(),0f x x x θθθ=<<∞+(ii) For half of the company’s policies θ=1, while for the other half θ=3.For a randomly selected policy, losses in Year 1 were 5.Determine the posterior probability that losses for this policy in Year 2 will exceed 8.(A) 0.11(B) 0.15(C) 0.19(D) 0.21(E) 0.2715. You are given total claims for two policyholders:Year1 2 3 4PolicyholderX 730 800 650 700Y 655 650 625 750Using the nonparametric empirical Bayes method, determine the Bühlmann credibilitypremium for Policyholder Y.(A) 655(B) 670(C) 687(D) 703(E) 71916. A particular line of business has three types of claims. The historical probability and thenumber of claims for each type in the current year are:Type HistoricalProbabilityNumber of Claimsin Current YearA 0.2744 112B 0.3512 180C 0.3744 138You test the null hypothesis that the probability of each type of claim in the current year is the same as the historical probability.Calculate the chi-square goodness-of-fit test statistic.(A) Less than 9(B) At least 9, but less than 10(C) At least 10, but less than 11(D) At least 11, but less than 12(E) At least 1217. Which of the following is false?(A) If the characteristics of a stochastic process change over time, then the process isnonstationary.(B) Representing a nonstationary time series by a simple algebraic model is often difficult.(C) Differences of a homogeneous nonstationary time series will always be nonstationary.(D) If a time series is stationary, then its mean, variance and, for any lag k, covariancemust also be stationary.(E) If the autocorrelation function for a time series is zero (or close to zero) for all lagsk>0, then no model can provide useful minimum mean-square-error forecasts offuture values other than the mean.18. The information associated with the maximum likelihood estimator of a parameter θ is 4n,where n is the number of observations.Calculate the asymptotic variance of the maximum likelihood estimator of 2θ.(A) 12n(B) 1n(C) 4n(D) 8n(E) 16n19. You are given:(i) The probability that an insured will have at least one loss during any year is p.(ii) The prior distribution for p is uniform on []0,0.5.(iii) An insured is observed for 8 years and has at least one loss every year.Determine the posterior probability that the insured will have at least one loss during Year 9.(A) 0.450(B) 0.475(C) 0.500(D) 0.550(E) 0.62520. At the beginning of each of the past 5 years, an actuary has forecast the annual claims for agroup of insureds. The table below shows the forecasts (X) and the actual claims (Y). Atwo-variable linear regression model is used to analyze the data.t X t Y t1 475 2542 254 4633 463 5154 515 5675 567 605You are given:(i) The null hypothesis is0:0,1Hαβ==.(ii) The unrestricted model fit yields ESS = 69,843.Which of the following is true regarding the F test of the null hypothesis?(A) The null hypothesis is not rejected at the 0.05 significance level.(B) The null hypothesis is rejected at the 0.05 significance level, but not at the 0.01 level.(C) The numerator has 3 degrees of freedom.(D) The denominator has 2 degrees of freedom.(E) TheF statistic cannot be determined from the information given.21-22. Use the following information for questions 21 and 22.For a survival study with censored and truncated data, you are given:Time (t) Number at Riskat Time t Failures at Time t1 30 52 27 93 32 64 25 55 20 4 21. The probability of failing at or before Time 4, given survival past Time 1, is31q.Calculate Greenwood’s approximation of the variance of 31 q.(A) 0.0067(B) 0.0073(C) 0.0080(D) 0.0091(E) 0.010521-22. (Repeated for convenience) Use the following information for questions 21 and 22.For a survival study with censored and truncated data, you are given:Time (t) Number at Riskat Time t Failures at Time t1 30 52 27 93 32 64 25 55 20 4 22. Calculate the 95% log-transformed confidence interval for H3b g, based on the Nelson-Aalenestimate.(A) (0.30,0.89)(B) (0.31,1.54)(C) (0.39,0.99)(D) (0.44,1.07)(E) (0.56,0.79)(i) Two risks have the following severity distributions:Amount of Claim Probability of ClaimAmount for Risk 1Probability of ClaimAmount for Risk 2250 0.5 0.72,500 0.3 0.260,000 0.2 0.1(ii) Risk 1 is twice as likely to be observed as Risk 2.A claim of 250 is observed.Determine the Bühlmann credibility estimate of the second claim amount from the same risk.