Math 370, Actuarial Problemsolving Spring 2008 A.J. Hildebrand. Practice Test, 1/28/2008 (with solutions)
|
|
- Randell Shaw
- 2 years ago
- Views:
Transcription
1 Math 370, Actuarial Problemsolving Spring 008 A.J. Hildebrand Practice Test, 1/8/008 (with solutions) About this test. This is a practice test made up of a random collection of 0 problems from past Course 1/P actuarial exams. Most of the problems have appeared on the Actuarial Problem sets passed out in class, but I have also included a number of additional problems. Topics covered. This test covers the topics of the entire actuarial exam syllubus, corresponding to Chapters 1 5 in Hogg/Tanis and Actuarial Problem Sets 1 5: General probability, discrete distributions, continuous distributions, multivariate distributions, and normal approximation and the Central Limit Theorem. Ordering of the problems. In order to mimick the conditions of the actual exam as closely as possible, the problems are in no particular order. Easy problems are mixed in with hard ones. In fact, I used a program to select the problems and to put them in random order, with no human intervention. Rules. This test is intended to simulate the real thing as closely as possible, so you should abide by the same rules. In particular, no notes, books, etc., and use calculators only for basic arithmetic operations. In particular, do not use calculators to compute integrals, derivatives, or to plot graphs. In the actuarial exam you are limited to calculators without such functions. It goes without saying that you shouldn t cheat. Don t copy an answer from your neighbor; if you do so, you are only cheating yourself, and you are defeating the purpose of this course. Time. You have :00 hours for this exam. This works out to 6 minutes per problem, which is the same as what you get in an actuarial exam. Use the time wisely, and don t let yourself get bogged down in a lengthy calculation. If you don t get a problem at first, move on to the next one. Answers/solutions. I will post an answer key and partial solutions on the course webpage, hildebr/370. 1
2 Math 370 A.J. Hildebrand Spring [Problem 1-5] 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 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) 0.8 Answer: B: 0.33 Problem 30/nov00 Hint/Solution: Let C denote collision coverage, and D disability coverage. We are given that (i) P (C) = P (D), (ii) C and D are independent, and (iii) P (C D) = 0.15, and we need to compute P (C D ). By independence, P (C D) = P (C)P (D). Using the independence along with the equations (i) and (iii) we get 0.15 = P (C D) = P (C)P (D) = P (D), so P (D) = and P (C) =. Hence P (C ) = and P (D ) =. Applying indendence again, we get P (C D ) = P (C )P (D ).. [Problem 4-103] A device runs until either of two components fails, at which point the device stops running. The joint density function of the lifetimes of the two components, both measured in hours, is { 1 f(x, y) = 7 (x + y) for 0 < x < 3 and 0 < y < 3, 0 otherwise. Calculate the probability that the device fails during its first hour of operation. (A) 0.04 (B) 0.41 (C) 0.44 (D) 0.59 (E) 0.96 Answer: B: 0.41 Problem 16/may03 Hint/Solution: The tricky part here is to properly interpret the statement the device fails during first hour, and identify the corresponding region in the xy-plane; once this is done, the problem amounts to computing a relatively simple double integral.
3 Math 370 Practice Test, 1/8/008 (with solutions) Spring [Problem 1-8] An insurance company pays hospital claims. The number of claims that include emergency room or operating room charges is 85% of the total number of claims. The number of claims that do not include emergency room charges is 5% of the total number of claims. The occurrence of emergency room charges is independent of the occurrence of operating room charges on hospital claims. Calculate the probability that a claim submitted to the insurance company includes operating room charges. (A) 0.10 (B) 0.0 (C) 0.5 (D) 0.40 (E) 0.80 Answer: D: 0.40 Problem 37/may03 Hint/Solution: The relevant events are operating room charges and emergency room charges ; we are given the probability of the union of these two events, and the probability of the complement of the second one, and we need to compute the probability of the first; for that, either use appropriate probability rules (e.g., the formula for the probability of a union), or determine the probability from a Venn diagram. 4. [Problem 3-4] The value, ν, of an appliance is given by ν(t) = e 7 0.t, where t denotes the number of years since purchase. If appliance fails within seven years of purchase, a warranty pays the owner the value of the appliance at the time of failure. After seven years, the warranty pays nothing. The time until failure of the appliance has an exponential distribution with mean 10. Calculate the expected payment from the warranty. (A) (B) (C) (D) (E) Answer: D: Problem 3/1999 Hint/Solution: Let X denote the payout under the warranty, and T the time until failure. Then X = e 7 0.T if T 7, and X = 0 otherwise. Since T has exponential density with mean 10, its density function is (1/10)e t/10 for t > 0. Thus, 5. [Problem 3-11] E(X) = 7 0 e 7 0.t 1 10 e t/10 dt = 0.1 e e 0.3t dt == 0.1 e 7 e = An insurance policy reimburses a loss up to a benefit limit of 10. The policyholder s loss, Y, follows a distribution with density function { y 3 for y > 1, f(x) = 0 otherwise. 3