(A) Less than 10,200(B) At least 10,200, but less than 10,400(C) At least 10,400, but less than 10,600(D) At least 10,600, but less than 10,800(E) At least 10,800(i) A sample x x x 1210,,,… is drawn from a distribution with probability density function:1211exp()exp(), 0[]x x x θθσσ−+−<<∞(ii)θσ>(iii) x x i i ==∑∑15050002 andEstimate θ by matching the first two sample moments to the corresponding population quantities.(A) 9(B) 10(C) 15(D) 20(E) 2125. You are given the following time-series model:115.028.0−−−++=t t t t y y εεWhich of the following statements about this model is false?(A) 10.4ρ=(B) 1,2,3,4,....k k ρρ<=(C) The model is ARMA(1,1).(D) The model is stationary.(E) The mean, µ, is 2.26. You are given a sample of two values, 5 and 9.You estimate Var(X ) using the estimator g (X 1, X 2) = 21().2i X X −∑Determine the bootstrap approximation to the mean square error of g .(A) 1(B) 2(C) 4(D) 8(E) 1627. You are given:(i) The number of claims incurred in a month by any insured has a Poisson distributionwith mean λ.(ii) The claim frequencies of different insureds are independent.(iii) The prior distribution is gamma with probability density function:()()6100100120efλλλλ−=(iv) Month Number of Insureds NumberofClaims1 100 62 150 83 200 114 300 ?Determine the Bühlmann-Straub credibility estimate of the number of claims in Month 4.(A) 16.7(B) 16.9(C) 17.3(D) 17.6(E) 18.028. You fit a Pareto distribution to a sample of 200 claim amounts and use the likelihood ratio testto test the hypothesis that 1.5α= and 7.8θ=.You are given:(i) The maximum likelihood estimates are α= 1.4 and θ = 7.6.(ii) The natural logarithm of the likelihood function evaluated at the maximum likelihoodestimates is −817.92.(iii) ()ln 7.8607.64i x +=∑Determine the result of the test.(A) Reject at the 0.005 significance level.(B) Reject at the 0.010 significance level, but not at the 0.005 level.(C) Reject at the 0.025 significance level, but not at the 0.010 level.(D) Reject at the 0.050 significance level, but not at the 0.025 level.(E) Do not reject at the 0.050 significance level.29. You are given:(i) The model is Y X i i i =+βε, i = 1, 2, 3.(ii)i X i Var εi b g11 12 2 93 316 (iii)The ordinary least squares residuals are εβi i i Y X =−, i = 1, 2, 3.Determine E X X X ,,ε12123d i.(A) 1.0(B) 1.8(C) 2.7(D) 3.7(E) 7.630. For a sample of 15 losses, you are given:(i)Interval Observed Number ofLosses(0, 2] 5(2, 5] 5(5, ∞) 5 (ii) Losses follow the uniform distribution on 0,θb g.Estimate θ by minimizing the function()231j jjjE OO=−∑, where j E is the expected number oflosses in the j th interval andjO is the observed number of losses in the j th interval.(A) 6.0(B) 6.4(C) 6.8(D) 7.2(E) 7.631. You are given:(i) The probability that an insured will have exactly one claim is θ.(ii) The prior distribution of θ has probability density function:πθθθb g=<<3201,A randomly chosen insured is observed to have exactly one claim.Determine the posterior probability that θ is greater than 0.60.(A) 0.54(B) 0.58(C) 0.63(D) 0.67(E) 0.7232. The distribution of accidents for 84 randomly selected policies is as follows:Number of Accidents Number of Policies0 321 262 123 74 45 26 1Total 84 Which of the following models best represents these data?binomial(A) Negativeuniform(B) Discrete(C) Poisson(D) Binomial(E) Either Poisson or Binomial33. A time series yt follows an ARIMA(1,1,1) model with φ107=., θ103=−. and σε210=..Determine the variance of the forecast error two steps ahead.(A) 1(B)5(C) 8(D)10(E) 12(i) Low-hazard risks have an exponential claim size distribution with mean θ. (ii) Medium-hazard risks have an exponential claim size distribution with mean 2θ. (iii) High-hazard risks have an exponential claim size distribution with mean 3θ. (iv) No claims from low-hazard risks are observed.(v) Three claims from medium-hazard risks are observed, of sizes 1, 2 and 3. (vi) One claim from a high-hazard risk is observed, of size 15.Determine the maximum likelihood estimate of θ.(A) 1(B) 2(C) 3(D) 4(E) 5(i)partial X =pure premium calculated from partially credible data(ii)partial E X µ⎡⎤=⎣⎦ (iii) Fluctuations are limited to ±k µ of the mean with probability P(iv) Z = credibility factorWhich of the following is equal to P ?(A) partial Pr k X k µµµµ⎡⎤−≤≤+⎣⎦(B) partial Pr +Z k Z X Z k µµ⎡⎤−≤≤⎣⎦(C) partial Pr +Z Z X Z µµµµ⎡⎤−≤≤⎣⎦(D) ()partial Pr 111k Z X Z k µ⎡⎤−≤+−≤+⎣⎦(E) ()partial Pr 1k Z X Z k µµµµµ⎡⎤−≤+−≤+⎣⎦36. For the model 1223344i i i i i Y X X X ββββε=++++, you are given:(i) N = 15(ii)(iii) ESS =28282.Calculate the standard error of 32ˆˆββ−.(A) 6.4(B) 6.8(C) 7.1(D) 7.5(E) 7.837. You are given:Assume a uniform distribution of claim sizes within each interval.Estimate E X X 2150c h g −∧.(A)Less than 200(B)At least 200, but less than 300(C)At least 300, but less than 400(D)At least 400, but less than 500(E)At least 50038. Which of the following statements about moving average models is false?(A) Both unweighted and exponentially weighted moving average (EWMA) models canbe used to forecast future values of a time series.(B) Forecasts using unweighted moving average models are determined by applying equalweights to a specified number of past observations of the time series.(C) Forecasts using EWMA models may not be true averages because the weights appliedto the past observations do not necessarily sum to one.(D) Forecasts using both unweighted and EWMA models are adaptive because theyautomatically adjust themselves to the most recently available data.(E) Using an EWMA model, the two-period forecast is the same as the one-periodforecast.39. You are given:(i) Each risk has at most one claim each year.(ii)Type of Risk Prior Probability Annual Claim ProbabilityI 0.7 0.1II 0.2 0.2III 0.1 0.4 One randomly chosen risk has three claims during Years 1-6.Determine the posterior probability of a claim for this risk in Year 7.(A) 0.22(B) 0.28(C) 0.33(D) 0.40(E) 0.4640. You are given the following about 100 insurance policies in a study of time to policysurrender:(i) The study was designed in such a way that for every policy that was surrendered, ar, is always equal to 100.new policy was added, meaning that the risk set,j(ii) Policies are surrendered only at the end of a policy year.(iii) The number of policies surrendered at the end of each policy year was observed to be:1 at the end of the 1st policy year2 at the end of the 2nd policy year3 at the end of the 3rd policy yearn at the end of the n th policy year(iv) The Nelson-Aalen empirical estimate of the cumulative distribution function at time n, F, is 0.542.)(ˆnWhat is the value of n?(A) 8(B) 9(C) 10(D) 11(E) 12**END OF EXAMINATION**Course 4, Fall 2003PRELIMINARY ANSWER KEYQuestion # Answer Question # Answer 1 A21 A 2 E22 D 3 E23 D 4 B24 D 5 D25 E 6 A26 D 7 C27 B 8 D28 C 9 E29 B 10 D30 E 11 C31 E 12 A32 A 13 E33 B 14 D34 B 15 C35 E 16 B36 C 17 C37 C 18 B38 C 19 A39 B 20 A40 E。
AMC美国数学竞赛2003AMC10A试题及标准答案解析
2003 AMC 10A1、What is the difference between the sum of the first even counting numbers and the sum of the first odd counting numbers?SolutionThe first even counting numbers are .The first odd counting numbers are .Thus, the problem is asking for the value of..Alternatively, using the sum of an arithmetic progression formula, we can write .2、Members of the Rockham Soccer League buy socks and T-shirts. Socks cost per pair and each T-shirt costs more than a pair of socks. Each member needs one pair of socks and a shirt for home games and another pair of socks and a shirt for away games. If the total cost is , how many members are in the League?SolutionSince T-shirts cost dollars more than a pair of socks, T-shirts cost dollars.Since each member needs pairs of socks and T-shirts, the total cost for member is dollars.Since dollars was the cost for the club, and was the cost per member, the number of members in the League is .3、A solid box is cm by cm by cm. A new solid is formed by removing a cube cm on a side from each corner of this box. What percent of the original volume is removed?SolutionThe volume of the original box isThe volume of each cube that is removed isSince there are corners on the box, cubes are removed.So the total volume removed is .Therefore, the desired percentage is .4、It takes Mary minutes to walk uphill km from her home to school, but it takes her only minutes to walk from school to home along the same route. What is her average speed, in km/hr, for the round trip?SolutionSince she walked km to school and km back home, her total distance is km.Since she spent minutes walking to school and minutes walking back home, her total time is