4 Math 370 A.J. Hildebrand Spring 008 What is the expected value of the benefit paid under the insurance policy? (A) 1.0 (B) 1.3 (C) 1.8 (D) 1.9 (E).0 Answer: D: 1.9 Hint/Solution: Y > 10. Thus, E(X) = 6. [Problem 4-15] Problem 34/may The benefit, X, is given by X = Y if Y 10, and by X = 10 if y y 3 dy + 10 ( 10 y 3 dy = 1 1 ) = A company offers a basic life insurance policy to its employees, as well as a supplemental life insurance policy. To purchase the supplemental policy, an employee must first purchase the basic policy. Let X denote the proportion of employees who purchase the basic policy, and Y the proportion of employees who purchase the supplemental policy. Let X and Y have the joint density function f(x, y) = (x + y) on the region where the density is positive. Given that 10% of the employees buy the basic policy, what is the probability that fewer than 5% buy the supplemental policy? (A) (B) (C) (D) (E) Answer: D: Problem 11/may00 Hint/Solution: We need to compute the conditional probability P (Y 0.05 X = 0.1). This is given by 0.05 y=0 h(y 0.1)dy, where h(y 0.1) is the conditional density of Y given that X = 0.1. Now, h(y 0.1) = f(0.1, y)/f X (0.1), where f(x, y) = (x + y) is the given density and f X (x) is the marginal density of X. The main difficulty of the problem is the proper interpretation of the phrase on the region where the joint density is positive. Obviously, since x and y both denote proportions, this region must be part of the unit square 0 x 1, 0 y 1. However, an additional restriction is imposed on x and y by the requirement that to purchase a supplemental policy, one must first purchase the basic policy. This means that the proportion of supplemental policy holders cannot be larger than the proportion of basic policy holders. Thus, x and y are restricted by the additional condition y x, and the region in question is the triangular region given by 0 x 1, 0 y x. Using these limits on x and y in the integrations, we then get f X (x) = y=x y=0 (x + y)dy = 3x (0 x 1), so f X (0.1) = 0.03 and h(y 0.1) = (0.1 + y)/0.03 = (00/3)(0.1 + y) for 0 y Finally, substituting this into the above formula for P (Y 0.05 X = 0.1) gives P (Y 0.05 X = 0.1) = (0.1 + y)dy = 00 ) ( =
5 Math 370 Practice Test, 1/8/008 (with solutions) Spring [Problem 4-54] An auto insurance policy will pay for damage to both the policyholder s car and the other driver s car in the event that the policyholder is responsible for an accident. The size of the payment for damage to the policyholder s car, X, has a marginal density function of 1 for 0 < x < 1. Given X = x, the size of the payment for damage to the other driver s car, Y, has conditional density of 1 for x < y < x + 1. If the policyholder is responsible for an accident, what is the probability that the payment for damage to the other driver s car will be greater than 0.500? (A) 3/8 (B) 1/ (C) 3/4 (D) 7/8 (E) 15/16 Answer: D: 7/8 Problem 34/nov01 Hint/Solution: The probability to compute is P (Y 0.5). We are given that X is uniform on [0, 1] and that, given X = x, Y is uniform on [x, x + 1]. Thus, f X (x) = 1 for 0 x 1 and h(y x) = 1 for x y x + 1, 0 x 1. Hence f(x, y) = f X (x)h(y x) = 1, 0 x 1, x y x + 1. In particular, this means that the joint density is uniform on the region (a parallelogram sketch!) 0 x 1, x y x + 1. Hence probabilities can be computed as ratios of areas. The entire parallelogram has area 1, and the part of this parallelogram on which y 0.5 has area 1 (1/)0.5 = 7/8 by elementary trigonometry. Hence P (Y 0.5) = (7/8)/1 = 7/8. 8. [Problem 3-0] A device that continuously measures and records seismic activity is placed in a remote region. The time, T, to failure of this device is exponentially distributed with mean 3 years. Since the device will not be monitored during its first two years of service, the time to discovery of its failure is X = max(t, ). Determine E(X). (A) e 6 (B) e /3 + 5e 4/3 (C) 3 (D) + 3e /3 (E) 5 Answer: D Problem 0/may01 Hint/Solution: Note that X = T if T and X = if 0 T <, Since T is exponentially distributed with mean 3, the density function of T is f(t) = (1/3)e t/3 5
6 Math 370 A.J. Hildebrand Spring 008 for t > 0. Thus, 9. [Problem 3-13] E(X) = E(max(T, )) == = 0 3 e t/3 dt + 0 = (1 e /3 ) te t/3 max(t, )(1/3)e t/3 dt t 3 e t/3 dt + e t/3 dt = (1 e /3 ) + e t/3 + 3e /3 = + 3e /3 The time, T, that a manufacturing system is out of operation has cumulative distribution function { 1 ( ) F (t) = t for t >, 0 otherwise. The resulting cost to the company is Y = T. Determine the density function of Y, for y > 4. Answer: (A) 4y (B) 8y 3/ (C) 8y 3 (D) 16y 1 (E) 104y 5 A Problem 3/may03 Hint/Solution: This is a routine application of the change of variables techniques; as always in these problems, you need to take a detour via the corresponding c.d.f. s to get the p.d.f. of the new variable. 10. [Problem 5-51] Let X and Y be the number of hours that a randomly selected person watches movies and sporting events, respectively, during a three-month period. The following information is known about X and Y : E(X) = 50, E(Y ) = 0, Var(X) = 50, Var(Y ) = 30, Cov(X, Y ) = 10. One hundred people are randomly selected and observed for these three months. Let T be the total number of hours that these one hundred people watch movies or sporting events during this three-month period. Approximate the value of P (T < 7100). (A) 0.6 (B) 0.84 (C) 0.87 (D) 0.9 (E) 0.97 Answer: B: 0.84 Problem 40/nov01 Hint/Solution: The total time that a single person spends watching is a random variable with the distribution of X + Y and mean and variance given by µ = E(X + Y ) = E(X) + E(Y ) = = 70, σ = Var(X + Y ) = Var(X) + Var(Y ) + Cov(X, Y ) = =