minutes = hours.Therefore her average speed in km/hr is5、Let and denote the solutions of . What is the value of ?SolutionUsing factoring:orSo and are and .Therefore the answer isOR we can use sum and product.6、Define to be for all real numbers and . Which of the following statements is not true?SolutionExamining statement C:when , but statement C says that it does for all . Therefore the statement that is not true is "for all " Alternatively, consider that the given "heart function" is actually the definition of the distance between two points. Examining all of the statements, only C is not necessarily true; if c is negative, the distance between and is the absolute value of , not itself, because distance is always nonnegative.7、How many non-congruent triangles with perimeter have integer side lengths?SolutionBy the triangle inequality, no one side may have a length greater than half the perimeter, which isSince all sides must be integers, the largest possible length of a side isTherefore, all such triangles must have all sides of length , , or . Since , at least one side must have a length of Thus, the remaining two sides have a combined length of . So, the remaining sides must be either and or and . Therefore, the number of triangles is .8、What is the probability that a randomly drawn positive factor ofis lessthan ?SolutionFor a positive number which is not a perfect square, exactly half of the positive factors will be less than .Since is not a perfect square, half of the positive factors of will be less than .Clearly, there are no positive factors of between and . Therefore half of the positive factors will be less than .So the answer is .9、SimplifySolution.Therefore:10、The polygon enclosed by t he solid lines in the figure consists of congruent squares joined edge-to-edge. One more congruent square is attached to an edge at one of the nine positions indicated. How many of the nine resulting polygons can be folded to form a cube with one face missing?SolutionLet the squares be labeled , , , and .When the polygon is folded, the "right" edge of square becomes adjacent to the "bottom edge" of square , and the "bottom" edge of square becomes adjacent to the "bottom" edge of square .So, any "new" square that is attached to those edges will prevent the polygon from becoming a cube with one face missing.Therefore, squares , , and will prevent the polygon from becoming a cube with one face missing.Squares , , , , , and will allow the polygon to become a cube with one face missing when folded.Thus the answer is .Another way to think of it is that a cube missing one face has of it's faces. Since the shape has faces already, we need another face. The only way to add anopther face is if the added square does not overlap any of the others. ,, and overlap, while 9 do not. The answer is11、The sum of the two -digit numbers and is . What is ?SolutionSince , , and are digits, , , .Therefore, .12、A point is randomly picked from inside the rectangle withvertices , , , and . What is the probability that ?SolutionThe rectangle has a width of and a height of .The area of this rectangle is .The line intersects the rectangle at and .The area which is the right isosceles triangle with side length that has vertices at , , and .The area of this triangle isTherefore, the probability that is13、The sum of three numbers is . The first is times the sum of the other two. The second is seven times the third. What is the product of all three?SolutionSolution 1Let the numbers be , , and in that order. The given tells us thatTherefore, the product of all three numbers is.Solution 2Alternatively, we can set up the system in matrix form:Or, in matrix formTo solve this matrix equation, we can rearrange it thus:Solving this matrix equation by using inverse matrices and matrix multiplication yieldsWhich means that , , and . Therefore,14、Let be the largest integer that is the product of exactly distinct prime numbers, , , and , where and are single digits. What is the sum of the digits of ?SolutionSince is a single digit prime number, the set of possible values of is .Since is a single digit prime number and is the units digit of the prime number , the set of possible values of is .Using these values for and , the set of possible values of isOut of this set, the prime