7 Math 370 Practice Test, 1/8/008 (with solutions) Spring 008 The total time, T, spent by 100 such people is a sum of 100 independent random variables, each of the type X + Y. By the CLT, T has approximately normal distribution with mean 100µ = = 7000 and standard deviation 100σ = = 100. Hence ( ) T P (T < 7100) = P < P (Z < 1) = Φ(1) = [Problem 4-51] Once a fire is reported to a fire insurance company, the company makes an initial estimate, X, of the amount it will pay to the claimant for the fire loss. When the claim is finally settled, the company pays an amount, Y, to the claimant. The company has determined that X and Y have the joint density function f(x, y) = x (x 1) y (x 1)/(x 1), x > 1, y > 1. Given that the initial claim estimated by the company is, determine the probability that the final settlement amount is between 1 and 3. (A) 1/9 (B) /9 (C) 1/3 (D) /3 (E) 8/9 Answer: E: 8/9 Problem 5/nov01 Hint/Solution: We need to compute P (1 Y 3 X = ). This is given by the integral ( ) 3 1 h(y ), where h(y ) is the conditional density of Y given X =. Now, h(y ) = f(, y)/f X (), where f(, y) = (1/)y 3 from the given joint density and f X () is the marginal density of X, at x =, i.e., f X () = y=1 f(, y)dy = y=1 1 y 3 dy = 1 4. Thus, h(y ) = y 3 for 1 < y <. Substituting this density into ( ) gives the answer: 1. [Problem 4-53] P (1 Y 3 X = ) = 3 1 y 3 dy = 8 9. A company is reviewing tornado damage claims under a farm insurance policy. Let X be the portion of a claim representing damage to the house and let Y be the portion of the same claim representing damage to the rest of the property. The joint density function of X and Y is { 6(1 (x + y)) for x > 0, y > 0, x + y < 1, f(x, y) = 0 otherwise. Determine the probability that the portion of a claim representing damage to the house is less than 0.. 7
8 Math 370 A.J. Hildebrand Spring 008 (A) (B) (C) (D) 0.51 (E) 0.50 Answer: C: Problem 5/may01 Hint/Solution: The probability to compute is P (X 0.), which can be computed as the integral over the joint density f(x, y) over the part of its range where x 0.. Now the range of f(x, y) is the triangle given by 0 x < 1, 0 y < 1 x (sketch!), and restricting x further by 0 x 0., we get the desired range of integration: 13. [Problem -8] P (X 0.) = = 0. 1 x x=0 0. x=0 y=0 6(1 (x + y))dxdy 3(1 x) dx = = 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) 1, 704 (D), 900 (E) 3, Answer: D:,900 Problem 9/1999 Hint/Solution: This would be a straightforward calculation except for the phrase given that the loss is greater than 0. The correct interpretation of this condition is to restrict consideration to those cases where the loss is greater than 0 and renormalize the probabilities for these cases so that they again add up to 1. Since the probability for a loss of 0 is 0.9, the remaining probabilities must add up to 0.1, so dividing each by 0.1 renormalizes them to have sum 1. Using these renormalized probabilities in the formula for an expectation gives the desired expectation under the above condition. 8
9 Math 370 Practice Test, 1/8/008 (with solutions) Spring [Problem 1-104] An auto insurance company has 10, 000 policyholders. Each policyholder is classified as (i) young or old; (ii) male or female; and (iii) married or single. Of these policyholders, 3000 are young, 4600 are male, and 7000 are married. The policyholders can also be classified as 130 young males, 3010 married males, and 1400 young married persons. Finally, 600 of the policyholders are young married males. How many of the company s policyholders are young, female, and single? (A) 80 (B) 43 (C) 486 (D) 880 (E) 896 Answer: D: 880 Problem 3/nov00 Hint/Solution: If we consider the sets young, male and married, we want to count the part of the young set that is outside the other two sets. This is easy by drawing a Venn diagram and using the given data. 15. [Problem 4-56] Two insurers provide bids on an insurance policy to a large company. The bids must be between 000 and 00. The company decides to accept the lower bid if the two bids differ by 0 or more. Otherwise, the company will consider the two bids further. Assume that the two bids are independent and are both uniformly distributed on the interval from 000 to 00. Determine the probability that the company considers the two bids further. (A) 0.10 (B) 0.19 (C) 0.0 (D) 0.41 (E) 0.60 Answer: B: 0.19 Problem 8/nov01 Hint/Solution: We need to compute P ( X Y < 0), where X and Y denote the two bids. We are given that X and Y are independent, with each having uniform distribution on [000, 00]. Thus, the joint distribution of X and Y is uniform on the square [000, 00] [000, 00]. The event X Y < 0 corresponds to a diagonal strip inside this square. Since the joint distribution is uniform over the whole square, the probability of this event is given by the ratio between the area of this strip and the area of the entire square. A simple geometric argument (sketch!) shows that the strip has area , so P ( X Y < 0) = ( )/00 = [Problem 3-15] 9
10 Math 370 A.J. Hildebrand Spring 008 The lifetime of a printer costing 00 is exponentially distributed with mean 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, 31 (B) 7, 358 (C) 7, 869 (D) 10, 56 (E) 1, 64 Answer: D: 10,56 Problem 3/may00 Hint/Solution: Let X denote the lifetime of the printer. Then the refund, Y, is given by Y = 00 if X 1, Y = 100 if 1 X, and Y = 0 otherwise. Thus, E(Y ) = 00P (X 1) + 100P (1 < X ). The probabilities P (... ) here are easily computed using that fact X has exponential distribution with mean, and thus c.d.f. F (x) = P (X x) = 1 e x/ for x 0. Thus, P (X 1) = 1 e 1/, P (1 < X ) = P (X ) P (X 1) = (1 e / ) (1 e 1/ ) = e 1/ e 1, and hence E(Y ) = = 10, [Problem 3-9] The monthly profit of Company I can be modeled by a continuous random variable with density function f. Company II has a monthly profit that is twice that of Company I. Determine the probability density function of the monthly profit of Company II. Answer: (A) 1 f ( ) x A Problem 3/nov00 (B) f ( ) x (C) f ( ) x (D) f(x) (E) f(x) Hint/Solution: This is an easy exercise in changing variables in density functions if you apply to correct procedure for that. Let X denote the profit for Company I and Y that for Company II. Then Y = X. Let f(x) and F (x) denote the p.d.f. and c.d.f. of X, and let g(x) and G(x) denote the corresponding functions for Y. Then we have G(x) = P (Y x) = P (X x) = P (X x/) = F (x/). Differentiating both sides with respect to x we get g(x) = G (x) = (1/)F (x/) = (1/)f(x/) since F (x) = f(x). Thus, the density of Y is (1/)f(x/). 18. [Problem 4-55] An insurance policy is written to cover a loss X where X has density function { 3 f(x) = 8 x for 0 x, 0 otherwise. The time (in hours) to process a claim of size x, where 0 x, is uniformly distributed on the interval from x to x. Calculate the probability that a randomly chosen claim on this policy is processed in three hours or more. 10
11 Math 370 Practice Test, 1/8/008 (with solutions) Spring 008 (A) 0.17 (B) 0.5 (C) 0.3 (D) 0.58 (E) 0.83 Answer: A: 0.17 Problem 3/may00 Hint/Solution: We need to compute P (Y 3), where Y denotes the time needed to process a claim. From the problem, we know that, given X = x, Y is uniformly distributed on [x, x]. Thus, the conditional density of Y given X = x is h(y x) = 1/x for x y x. Using this density along with the given (ordinary) density of X, we can compute the joint density of X and Y : f(x, y) = h(y x)f X (x) = 1 x 3 8 x = 3 x, 0 x, x y x. 8 A careful sketch of the region given here shows that the part of it that satisfies y 3 is described by the inequalities 3/ x, 3 y x. Hence, P (Y 3) = x=3/ x y=3 38xdydx = x=3/ and calculating the latter integral we get the answer [Problem 5-3] 3 x(x 3)dx. 8 In an analysis of healthcare data, ages have been rounded to the nearest multiple of 5 years. The difference between the true age and the rounded age is assumed to be uniformly distributed on the interval from.5 years to.5 years. The healthcare data are based on a random sample of 48 people. What is the approximate probability that the mean of the rounded ages is within 0.5 years of the mean of the true ages? (A) 0.14 (B) 0.38 (C) 0.57 (D) 0.77 (E) 0.88 Answer: D: 0.77 Problem 19/may00 Hint/Solution: We need to compute P ( X 0.5), where X is the mean of a random sample of size 48 from a uniform distribution on [.5,.5]. Obviously this distribution has mean 0, and an easy computation (or using the formula σ = (b a) /1 for the variance of a uniform distribution on an interval [a, b]) shows that the standard deviation is By the CLT, X has approximately normal distribution with mean 0 and standard deviation 1.44/ 48 = 0.078, so ( X P ( X 0.5) = P ) [Problem 3-7] P ( Z 1.) = Φ(1.) Φ( 1.) = 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. 11
12 Math 370 A.J. Hildebrand Spring 008 (A) 0.07 (B) 0.9 (C) 0.38 (D) 0.4 (E) 0.57 Answer: D: 0.4 Problem 4/may03 Hint/Solution: The general form of the c.d.f. of an exponential distribution is F (x) = 1 e x/θ for x > 0. Since the median is 4 hours, we have F (4) = 1/. This allows us to determine the parameter θ: 1 e 4/θ = 1/, so θ = 4/ ln. The probability that the component will work for at least 5 hours is given by 1 F (5) = e 5/θ = e (5/4) ln =
Math 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 3 Solutions
Math 37/48, Spring 28 Prof. A.J. Hildebrand Actuarial Exam Practice Problem Set 3 Solutions About this problem set: These are problems from Course /P actuarial exams that I have collected over the years,
Math 370, Spring 2008 Prof. A.J. Hildebrand. Practice Test 2 Solutions
Math 370, Spring 008 Prof. A.J. Hildebrand Practice Test Solutions About this test. This is a practice test made up of a random collection of 5 problems from past Course /P actuarial exams. Most of the
Math 370, Spring 2008 Prof. A.J. Hildebrand. Practice Test 2
Math 370, Spring 2008 Prof. A.J. Hildebrand Practice Test 2 About this test. This is a practice test made up of a random collection of 15 problems from past Course 1/P actuarial exams. Most of the problems
Math 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 5 Solutions
Math 370/408, Spring 2008 Prof. A.J. Hildebrand Actuarial Exam Practice Problem Set 5 Solutions About this problem set: These are problems from Course 1/P actuarial exams that I have collected over the
Math 370, Spring 2008 Prof. A.J. Hildebrand. Practice Test 1 Solutions
Math 70, Spring 008 Prof. A.J. Hildebrand Practice Test Solutions About this test. This is a practice test made up of a random collection of 5 problems from past Course /P actuarial exams. Most of the
Math 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 2 Solutions
Math 70/408, Spring 2008 Prof. A.J. Hildebrand Actuarial Exam Practice Problem Set 2 Solutions About this problem set: These are problems from Course /P actuarial exams that I have collected over the years,
SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS
SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS Copyright 005 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study
SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS
SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS Copyright 007 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study
SOCIETY OF ACTUARIES EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS
SOCIETY OF ACTUARIES EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS Copyright 015 by the Society of Actuaries Some of the questions in this study note are taken from past examinations. Some of the questions
Math 370, Actuarial Problemsolving Spring 2008 A.J. Hildebrand. Problem Set 1 (with solutions)
Math 370, Actuarial Problemsolving Spring 2008 A.J. Hildebrand Problem Set 1 (with solutions) About this problem set: These are problems from Course 1/P actuarial exams that I have collected over the years,
1. A survey of a group s viewing habits over the last year revealed the following
1. A survey of a group s viewing habits over the last year revealed the following information: (i) 8% watched gymnastics (ii) 9% watched baseball (iii) 19% watched soccer (iv) 14% watched gymnastics and
STA 256: Statistics and Probability I
Al Nosedal. University of Toronto. Fall 2014 1 2 3 4 5 My momma always said: Life was like a box of chocolates. You never know what you re gonna get. Forrest Gump. Experiment, outcome, sample space, and
UNIVERSITY of TORONTO. Faculty of Arts and Science
UNIVERSITY of TORONTO Faculty of Arts and Science AUGUST 2005 EXAMINATION AT245HS uration - 3 hours Examination Aids: Non-programmable or SOA-approved calculator. Instruction:. There are 27 equally weighted
SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS
SOCIETY OF ACTUARIES/CASUALTY ACTUARIAL SOCIETY EXAM P PROBABILITY EXAM P SAMPLE QUESTIONS Copyright 5 by the Society of Actuaries and the Casualty Actuarial Society Some of the questions in this study
Math 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 1
Math 370/408, Spring 2008 Prof. A.J. Hildebrand Actuarial Exam Practice Problem Set 1 About this problem set: These are problems from Course 1/P actuarial exams that I have collected over the years, grouped
Joint distributions Math 217 Probability and Statistics Prof. D. Joyce, Fall 2014
Joint distributions Math 17 Probability and Statistics Prof. D. Joyce, Fall 14 Today we ll look at joint random variables and joint distributions in detail. Product distributions. If Ω 1 and Ω are sample
EDUCATION AND EXAMINATION COMMITTEE SOCIETY OF ACTUARIES RISK AND INSURANCE. Copyright 2005 by the Society of Actuaries
EDUCATION AND EXAMINATION COMMITTEE OF THE SOCIET OF ACTUARIES RISK AND INSURANCE by Judy Feldman Anderson, FSA and Robert L. Brown, FSA Copyright 25 by the Society of Actuaries The Education and Examination
Math 431 An Introduction to Probability. Final Exam Solutions
Math 43 An Introduction to Probability Final Eam Solutions. A continuous random variable X has cdf a for 0, F () = for 0 <
Continuous Probability Distributions (120 Points)
Economics : Statistics for Economics Problem Set 5 - Suggested Solutions University of Notre Dame Instructor: Julio Garín Spring 22 Continuous Probability Distributions 2 Points. A management consulting
Definition: Suppose that two random variables, either continuous or discrete, X and Y have joint density
HW MATH 461/561 Lecture Notes 15 1 Definition: Suppose that two random variables, either continuous or discrete, X and Y have joint density and marginal densities f(x, y), (x, y) Λ X,Y f X (x), x Λ X,
Chapter 4 Expected Values
Chapter 4 Expected Values 4. The Expected Value of a Random Variables Definition. Let X be a random variable having a pdf f(x). Also, suppose the the following conditions are satisfied: x f(x) converges
3. The Economics of Insurance
3. The Economics of Insurance Insurance is designed to protect against serious financial reversals that result from random evens intruding on the plan of individuals. Limitations on Insurance Protection
Joint Exam 1/P Sample Exam 1
Joint Exam 1/P Sample Exam 1 Take this practice exam under strict exam conditions: Set a timer for 3 hours; Do not stop the timer for restroom breaks; Do not look at your notes. If you believe a question
MT426 Notebook 3 Fall 2012 prepared by Professor Jenny Baglivo. 3 MT426 Notebook 3 3. 3.1 Definitions... 3. 3.2 Joint Discrete Distributions...
MT426 Notebook 3 Fall 2012 prepared by Professor Jenny Baglivo c Copyright 2004-2012 by Jenny A. Baglivo. All Rights Reserved. Contents 3 MT426 Notebook 3 3 3.1 Definitions............................................
Math 461 Fall 2006 Test 2 Solutions
Math 461 Fall 2006 Test 2 Solutions Total points: 100. Do all questions. Explain all answers. No notes, books, or electronic devices. 1. [105+5 points] Assume X Exponential(λ). Justify the following two
Contents. 3 Survival Distributions: Force of Mortality 37 Exercises Solutions... 51
Contents 1 Probability Review 1 1.1 Functions and moments...................................... 1 1.2 Probability distributions...................................... 2 1.2.1 Bernoulli distribution...................................
Continuous Distributions
MAT 2379 3X (Summer 2012) Continuous Distributions Up to now we have been working with discrete random variables whose R X is finite or countable. However we will have to allow for variables that can take
4. Joint Distributions of Two Random Variables
4. Joint Distributions of Two Random Variables 4.1 Joint Distributions of Two Discrete Random Variables Suppose the discrete random variables X and Y have supports S X and S Y, respectively. The joint
Recitation 4. 24xy for 0 < x < 1, 0 < y < 1, x + y < 1 0 elsewhere
Recitation. Exercise 3.5: If the joint probability density of X and Y is given by xy for < x
University of California, Berkeley, Statistics 134: Concepts of Probability
University of California, Berkeley, Statistics 134: Concepts of Probability Michael Lugo, Spring 211 Exam 2 solutions 1. A fair twenty-sided die has its faces labeled 1, 2, 3,..., 2. The die is rolled