values areTherefore the possible values of are:The largest possible value of is .So, the sum of the digits of is15、What is the probability that an integer in the set is divisible by and not divisible by ?SolutionThere are integers in the set.Since every 2nd integer is divisible by , there are integers divisible by in the set.To be divisible by both and , a number must be divisible by .Since every 6th integer is divisible by , there are integers divisible by both and in the set.So there are integers in this set that are divisible by and not divisible by .Therefore, the desired probability is16、What is the units digit of ?SolutionSince :Therefore, the units digit is17、The number of inches in the perimeter of an equilateral triangle equals the number of square inches in the area of its circumscribed circle. What is the radius, in inches, of the circle?SolutionLet be the length of a side of the equilateral triangle and let be the radius of the circle.In a circle with a radius the side of an inscribed equilateral triangle is .So .The perimeter of the triangle isThe area of the circle isSo:18、What is the sum of the reciprocals of the roots of the equationSolutionMultiplying both sides by :Let the roots be and .The problem is asking forBy Vieta's formulas:So the answer is .19、A semicircle of diameter sits at the top of a semicircle of diameter , as shown. The shaded area inside the smaller semicircle and outside the larger semicircle is called a lune. Determine the area of this lune.SolutionThe shaded area is equal to the area of the smaller semicircle minus the area of a sector of the larger circle plus the area of a triangle formed by two radii of the larger semicircle and the diameter of the smaller semicircle.The area of the smaller semicircle is .Since the radius of the larger semicircle is equal to the diameter of the smaller semicircle, the triangle is an equilateral triangle and the sector measures .The area of the sector of the larger semicircle is . The area of the triangle isSo the shaded area is20、A base-three-digit number is selected at random. Which of the following is closest to the probability that the base-representation and the base-representation of are both three-digit numerals?SolutionTo be a three digit number in base-10:Thus there are three-digit numbers in base-10To be a three-digit number in base-9:To be a three-digit number in base-11:So,Thus, there are base-10 three-digit numbers that are three digit numbers in base-9 and base-11.Therefore the desired probability is .21、Pat is to select six cookies from a tray containing only chocolate chip, oatmeal, and peanut butter cookies. There are at least six of each of these three kinds of cookies on the tray. How many different assortments of six cookies can be selected?SolutionSolution 1Let the ordered triplet represent the assortment of chocolate chip cookies, oatmeal cookies, and peanut butter cookies.Using casework:Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookie:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are:assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are: assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The assortments are: assortments.Pat selects chocolate chip cookies:Pat needs to select more cookies that are either oatmeal or peanut butter.The only assortment is: assortment.The total number of assortments of cookies that can be collected isSolution 2It is given that it is possible to select at least 6 of each. Therefore, we can make a bijection to the number of ways to divide the six choices into three categories, since it is assumed that their order is unimportant. Using the ball and urns formula, the number of ways to do this is22、In rectangle , we have , , is on with, is on with , line intersects line at , and is on line with . Find the length .SolutionS olution 1(Opposite angles are equal).(Both are 90 degrees).(Alt. Interior Angles are congruent).Therefore and are similar. and are also similar.is 9, therefore must equal 5. Similarly, must equal 3. Because and are similar, the ratio of and , must also hold true for and . , so is of . By Pythagorean theorem, ..So ..Therefore .Solution 2Since is a rectangle, .Since is a rectangle and ,.Since is a rectangle, .So, is a transversal, and .This is sufficient to prove that and .Using ratios:Since can't have 2 different lengths, both expressions for must be equal.Solution 3Since is a rectangle, , , and . From the Pythagorean Theorem, .LemmaStatement:Proof: , obviously.Since two angles of the triangles are equal, the third angles must equal each other. Therefore, the triangles are similar.Let .Also, , thereforeWe can multiply both sides by to get that is twice of 10, orSolution 4We extend BC such that it intersects GF at X. Since ABCD is a rectangle, it follows that CD=8, therefore, XF=8. Let GX=y. From the similarity of triangles GCH and GEA, we have the ratio 3:5 (as CH=9-6=3, and EA=9-4=5). GX and GF are the altitudes of GCH andGEA, respectively. Thus, y:y+8 = 3:5, from which we have y=12, thus GF=y+8=12+8=20. B.23、A large equilateral triangle is constructed by using toothpicks to create rows of small equilateral triangles. For example, in the figure we have rows of small congruent equilateral triangles, with small triangles in the base row. How many toothpicks would be needed to construct a large equilateral triangle if the base row of the triangle consists of small equilateral triangles?SolutionSolution 1There are small equilateral triangles.Each small equilateral triangle needs toothpicks to make it.But, each toothpick that isn't one of the toothpicks on the outside of the large equilateral triangle is a side for small equilateral triangles.So, the number of toothpicks on the inside of the large equilateral triangle isTherefore the total number of toothpicks isSolution 2We see that the bottom row of small triangles is formed from downward-facing triangles and upward-facing triangles. Since each downward-facing triangle uses three distinct toothpicks, andsince the total number of downward-facing triangles is, we have that the total number of toothpicks is24、Sally has five red cards numbered through and four blue cards numbered through . She stacks the cards so that the colors alternate and so that the number on each red card divides evenly into the number on each neighboring blue card. What is the sum of the numbers on the middle three cards?SolutionLet and designate the red card numbered and the blue card numbered , respectively.is the only blue card that evenly divides, so must be at one end of the stack and must be the card next to it.is the only other red card that evenly divides , so must be the other card next to .is the only blue card that evenly divides, so must be at the other end of the stack and must be the card next to it.is the only other red card that evenly divides , so must be the other card next to .doesn't evenly divide , so must be next to , must be next to , and must be in the middle.This yields the following arrangement from top to bottom:Therefore, the sum of the numbers on the middle three cards is.25、Let be a -digit number, and let and be the quotient and remainder, respectively, when is divided by . For how many values of is divisible by ?SolutionSolution 1When a -digit number is divided by , the first digits become the quotient, , and the last digits become the remainder, .Therefore, can be any integer from to inclusive, and can be any integer from to inclusive.For each of the possible values of , there are at least possible values of such that .Since there is "extra" possible value of that is congruent to , each of the values of that are congruent tohave more possible value of such that .Therefore, the number of possible values of such that is .Solution 2Let equal , where through are digits. Therefore,We now take :The divisor trick for 11 is as follows:"Let be an digit integer. Ifis divisible by , then is also divisible by ."Therefore, the five digit number is divisible by 11. The 5-digit multiples of 11 range from to . There aredivisors of 11 between those inclusive..- Solution 3Since q is a quotient and r is a remainder when n is divided by 100. So we have . Since we are counting choices where q+r is divisible by 11, we have for some k. This means that n is the sum of two multiples of 11 and would thus itself be a divisor of 11. Then we can count all the four digit divisors of 11 as in Solution 2. (This solution is essentially the same as solution 2, but it does not necessarily involve mods and so could potentially be faster.)NotesThe part labeled "divisor trick" actually follows from the same observation we made in the previous step: , therefore and for all . For a digit numberwe get, as claimed.Also note that in the "divisor trick" we actually want to assign the signs backwards - if we make sure that the last sign is a , the result will have the same remainder modulo as the original number.。