Joint Probability Distributions and Random Samples (Devore Chapter Five)
Joint Probability Distributions and Random Samples (Devore Chapter Five) 1016-345-01 Probability and Statistics for Engineers Winter 2010-2011 Contents 1 Joint Probability Distributions 1 1.1 Two Discrete
MULTIVARIATE PROBABILITY DISTRIBUTIONS
MULTIVARIATE PROBABILITY DISTRIBUTIONS. PRELIMINARIES.. Example. Consider an experiment that consists of tossing a die and a coin at the same time. We can consider a number of random variables defined
Bivariate Distributions
Chapter 4 Bivariate Distributions 4.1 Distributions of Two Random Variables In many practical cases it is desirable to take more than one measurement of a random observation: (brief examples) 1. What is
MAS108 Probability I
1 QUEEN MARY UNIVERSITY OF LONDON 2:30 pm, Thursday 3 May, 2007 Duration: 2 hours MAS108 Probability I Do not start reading the question paper until you are instructed to by the invigilators. The paper
( ) = 1 x. ! 2x = 2. The region where that joint density is positive is indicated with dotted lines in the graph below. y = x
Errata for the ASM Study Manual for Exam P, Eleventh Edition By Dr. Krzysztof M. Ostaszewski, FSA, CERA, FSAS, CFA, MAAA Web site: http://www.krzysio.net E-mail: krzysio@krzysio.net Posted September 21,
3. Continuous Random Variables
3. Continuous Random Variables A continuous random variable is one which can take any value in an interval (or union of intervals) The values that can be taken by such a variable cannot be listed. Such
Solution. Let us write s for the policy year. Then the mortality rate during year s is q 30+s 1. q 30+s 1
Solutions to the May 213 Course MLC Examination by Krzysztof Ostaszewski, http://wwwkrzysionet, krzysio@krzysionet Copyright 213 by Krzysztof Ostaszewski All rights reserved No reproduction in any form
Statistics 100A Homework 7 Solutions
Chapter 6 Statistics A Homework 7 Solutions Ryan Rosario. A television store owner figures that 45 percent of the customers entering his store will purchase an ordinary television set, 5 percent will purchase
Random Variables. Chapter 2. Random Variables 1
Random Variables Chapter 2 Random Variables 1 Roulette and Random Variables A Roulette wheel has 38 pockets. 18 of them are red and 18 are black; these are numbered from 1 to 36. The two remaining pockets
ECE302 Spring 2006 HW5 Solutions February 21, 2006 1
ECE3 Spring 6 HW5 Solutions February 1, 6 1 Solutions to HW5 Note: Most of these solutions were generated by R. D. Yates and D. J. Goodman, the authors of our textbook. I have added comments in italics
e.g. arrival of a customer to a service station or breakdown of a component in some system.
Poisson process Events occur at random instants of time at an average rate of λ events per second. e.g. arrival of a customer to a service station or breakdown of a component in some system. Let N(t) be
ST 371 (VIII): Theory of Joint Distributions
ST 371 (VIII): Theory of Joint Distributions So far we have focused on probability distributions for single random variables. However, we are often interested in probability statements concerning two or
Microeconomic Theory: Basic Math Concepts
Microeconomic Theory: Basic Math Concepts Matt Van Essen University of Alabama Van Essen (U of A) Basic Math Concepts 1 / 66 Basic Math Concepts In this lecture we will review some basic mathematical concepts
MTH135/STA104: Probability
MTH135/STA14: Probability Homework # 8 Due: Tuesday, Nov 8, 5 Prof Robert Wolpert 1 Define a function f(x, y) on the plane R by { 1/x < y < x < 1 f(x, y) = other x, y a) Show that f(x, y) is a joint probability
5. Continuous Random Variables
5. Continuous Random Variables Continuous random variables can take any value in an interval. They are used to model physical characteristics such as time, length, position, etc. Examples (i) Let X be
Important Probability Distributions OPRE 6301
Important Probability Distributions OPRE 6301 Important Distributions... Certain probability distributions occur with such regularity in real-life applications that they have been given their own names.
Probability & Statistics Primer Gregory J. Hakim University of Washington 2 January 2009 v2.0
Probability & Statistics Primer Gregory J. Hakim University of Washington 2 January 2009 v2.0 This primer provides an overview of basic concepts and definitions in probability and statistics. We shall
Solution Using the geometric series a/(1 r) = x=1. x=1. Problem For each of the following distributions, compute
Math 472 Homework Assignment 1 Problem 1.9.2. Let p(x) 1/2 x, x 1, 2, 3,..., zero elsewhere, be the pmf of the random variable X. Find the mgf, the mean, and the variance of X. Solution 1.9.2. Using the
Lecture 6: Discrete & Continuous Probability and Random Variables
Lecture 6: Discrete & Continuous Probability and Random Variables D. Alex Hughes Math Camp September 17, 2015 D. Alex Hughes (Math Camp) Lecture 6: Discrete & Continuous Probability and Random September
Math 151. Rumbos Spring 2014 1. Solutions to Assignment #22
Math 151. Rumbos Spring 2014 1 Solutions to Assignment #22 1. An experiment consists of rolling a die 81 times and computing the average of the numbers on the top face of the die. Estimate the probability
Lecture Notes 1. Brief Review of Basic Probability
Probability Review Lecture Notes Brief Review of Basic Probability I assume you know basic probability. Chapters -3 are a review. I will assume you have read and understood Chapters -3. Here is a very
MA 485-1E, Probability (Dr Chernov) Final Exam Wed, Dec 10, 2003 Student s name Be sure to show all your work. Full credit is given for 100 points.
MA 485-1E, Probability (Dr Chernov) Final Exam Wed, Dec 10, 2003 Student s name Be sure to show all your work. Full credit is given for 100 points. 1. (10 pts) Assume that accidents on a 600 miles long
Examination 110 Probability and Statistics Examination
Examination 0 Probability and Statistics Examination Sample Examination Questions The Probability and Statistics Examination consists of 5 multiple-choice test questions. The test is a three-hour examination
( ) is proportional to ( 10 + x)!2. Calculate the
PRACTICE EXAMINATION NUMBER 6. An insurance company eamines its pool of auto insurance customers and gathers the following information: i) All customers insure at least one car. ii) 64 of the customers
Senior Secondary Australian Curriculum
Senior Secondary Australian Curriculum Mathematical Methods Glossary Unit 1 Functions and graphs Asymptote A line is an asymptote to a curve if the distance between the line and the curve approaches zero
Continuous Random Variables and Probability Distributions. Stat 4570/5570 Material from Devore s book (Ed 8) Chapter 4 - and Cengage
4 Continuous Random Variables and Probability Distributions Stat 4570/5570 Material from Devore s book (Ed 8) Chapter 4 - and Cengage Continuous r.v. A random variable X is continuous if possible values
ECE302 Spring 2006 HW7 Solutions March 11, 2006 1
ECE32 Spring 26 HW7 Solutions March, 26 Solutions to HW7 Note: Most of these solutions were generated by R. D. Yates and D. J. Goodman, the authors of our textbook. I have added comments in italics where
1. Let A, B and C are three events such that P(A) = 0.45, P(B) = 0.30, P(C) = 0.35,
1. Let A, B and C are three events such that PA =.4, PB =.3, PC =.3, P A B =.6, P A C =.6, P B C =., P A B C =.7. a Compute P A B, P A C, P B C. b Compute P A B C. c Compute the probability that exactly
Worked examples Multiple Random Variables
Worked eamples Multiple Random Variables Eample Let X and Y be random variables that take on values from the set,, } (a) Find a joint probability mass assignment for which X and Y are independent, and
Cork Institute of Technology. CIT Mathematics Examination, Paper 2 Sample Paper A
Cork Institute of Technology CIT Mathematics Examination, 2015 Paper 2 Sample Paper A Answer ALL FIVE questions. Each question is worth 20 marks. Total marks available: 100 marks. The standard Formulae
Random Variables, Expectation, Distributions
Random Variables, Expectation, Distributions CS 5960/6960: Nonparametric Methods Tom Fletcher January 21, 2009 Review Random Variables Definition A random variable is a function defined on a probability
6. Distribution and Quantile Functions
Virtual Laboratories > 2. Distributions > 1 2 3 4 5 6 7 8 6. Distribution and Quantile Functions As usual, our starting point is a random experiment with probability measure P on an underlying sample spac
Area Between Curves. The idea: the area between curves y = f(x) and y = g(x) (if the graph of f(x) is above that of g(x)) for a x b is given by
MATH 42, Fall 29 Examples from Section, Tue, 27 Oct 29 1 The First Hour Area Between Curves. The idea: the area between curves y = f(x) and y = g(x) (if the graph of f(x) is above that of g(x)) for a x
ALGEBRA I (Common Core) Wednesday, June 17, :15 to 4:15 p.m., only
ALGEBRA I (COMMON CORE) The University of the State of New York REGENTS HIGH SCHOOL EXAMINATION ALGEBRA I (Common Core) Wednesday, June 17, 2015 1:15 to 4:15 p.m., only Student Name: School Name: The possession
MATH SOLUTIONS TO PRACTICE FINAL EXAM. (x 2)(x + 2) (x 2)(x 3) = x + 2. x 2 x 2 5x + 6 = = 4.
MATH 55 SOLUTIONS TO PRACTICE FINAL EXAM x 2 4.Compute x 2 x 2 5x + 6. When x 2, So x 2 4 x 2 5x + 6 = (x 2)(x + 2) (x 2)(x 3) = x + 2 x 3. x 2 4 x 2 x 2 5x + 6 = 2 + 2 2 3 = 4. x 2 9 2. Compute x + sin
Overview of Monte Carlo Simulation, Probability Review and Introduction to Matlab
Monte Carlo Simulation: IEOR E4703 Fall 2004 c 2004 by Martin Haugh Overview of Monte Carlo Simulation, Probability Review and Introduction to Matlab 1 Overview of Monte Carlo Simulation 1.1 Why use simulation?
Sums of Independent Random Variables
Chapter 7 Sums of Independent Random Variables 7.1 Sums of Discrete Random Variables In this chapter we turn to the important question of determining the distribution of a sum of independent random variables
Algebra 2 Final Exam Review Ch 7-14
Algebra 2 Final Exam Review Ch 7-14 Short Answer 1. Simplify the radical expression. Use absolute value symbols if needed. 2. Simplify. Assume that all variables are positive. 3. Multiply and simplify.
Math 425 (Fall 08) Solutions Midterm 2 November 6, 2008
Math 425 (Fall 8) Solutions Midterm 2 November 6, 28 (5 pts) Compute E[X] and Var[X] for i) X a random variable that takes the values, 2, 3 with probabilities.2,.5,.3; ii) X a random variable with the
Jointly Distributed Random Variables
Jointly Distributed Random Variables COMP 245 STATISTICS Dr N A Heard Contents 1 Jointly Distributed Random Variables 1 1.1 Definition......................................... 1 1.2 Joint cdfs..........................................
SOLUTIONS. f x = 6x 2 6xy 24x, f y = 3x 2 6y. To find the critical points, we solve
SOLUTIONS Problem. Find the critical points of the function f(x, y = 2x 3 3x 2 y 2x 2 3y 2 and determine their type i.e. local min/local max/saddle point. Are there any global min/max? Partial derivatives
Aggregate Loss Models
Aggregate Loss Models Chapter 9 Stat 477 - Loss Models Chapter 9 (Stat 477) Aggregate Loss Models Brian Hartman - BYU 1 / 22 Objectives Objectives Individual risk model Collective risk model Computing
Introduction. The Aims & Objectives of the Mathematical Portion of the IBA Entry Test
Introduction The career world is competitive. The competition and the opportunities in the career world become a serious problem for students if they do not do well in Mathematics, because then they are
Solutions to Homework 10
Solutions to Homework 1 Section 7., exercise # 1 (b,d): (b) Compute the value of R f dv, where f(x, y) = y/x and R = [1, 3] [, 4]. Solution: Since f is continuous over R, f is integrable over R. Let x
STAT 430/510 Probability Lecture 14: Joint Probability Distribution, Continuous Case
STAT 430/510 Probability Lecture 14: Joint Probability Distribution, Continuous Case Pengyuan (Penelope) Wang June 20, 2011 Joint density function of continuous Random Variable When X and Y are two continuous
Chapters 5. Multivariate Probability Distributions
Chapters 5. Multivariate Probability Distributions Random vectors are collection of random variables defined on the same sample space. Whenever a collection of random variables are mentioned, they are
sheng@mail.ncyu.edu.tw 1 Content Introduction Expectation and variance of continuous random variables Normal random variables Exponential random variables Other continuous distributions The distribution
Simplify the rational expression. Find all numbers that must be excluded from the domain of the simplified rational expression.