SOA真题November2003Course5
SOA真题November2003Course5SOA真题November2003Course5SOA真题November2003Course5course 5morning sessionapplication of basic actuarial principlessection a-written answercourse 5: fall 2003 - 2 - go on to next pagemorning session**beginning of examination 5**morning session1. (5 points) a large employer is considering offering a private pension plan.(a) describe the reasons for offering such a plan.(b) describe the process involved in designing and implementing such a plan.2. (5 points)(a) for basic group term life insurance, briefly describe each of the following items:(i) typical plan designs offered(ii) eligibility provisions(iii) continuity of coverage provisions(b) briefly describe how supplemental group term life insurance is different frombasic group term life insurance with respect to:(i) typical plan designs offered(ii) eligibility provisions(iii) continuity of coverage provisionscourse 5: fall 2003 - 3 - go on to next pagemorning session3. (5 points)(a) describe the reasons a life insurance company may reinsure its risk.(b) abc life insurance company has a 40% quota share reinsurance treaty on a firstdollar basis. its retention limit is $500,000 per policy.policy 1 policy 2net amount at risk $750,000 xamount retained r yamount reinsured on a first dollar basis s zamount reinsured on an excess basis t $100,000calculate the missing values in the table above.show all work.4. (5 points) explain the u.s. laws and regulations with respect to market conduct thatapply to a life insurance company and its agents.5. (6 points) mary and john, respectively 45 and 42 years old, are considering the purchaseof a non-participating whole life, joint last-to-die policy, paid-up at first death with:a 5 year term individual rider life insurance convertible and renewable upto age 65 for mary and john;a critical illness rider covering 50 illnesses for mary and john;a disability income rider providing a lifetime benefit with a 2 weekwaiting period for mary and john.(a) describe briefly the policy and its riders.(b) describe alternatives to each coverage that could reduce the cost to mary andjohn.course 5: fall 2003 - 4 - go on to next pagemorning session6. (7 points) for a defined benefit pension plan, you are given:pension plan formula:1.5% of final year’s salary for each year of service up to 10 years, plus2.0% of final year’s salary for each year of service after 10 years.interest rate 6%salary growth rate 4%pre-retirement decrements noneassumed retirement age 6512a&&65 12assets at 1/1/2003 300,000assets at 1/1/2004 320,000contribution made on 12/31/2003 5,000funding method projected unit creditemployee age at hire age on 1/1/2003 salary on 1/1/2003a 30 40 30,000b 30 60 50,000(a) calculate the unfunded accrued liability at 1/1/2003.(b) the actual accrued liability on 1/1/2004 is 350,000.calculate the total experience gain/loss as of that date.show all work.course 5: fall 2003 - 5 - go on to next pagemorning session7. (5 points) for a property and casualty insurance policy issued january 1, 2000, you aregiven:effective date rate changemay 1, 2000 +5%november 1, 2000 +10%calendar yearearned premiumexpected effective incurred losses, trendedand developed through december 31, 20022000 120,000 100,0002001 130,000 110,0002002 140,000 120,000expense ratio: 30%present average manual rate: 45assume all policies have a one-year term and the premium is uniformlydistributed.calculate the indicated average gross rate as of january 1, 2003.show all work.SOA真题November2003Course5 相关内容:。
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Which of the following represents g?
(A)
15 y for 0 < y < 1 g ( y) = otherwise 0 15 y 2 for x 2 < y < x g ( y) = 2 0 otherwise 15 y 2 for 0 < y < 1 g ( y) = 2 0 otherwise
(A) (B) (C) (D) (E)
0.07 0.29 0.38 0.42 0.57
May 2003
9
Course 1
5.
An insurance company examines its pool of auto insurance customers and gathers the following information:
Course 1
10
Form 03A
6.