MAC 1105 Final Review Simplify the rational expression. Find all numbers that must be excluded from the domain of the simplified rational expression. 1) 8x 2-49x + 6 x - 6 A) 1, x 6 B) 8x - 1, x 6 x -
MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Math 110 Review for Final Examination 2012 Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Match the equation to the correct graph. 1) y = -
Topic 4: Multivariate random variables. Multiple random variables
Topic 4: Multivariate random variables Joint, marginal, and conditional pmf Joint, marginal, and conditional pdf and cdf Independence Expectation, covariance, correlation Conditional expectation Two jointly
Conditional expectation
A Conditional expectation A.1 Review of conditional densities, expectations We start with the continuous case. This is sections 6.6 and 6.8 in the book. Let X, Y be continuous random variables. We defined
Sucesiones repaso Klasse 12 [86 marks]
Sucesiones repaso Klasse [8 marks] a. Consider an infinite geometric sequence with u = 40 and r =. (i) Find. u 4 (ii) Find the sum of the infinite sequence. (i) correct approach () e.g. u 4 = (40) u 4
Chapter 4 Lecture Notes
Chapter 4 Lecture Notes Random Variables October 27, 2015 1 Section 4.1 Random Variables A random variable is typically a real-valued function defined on the sample space of some experiment. For instance,
Continuous Random Variables
Probability 2 - Notes 7 Continuous Random Variables Definition. A random variable X is said to be a continuous random variable if there is a function f X (x) (the probability density function or p.d.f.)
4. Joint Distributions
Virtual Laboratories > 2. Distributions > 1 2 3 4 5 6 7 8 4. Joint Distributions Basic Theory As usual, we start with a random experiment with probability measure P on an underlying sample space. Suppose
Expectation Discrete RV - weighted average Continuous RV - use integral to take the weighted average
PHP 2510 Expectation, variance, covariance, correlation Expectation Discrete RV - weighted average Continuous RV - use integral to take the weighted average Variance Variance is the average of (X µ) 2
The Big 50 Revision Guidelines for S1
The Big 50 Revision Guidelines for S1 If you can understand all of these you ll do very well 1. Know what is meant by a statistical model and the Modelling cycle of continuous refinement 2. Understand
Econ 132 C. Health Insurance: U.S., Risk Pooling, Risk Aversion, Moral Hazard, Rand Study 7
Econ 132 C. Health Insurance: U.S., Risk Pooling, Risk Aversion, Moral Hazard, Rand Study 7 C2. Health Insurance: Risk Pooling Health insurance works by pooling individuals together to reduce the variability
Cork Institute of Technology. CIT Mathematics Examination, Paper 1 Sample Paper A
Cork Institute of Technology CIT Mathematics Examination, 2015 Paper 1 Sample Paper A Answer ALL FIVE questions. Each question is worth 20 marks. Total marks available: 100 marks. The standard Formulae
Lesson List
2016-2017 Lesson List Table of Contents Operations and Algebraic Thinking... 3 Number and Operations in Base Ten... 6 Measurement and Data... 9 Number and Operations - Fractions...10 Financial Literacy...14
Continuous Random Variables
Continuous Random Variables COMP 245 STATISTICS Dr N A Heard Contents 1 Continuous Random Variables 2 11 Introduction 2 12 Probability Density Functions 3 13 Transformations 5 2 Mean, Variance and Quantiles
Change of Continuous Random Variable
Change of Continuous Random Variable All you are responsible for from this lecture is how to implement the Engineer s Way (see page 4) to compute how the probability density function changes when we make
6.4 Logarithmic Equations and Inequalities
6.4 Logarithmic Equations and Inequalities 459 6.4 Logarithmic Equations and Inequalities In Section 6.3 we solved equations and inequalities involving exponential functions using one of two basic strategies.
MATH 201. Final ANSWERS August 12, 2016
MATH 01 Final ANSWERS August 1, 016 Part A 1. 17 points) A bag contains three different types of dice: four 6-sided dice, five 8-sided dice, and six 0-sided dice. A die is drawn from the bag and then rolled.
MATH 10: Elementary Statistics and Probability Chapter 5: Continuous Random Variables
MATH 10: Elementary Statistics and Probability Chapter 5: Continuous Random Variables Tony Pourmohamad Department of Mathematics De Anza College Spring 2015 Objectives By the end of this set of slides,
A Pricing Model for Underinsured Motorist Coverage
A Pricing Model for Underinsured Motorist Coverage by Matthew Buchalter ABSTRACT Underinsured Motorist (UIM) coverage, also known as Family Protection coverage, is a component of most Canadian personal
Math 120 Final Exam Practice Problems, Form: A
Math 120 Final Exam Practice Problems, Form: A Name: While every attempt was made to be complete in the types of problems given below, we make no guarantees about the completeness of the problems. Specifically,
Institute of Actuaries of India Subject CT3 Probability and Mathematical Statistics
Institute of Actuaries of India Subject CT3 Probability and Mathematical Statistics For 2015 Examinations Aim The aim of the Probability and Mathematical Statistics subject is to provide a grounding in
MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education)
MATH 095, College Prep Mathematics: Unit Coverage Pre-algebra topics (arithmetic skills) offered through BSE (Basic Skills Education) Accurately add, subtract, multiply, and divide whole numbers, integers,
7 CONTINUOUS PROBABILITY DISTRIBUTIONS
7 CONTINUOUS PROBABILITY DISTRIBUTIONS Chapter 7 Continuous Probability Distributions Objectives After studying this chapter you should understand the use of continuous probability distributions and the