Let X and Y be continuous random variables with joint density function
8 xy f ( x, y ) = 3 0
for 0 ≤ x ≤ 1, x ≤ y ≤ 2 x otherwise.
Let X be a continuous random variable with density function
x for − 2 ≤ x ≤ 4 f ( x ) = 10 0 otherwise.
Calculate the expected value of X.
(A)
1 5 3 5
Calculate g′(3).
(A)
−34e −3 −29e −3 −5e −3 −4e −3 63e −3
(B) (C) (D) (E)
May 2003
19
Course 1
15.
An insurance policy pays a total medical benefit consisting of two parts for each claim. Let X represent the part of the benefit that is paid to the surgeon, and let Y represent the part that is paid to the hospital. The variance of X is 5000, the variance of Y is 10,000, and the variance of the total benefit, X + Y , is 17,000.
Calculate the probability that a randomly selected customer insures exactly one car and that car is not a sports car.
(A) (B) (C) (D) (E)
0.13 0.21 0.24 0.25 0.30
3/ 2 1/ 2 2 15 y (1 − y ) for x < y < x g ( y) = otherwise 0 3/ 2 1/ 2 15 y (1 − y ) for 0 < y < 1 g ( y) = otherwise 0
(B)
(C)
(D)
(E)
May 2003
Calculate the probability that the driver was age 16-20.
(A) (B) (C) (D) (E)
0.13 0.16 0.19 0.23 0.40
May 2003
13
Course 1
9.
An insurance company determines it cannot write medical malpractice insurance profitably and stops selling the coverage. In spite of this action, the company will have to pay claims for many years on existing medical malpractice policies.
Age of Driver 16-20 21-30 31-65 66-99
Probability of Accident 0.06 0.03 0.02 0.04
Portion of Company’s Insured Drivers 0.08 0.15 0.49 0.28
A randomly selected driver that the company insures has an accident.
May 2003 - Course 1
1.
A survey of a group’s viewing habits over the last year revealed the following information:
(i) (ii) (iii) (iv) (v) (vi) (vii)
28% watched gymnastics 29% watched baseball 19% watched soccer 14% watched gymnastics and baseball 12% watched baseball and soccer 10% watched gymnastics and soccer 8% watched all three sports.
4.
The time to failure of a component in an electronic device has an exponential distribution with a median of four hours.
Calculate the probability that the component will work without failing for at least five hours.
Course 1
18
Form 03A
14.
Let f be a differentiable function such that:
f ( x + h ) − f ( x ) = 3 x 2 h + 3 xh 2 + h3 + 2h for all x and h f (0) = 1
Let g ( x ) = e − x f ( x ) .
(B) (C) (D)
1
28 15 12 5
(E)
May 2003
17
Course 1
13.
A charity receives 2025 contributions. Contributions are assumed to be independent and identically distributed with mean 3125 and standard deviation 250.
Calculate the total medical malpractice payments that the company pays in all years after it stops selling the coverage.
(A) (B) (C) (D) (E)
75 150 240 300 360
The company pays 60 for medical malpractice claims the year after it stops selling the coverage. Each subsequent year’s payments are 20% less than those of the previous year.
c f ( x ) − dg ( x ) . x →0 f ( x ) − g ( x )
(A) (B) (C) (D) (E)
0 cf ′ ( 0 ) − dg ′ ( 0 ) f ′ ( 0) − g′ ( 0)
f ′ ( 0) − g′ ( 0) c−d c+d
Course 1
8
Form 03A
Each of the graphs below contains two curves.
Identify the graph containing a curve representing a function y = f ( x ) and a curve representing its second derivative y = f ′′ ( x ) .
Course 1
14
Form 03A
10.
Let X and Y be continuous random variables with joint density function
15 y for x 2 ≤ y ≤ x f ( x, y ) = otherwise. 0
Let g be the marginal density function of Y.
∫ f ( 2 x ) dx .
0
2
(A) (B) (C) (D) (E)
32
3 4 6 8
Course 1
12
Form 03A
8.
An auto insurance company insures drivers of all ages. An actuary compiled the following statistics on the company’s insured drivers: