# Math 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 2 Solutions

Save this PDF as:

Size: px
Start display at page:

Download "Math 370/408, Spring 2008 Prof. A.J. Hildebrand. Actuarial Exam Practice Problem Set 2 Solutions"

## Transcription

1 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, grouped by topic, and ordered by difficulty. All of the exams have appeared on the SOA website at some point in the past, though most of them are no longer there. The exams are copy-righted by the SOA. Note on the topics covered: This problem set covers discrete random variables (Chapter 2 of Hogg/Tanis). Note on the ordering of the problems: The problems are loosely grouped by topic, and very roughly ordered by difficulty (though that ordering is very subjective). The last group of problems (those with triple digit numbers) are harder, either because they are conceptually more difficult, or simply because they require lengthy computations and may take two or three times as long to solve than an average problem. Answers and solutions: I will post an answer key and hints/solutions on the course webpage, hildebr/70.

2 Math 70/408 Spring [2-] An insurance policy pays an individual 00 per day for up to days of hospitalization and 25 per day for each day of hospitalization thereafter. The number of days of hospitalization, X, is a discrete random variable with probability function { 6 k P (X = k) = 5 for k =, 2,, 4, 5, 0 otherwise. Calculate the expected payment for hospitalization under this policy. (A) 85 (B) 6 (C) 68 (D) 2 (E) 255 Answer: D: 2 Solution: A fairly easy exercise in discrete densities; just make a table of the total payment for each value of k, along with the associated probability P (X = k), and then compute the expected value of this payment. 2. [2-2] An actuary determines that the claim size for a certain class of accidents is a random variable, X, with moment generating function M X (t) = ( 2500t) 4. Determine the standard deviation of the claim size for this class of accidents. (A), 40 (B) 5, 000 (C) 8, 660 (D) 0, 000 (E), 80 Answer: B: 5,000 Solution: Use the formulas M (0) = E(X), M (0) = E(X 2 ).. [2-] An actuary has discovered that policyholders are three times as likely to file two claims as to file four claims. If the number of claims filed has a Poisson distribution, what is the variance of the number of claims filed? Answer: D: 2 (A) / (B) (C) 2 (D) 2 (E) 4 Solution: Let X denote the number of claims. We are given that X has Poisson distribution and that P (X = 2) = P (X = 4). Substitute the formula for a Poisson p.m.f., P (X = x) = e λ λ x /x! into this equation to determine λ. Then use the fact that σ 2 = λ for a Poisson distribution with parameter λ. 4. [2-4] In modeling the number of claims filed by an individual under an automobile policy during a three-year period, an actuary makes the simplifying assumption that for all integers n 0, p n+ = (/5)p n, where p n represents the probability that the policyholder files n claims during the period. Under this assumption, what is the probability that a policyholder files more than one claim during the period? 2

3 Math 70/408 Spring 2008 (A) 0.04 (B) 0.6 (C) 0.20 (D) 0.80 (E) 0.96 Answer: A: 0.04 Solution: The given relation for p n implies that p n = (/5)p n = (/5) 2 p n 2 = = (/5) n p 0 for n = 0,, 2,... Since these probabilities must add up to, we have = n=0 (/5)n p 0 = ( /5) p 0 = (5/4)p 0 (using the formula for the sum of a geometric series), so p 0 = 4/5. The probability to compute is p 0 p = 4/5 4/25 = [2-5] As part of the underwriting process for insurance, each prospective policyholder is tested for high blood pressure. Let X represent the number of tests completed when the first person with high blood pressure is found. The expected value of X is 2.5. Calculate the probability that the sixth person tested is the first one with high blood pressure. (A) (B) 0.05 (C) (D) 0.6 (E) 0.94 Answer: B: 0.05 Solution: By the formulas for a geometric distribution, µ = /p, so p = /µ = /2.5, and P (X = 6) = ( p) 5 p = (.5/2.5) 5 (/2.5) = [2-6] Let X be a random variable with moment generating function Calculate the variance of X. Answer: A: 2 ( ) 2 + e t 9 M(t) =, < t < +. (A) 2 (B) (C) 8 (D) 9 (E) Solution: Use the formulas E(X) = M (0) and E(X 2 ) = M (0). Be careful when computing the derivatives of M(t), using the chain rule! 7. [2-7] A small commuter plane has 0 seats. The probability that any particular passenger will not show up for a flight is 0.0, independent of other passengers. The airline sells 2 tickets for the flight. Calculate the probability that more passengers show up for the flight than there are seats available. (A) (B) 0.04 (C) (D) 0.22 (E) Answer: E: Solution: Consider the 2 tickets as 2 Bernoulli (success/failure) trials with success meaning that the passenger holding the ticket shows up (so that p = 0.9. The probability to compute is then that of getting or 2 successes in 2 such trials, which is given by the binomial distribution. 8. [2-8] The distribution of loss due to fire damage to a warehouse is:

4 Math 70/408 Spring 2008 Amount of loss Probability , , , , Given that a loss is greater than zero, calculate the expected amount of the loss. (A) 290 (B) 22 (C), 704 (D) 2, 900 (E) 2, 222 Answer: D: 2,900 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. Since the probability for a loss of 0 is 0.9, the remaining probabilities must add up to 0., so dividing each by 0. renormalizes them to have sum. Using these renormalized probabilities in the formula for an expectation gives the desired expectation under the above condition. 9. [2-9] Let X, X 2, X be a random sample from a discrete distribution with probability function for x = 0, 2 p(x) = for x =, 0 otherwise. Determine the moment generating function, M(t), of Y = X X 2 X. (A) et (B) + 2e t (C) ( + 2 et) (D) et (E) + 2 et Answer: A Solution: This is easier than it looks at first glance, since Y = X X 2 X takes on only values 0 and, and Y = occurs if and only if all of X, X 2, X are equal to. The latter occurs with probability (2/) = 8/27, P (Y = ) = 8/27 and P (Y = 0) = 8/27 = 9/27, and therefore M(t) = e 0t P (Y = 0) + e t P (Y = ) = (9/27) + e t (8/27). 0. [2-5] 4

5 Math 70/408 Spring 2008 The number of injury claims per month is modeled by a random variable N with P (N = n) =, where n 0. (n + )(n + 2) Determine the probability of at least one claim during a particular month, given that there have been at most four claims during that month. Answer: B: 2/5 (A) / (B) 2/5 (C) /2 (D) /5 (E) 5/6 Solution: This is a very easy problem: We need to compute P (N N 4), which is the same as P ( N 4)/P (N 4). Now, P ( N 4) = P (N 4) = 4 n= 4 n=0 P (N = n) = =, P (N = n) = = 5 6. Dividing the first probability by the second gives the answer: (/)/(5/6) = 2/5.. [2-52] An insurance company determines that N, the number of claims received in a week, is a random variable with P (N = n) = 2 n, where n 0. The company also determines that the number of claims received in a given week is independent of the number of claims received in any other week. Determine the probability that exactly seven claims will be received during a given two-week period. (A) 256 (B) 28 (C) 7 52 (D) 64 (E) 2 Answer: D: /64 Solution: Let N and N 2, denote the number of claims during the first, respectively second, week. The total number of claims received during the two-week period then is N + N 2, and so we need to compute P (N + N 2 = 7). By breaking this probability down into cases according to the values of (N, N 2 ), and using the independence of N and N 2 and the given distribution, we get 2. [2-5] P (N + N 2 = 7) = = 7 P (N = n, N 2 = 7 n) = n=0 7 n=0 2 9 = = n 2 (7 n) A company prices its hurricane insurance using the following assumptions: n=0 (i) In any calendar year, there can be at most one hurricane. (ii) In any calendar year, the probability of a hurricane is (iii) The number of hurricanes in any calendar year is independent of the number of hurricanes in any other calendar year. 5

6 Math 70/408 Spring 2008 Using the company s assumptions, calculate the probability that there are fewer than hurricanes in a 20-year period. (A) 0.06 (B) 0.9 (C) 0.8 (D) 0.62 (E) 0.92 Answer: E: 0.92 Solution: By the given assumptions the occurrence of hurricanes can be modeled as success/failure trials, with success meaning that a hurricane occurs in a given year and occurring with probability p = Thus, the probability that there are fewer than hurricanes in a 20-year period is equal to the probability of having less than successes in 20 success/failure trials with p = By the binomial distribution, this probability is. [2-54] 2 x=0 ( ) x x x = = An insurance policy on an electrical device pays a benefit of 4000 if the device fails during the first year. The amount of the benefit decreases by 000 each successive year until it reaches 0. If the device has not failed by the beginning of any given year, the probability of failure during that year is 0.4. What is the expected benefit under this policy? (A) 224 (B) 2400 (C) 2500 (D) 2667 (E) 2694 Answer: E: 2694 Solution: Let X denote the year in which the device fails, and Y the benefit. Then 4000 if X =, 000 if X = 2, Y = 2000 if X =, 000 if X = 4, 0 if X 5. We need to compute E(Y ). This is easy once we know the probabilities P (X = n) for n =, 2,, 4. Now P (X = n) is the probability that the device is working during years, 2,..., n and fails during year n. If we assume that these probabilities behave like success/failure probabilities in a sequence of independent success/failure trials, then we would have P (X = n) = ( p) n p, with p, the failure probability in a given year, equal to 0.4, so P (X = n) = 0.6 n 0.4, n =, 2,... (*) With this assumption we get E(Y ) = 4, , , , = which is the correct answer. The above argument, while leading to the correct answer, requires additional justification since the above assumption, while natural, is not directly stated in the problem. Instead the assumption made in the problem is contained in the phrase: If the device has not failed by the beginning of any given year, the probability of failure during that year is 0.4. In terms of 6

7 Math 70/408 Spring 2008 the random variable X (the year of failure), this translates into P (X = n X n) = 0.4, or, equivalently ( ) P (X = n)/p (X n) = 0.4. From this we can derive the probabilities ( ): Setting n =, 2,,... in ( ), we get P (X = ) = 0.4P (X ) = 0.4 = 0.4, P (X = 2) = 0.4P (X 2) = 0.4( P (X = )) = 0.4( 0.4) = , n =, argument gives P (X = ) = 0.4( ) = , etc. 4. [2-55] An insurance company sells a one-year automobile policy with a deductible of 2. The probability that the insured will incur a loss is If there is a loss, the probability of a loss of amount N is K/N, for N =,..., 5 and K a constant. These are the only possible loss amounts and no more than one loss can occur. Determine the net premium for this policy. (A) 0.0 (B) (C) (D) 0.0 (E) 0.50 Answer: A: 0.0 Solution: The net premium is the premium (per policy) at which the company breaks even. This is equal to the expected insurance payout (per policy). Taking into account the deductible, the payout is if N = ; 2 if N = 4; if N = 5; and 0 in all other cases. Thus, the expected payout is P (N = ) + 2P (N = 4) + P (N = 5) = K + 2K 4 + K 5 = 4 0 K. To determine the constant K, we use the fact that the (overall) probability of a loss is Thus, the sum of the probabilities for the possible loss amounts has to equal 0.05: 0.05 = 5 N= K N = 7 60 K. Hence K = /7, and the above expectation betcomes (4/0)(/7) = [2-56] A probability distribution of the claim sizes for an auto insurance policy is given in the table below: Claim size Probability

8 Math 70/408 Spring 2008 What percentage of the claims are within one standard deviation of the mean claim size? (A) 45% (B) 55% (C) 68% (D) 85% (E) 00% Answer: A: 0.45 Solution: A routine, but rather lengthy computation: First find µ = E(X), E(X 2 ), Var(X), and σ. Then add up the probabilities for those x-values that satisfy x µ σ. 6. [2-60] Two life insurance policies, each with a death benefit of 0, 000 and a one-time premium of 500, are sold to a couple, one for each person. The policies will expire at the end of the tenth year. The probability that only the wife will survive at least ten years is 0.025, the probability that only the husband will survive at least ten years is 0.0, and the probability that both of them will survive at least ten years is What is the expected excess of premiums over claims, given that the husband survives at least ten years? Answer: E: 897 (A) 50 (B) 85 (C) 97 (D) 870 (E) 897 Solution: Let X denote the excess of premiums over claims. Then we have to compute E(X husband survives), the conditional expectation of X given that the husband survives. We have two cases to consider: (a) Only the husband survives. This occurs with probability 0.0. In this case the claims are 0000, while the premiums collected are = 000. Thus, X = = (b) Both husband and wife survive. This occurs with probability In this case the claims are 0, so with the premiums being as before 000, we have X = = 000. The probability that the husband survives is the sum of the above probabilities, i.e., = Hence, the desired conditional expectation is 7. [2-0] E(X husband survives) = ( 9000)(0.0) + (000)(0.96) 0.97 = A company establishes a fund of 20 from which it wants to pay an amount, C, to any of its 20 employees who achieve a high performance level during the coming year. Each employee has a 2% chance of achieving a high performance level during the coming year, independent of any other employee. Determine the maximum value of C for which the probability is less than % that the fund will be inadequate to cover all payments for high performance. Answer: D: 60 (A) 24 (B) 0 (C) 40 (D) 60 (E) 20 Solution: At first glance this seems to be a Central Limit Theorem problem, but it actually is just a trial-and-error calculation with the binomial distribution. In order to determine C we need to find the minimal integer n such that the probability that there are more than n high performance employees is below 0.0. The maximal value of C then is 20/n. 8

9 Math 70/408 Spring 2008 Let X denote the number of high performance employees. Then X has binomial distribution with n = 20 and p = 0.02, and we can calculate: P (X = 0) = = 0.668, P (X = ) = = 0.272, P (X = 2) = ( ) = Since the sum of all three of these probabilities, = 0.99, exceeds 0.99, but the sum of the first two alone, = 0.94, does not, n = 2 is the minimal integer such that P (X > 2) is less than 0.0. Hence C = 20/2 = [2-02] A company takes out an insurance policy to cover accidents that occur at its manufacturing plant. The probability that one or more accidents will occur during any given month is /5. The number of accidents that occur in any given month is independent of the number of accidents that occur in all other months. Calculate the probability that there will be at least four months in which no accidents occur before the fourth month in which at least one accident occurs. (A) 0.0 (B) 0.2 (C) 0.2 (D) 0.29 (E) 0.4 Answer: D: 0.29 Solution: The given situation can be modelled as a series of independent success/failure trials, with the trials corresponding to the months, and success meaning that one or more accidents have occurred in that month, so that the success probability is p = /5. Thus, in the language of this model, we seek the probability that there will be at least 4 failures before the 4-th success occurs. Now the event at least 4 failures occur before the 4-th success is exactly equivalent to 4-th success occurs at trial 8 or later (think about it!). We compute the probability of the latter event by considering its complement, 4-th success occurs at trial 7 or earlier. The latter event splits into four disjoint events, A n = success in trial n, exactly successes in trials,..., n, n = 4, 5, 6, 7. Now ( ) n P (A n ) = p P ( successes in n trials) = p p ( p) n 4 ( ) n = (/5) 4 (2/5) n 4. Hence, the probability we seek is 9. [2-0] 7 P (A n ) = n=4 ( ) ( ) ( ) = ( ) = ( ) A company buys a policy to insure its revenue in the event of major snowstorms that shut down business. The policy pays nothing for the first such snowstorm of the year and 0, 000 for each one therefore, until the end of the year. The number of major snowstorms per year that shut down business is assumed to have Poisson distribution with mean.5. What is the expected amount paid to the company under this policy during a one-year period? (A) 2, 769 (B) 5, 000 (C) 7, 2 (D) 8, 47 (E) 0, 578 9

10 Math 70/408 Spring 2008 Answer: C: 7,2 Solution: The main difficulty in this problem is that it requires a slightly tricky manipulation of the exponential series. Let X denote the number of snowstorms, and let Y denote the insurance payout, measured in units of 0, 000. We need to calculate E(Y ). From the given information, X is Poisson distributed with mean.5 and Y is related to X by { X if X =, 2,..., Y = 0 if X = 0. Hence, E(Y ) = (x )P (X = x) = (x )e x= = e.5 ( x= x.5x x! x= ).5 x x= = e ( x (x )! x= = e.5 (.5e.5 e.5 + ) = 0.5 e.5 = x! ( Since the units of Y are 0, 000, we get 7, 2 as answer. 20. [2-04] x=0.5 x x!.5.5x A study is being conducted in which the health of two independent groups of ten policyholders is being monitored over a one-year period of time. Individual participants in the study drop out before the end of the study with probability 0.2 (independently of the other participants). What is the probability that at least 9 participants complete the study in one of the two groups, but not in both groups? (A) (B) 0.92 (C) 0.25 (D) 0.76 (E) Answer: E: Solution: This is a two stage problem. For the first stage, focus on a single group, and consider the probability, say q, that in this group at least 9 participants complete the study. Then q is the probability that in 0 success/failure trials with success probability p = 0.8 there are at least 9 successes, so q can be computed using the binomial distribution: q = ( 0 9 ) = For the second stage, we consider two such groups, and compute the probability that exactly one of them is successful in the above sense. This probability is given by 2q( q) = , as can be seen either by considering separately the cases group successful, group 2 unsuccessful and group unsuccessful, group 2 successful, or by another application of the success/trial model, this time with 2 trials and success probability q. x! )) 0

### 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 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 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 370/408, 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

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, Actuarial Problemsolving Spring 2008 A.J. Hildebrand. Practice Test, 1/28/2008 (with solutions)

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

### 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,

### 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

### 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

### 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

### ECE302 Spring 2006 HW4 Solutions February 6, 2006 1

ECE302 Spring 2006 HW4 Solutions February 6, 2006 1 Solutions to HW4 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

### 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

### ECE302 Spring 2006 HW3 Solutions February 2, 2006 1

ECE302 Spring 2006 HW3 Solutions February 2, 2006 1 Solutions to HW3 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

### Chapter 3: DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS

Chapter 3: DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Part 4: Geometric Distribution Negative Binomial Distribution Hypergeometric Distribution Sections 3-7, 3-8 The remaining discrete random

### Premium Calculation. Lecture: Weeks 12-14. Lecture: Weeks 12-14 (STT 455) Premium Calculation Fall 2014 - Valdez 1 / 31

Premium Calculation Lecture: Weeks 12-14 Lecture: Weeks 12-14 (STT 455) Premium Calculation Fall 2014 - Valdez 1 / 31 Preliminaries Preliminaries An insurance policy (life insurance or life annuity) is

TABLE OF CONTENTS 1. Survival A. Time of Death for a Person Aged x 1 B. Force of Mortality 7 C. Life Tables and the Deterministic Survivorship Group 19 D. Life Table Characteristics: Expectation of Life

### Premium Calculation. Lecture: Weeks 12-14. Lecture: Weeks 12-14 (Math 3630) Annuities Fall 2015 - Valdez 1 / 32

Premium Calculation Lecture: Weeks 12-14 Lecture: Weeks 12-14 (Math 3630) Annuities Fall 2015 - Valdez 1 / 32 Preliminaries Preliminaries An insurance policy (life insurance or life annuity) is funded

### 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

### 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

### 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

### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.

Ch. 4 Discrete Probability Distributions 4.1 Probability Distributions 1 Decide if a Random Variable is Discrete or Continuous 1) State whether the variable is discrete or continuous. The number of cups

### 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.

### Homework 4 - KEY. Jeff Brenion. June 16, 2004. Note: Many problems can be solved in more than one way; we present only a single solution here.

Homework 4 - KEY Jeff Brenion June 16, 2004 Note: Many problems can be solved in more than one way; we present only a single solution here. 1 Problem 2-1 Since there can be anywhere from 0 to 4 aces, the

### 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

### 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,

### 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

### 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

### November 2012 Course MLC Examination, Problem No. 1 For two lives, (80) and (90), with independent future lifetimes, you are given: k p 80+k

Solutions to the November 202 Course MLC Examination by Krzysztof Ostaszewski, http://www.krzysio.net, krzysio@krzysio.net Copyright 202 by Krzysztof Ostaszewski All rights reserved. No reproduction in

### Mathematics of Risk. Introduction. Case Study #1 Personal Auto Insurance Pricing. Mathematical Concepts Illustrated. Background

Mathematics of Risk Introduction There are many mechanisms that individuals and organizations use to protect themselves against the risk of financial loss. Government organizations and public and private

### 2. Discrete random variables

2. Discrete random variables Statistics and probability: 2-1 If the chance outcome of the experiment is a number, it is called a random variable. Discrete random variable: the possible outcomes can be

### Chapter 5. Discrete Probability Distributions

Chapter 5. Discrete Probability Distributions Chapter Problem: Did Mendel s result from plant hybridization experiments contradicts his theory? 1. Mendel s theory says that when there are two inheritable

### Chapter 3: DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS. Part 3: Discrete Uniform Distribution Binomial Distribution

Chapter 3: DISCRETE RANDOM VARIABLES AND PROBABILITY DISTRIBUTIONS Part 3: Discrete Uniform Distribution Binomial Distribution Sections 3-5, 3-6 Special discrete random variable distributions we will cover

### LECTURE 16. Readings: Section 5.1. Lecture outline. Random processes Definition of the Bernoulli process Basic properties of the Bernoulli process

LECTURE 16 Readings: Section 5.1 Lecture outline Random processes Definition of the Bernoulli process Basic properties of the Bernoulli process Number of successes Distribution of interarrival times The

### Math 408, Actuarial Statistics I, Spring 2008. Solutions to combinatorial problems

, Spring 2008 Word counting problems 1. Find the number of possible character passwords under the following restrictions: Note there are 26 letters in the alphabet. a All characters must be lower case

### COURSE 1 REVISED SAMPLE EXAM

COURSE REVISED SAMPLE EXAM A table of values for the normal distribution will be provided with the Course Exam. Revised August 999 Problem # A marketing survey indicates that 60% of the population owns

### 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

### May 2012 Course MLC Examination, Problem No. 1 For a 2-year select and ultimate mortality model, you are given:

Solutions to the May 2012 Course MLC Examination by Krzysztof Ostaszewski, http://www.krzysio.net, krzysio@krzysio.net Copyright 2012 by Krzysztof Ostaszewski All rights reserved. No reproduction in any

### The Binomial Probability Distribution

The Binomial Probability Distribution MATH 130, Elements of Statistics I J. Robert Buchanan Department of Mathematics Fall 2015 Objectives After this lesson we will be able to: determine whether a probability

### 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...................................

### 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

### SOCIETY OF ACTUARIES. EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE QUESTIONS

SOCIETY OF ACTUARIES EXAM MLC Models for Life Contingencies EXAM MLC SAMPLE QUESTIONS The following questions or solutions have been modified since this document was prepared to use with the syllabus effective

### ACMS 10140 Section 02 Elements of Statistics October 28, 2010. Midterm Examination II

ACMS 10140 Section 02 Elements of Statistics October 28, 2010 Midterm Examination II Name DO NOT remove this answer page. DO turn in the entire exam. Make sure that you have all ten (10) pages of the examination

### **BEGINNING OF EXAMINATION**

November 00 Course 3 Society of Actuaries **BEGINNING OF EXAMINATION**. You are given: R = S T µ x 0. 04, 0 < x < 40 0. 05, x > 40 Calculate e o 5: 5. (A) 4.0 (B) 4.4 (C) 4.8 (D) 5. (E) 5.6 Course 3: November

### 0 x = 0.30 x = 1.10 x = 3.05 x = 4.15 x = 6 0.4 x = 12. f(x) =

. A mail-order computer business has si telephone lines. Let X denote the number of lines in use at a specified time. Suppose the pmf of X is as given in the accompanying table. 0 2 3 4 5 6 p(.0.5.20.25.20.06.04

### 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 <

### Probability and Statistics Prof. Dr. Somesh Kumar Department of Mathematics Indian Institute of Technology, Kharagpur

Probability and Statistics Prof. Dr. Somesh Kumar Department of Mathematics Indian Institute of Technology, Kharagpur Module No. #01 Lecture No. #15 Special Distributions-VI Today, I am going to introduce

### Tenth Problem Assignment

EECS 40 Due on April 6, 007 PROBLEM (8 points) Dave is taking a multiple-choice exam. You may assume that the number of questions is infinite. Simultaneously, but independently, his conscious and subconscious

### MATH 3630 Actuarial Mathematics I Class Test 2 Wednesday, 17 November 2010 Time Allowed: 1 hour Total Marks: 100 points

MATH 3630 Actuarial Mathematics I Class Test 2 Wednesday, 17 November 2010 Time Allowed: 1 hour Total Marks: 100 points Please write your name and student number at the spaces provided: Name: Student ID:

### RANDOM VARIABLES MATH CIRCLE (ADVANCED) 3/3/2013. 3 k) ( 52 3 )

RANDOM VARIABLES MATH CIRCLE (ADVANCED) //0 0) a) Suppose you flip a fair coin times. i) What is the probability you get 0 heads? ii) head? iii) heads? iv) heads? For = 0,,,, P ( Heads) = ( ) b) Suppose

### JANUARY 2016 EXAMINATIONS. Life Insurance I

PAPER CODE NO. MATH 273 EXAMINER: Dr. C. Boado-Penas TEL.NO. 44026 DEPARTMENT: Mathematical Sciences JANUARY 2016 EXAMINATIONS Life Insurance I Time allowed: Two and a half hours INSTRUCTIONS TO CANDIDATES:

### Further Topics in Actuarial Mathematics: Premium Reserves. Matthew Mikola

Further Topics in Actuarial Mathematics: Premium Reserves Matthew Mikola April 26, 2007 Contents 1 Introduction 1 1.1 Expected Loss...................................... 2 1.2 An Overview of the Project...............................

### Fifth Problem Assignment

EECS 40 PROBLEM (24 points) Discrete random variable X is described by the PMF { K x p X (x) = 2, if x = 0,, 2 0, for all other values of x Due on Feb 9, 2007 Let D, D 2,..., D N represent N successive

### WHERE DOES THE 10% CONDITION COME FROM?

1 WHERE DOES THE 10% CONDITION COME FROM? The text has mentioned The 10% Condition (at least) twice so far: p. 407 Bernoulli trials must be independent. If that assumption is violated, it is still okay

### ACMS 10140 Section 02 Elements of Statistics October 28, 2010 Midterm Examination II Answers

ACMS 10140 Section 02 Elements of Statistics October 28, 2010 Midterm Examination II Answers Name DO NOT remove this answer page. DO turn in the entire exam. Make sure that you have all ten (10) pages

### 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

### Statistics 100A Homework 4 Solutions

Chapter 4 Statistics 00A Homework 4 Solutions Ryan Rosario 39. A ball is drawn from an urn containing 3 white and 3 black balls. After the ball is drawn, it is then replaced and another ball is drawn.

### Master s Theory Exam Spring 2006

Spring 2006 This exam contains 7 questions. You should attempt them all. Each question is divided into parts to help lead you through the material. You should attempt to complete as much of each problem

Premium Calculation - continued Lecture: Weeks 1-2 Lecture: Weeks 1-2 (STT 456) Premium Calculation Spring 2015 - Valdez 1 / 16 Recall some preliminaries Recall some preliminaries An insurance policy (life

### Probability and Expected Value

Probability and Expected Value This handout provides an introduction to probability and expected value. Some of you may already be familiar with some of these topics. Probability and expected value are

### Contents. TTM4155: Teletraffic Theory (Teletrafikkteori) Probability Theory Basics. Yuming Jiang. Basic Concepts Random Variables

TTM4155: Teletraffic Theory (Teletrafikkteori) Probability Theory Basics Yuming Jiang 1 Some figures taken from the web. Contents Basic Concepts Random Variables Discrete Random Variables Continuous Random

### Heriot-Watt University. BSc in Actuarial Mathematics and Statistics. Life Insurance Mathematics I. Extra Problems: Multiple Choice

Heriot-Watt University BSc in Actuarial Mathematics and Statistics Life Insurance Mathematics I Extra Problems: Multiple Choice These problems have been taken from Faculty and Institute of Actuaries exams.

### The number of phone calls to the attendance office of a high school on any given school day A) continuous B) discrete

Exam Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Provide an appropriate response. 1) State whether the variable is discrete or continuous.

### CHAPTER 6: Continuous Uniform Distribution: 6.1. Definition: The density function of the continuous random variable X on the interval [A, B] is.

Some Continuous Probability Distributions CHAPTER 6: Continuous Uniform Distribution: 6. Definition: The density function of the continuous random variable X on the interval [A, B] is B A A x B f(x; A,

### Some Observations on Variance and Risk

Some Observations on Variance and Risk 1 Introduction By K.K.Dharni Pradip Kumar 1.1 In most actuarial contexts some or all of the cash flows in a contract are uncertain and depend on the death or survival

### P(X = x k ) = 1 = k=1

74 CHAPTER 6. IMPORTANT DISTRIBUTIONS AND DENSITIES 6.2 Problems 5.1.1 Which are modeled with a unifm distribution? (a Yes, P(X k 1/6 f k 1,...,6. (b No, this has a binomial distribution. (c Yes, P(X k

### We hope you will find this sample a useful supplement to your existing educational materials, and we look forward to receiving your comments.

To Teachers of Mathematics: Thank you for visiting the BeAnActuary.org booth at the annual meeting of the National Council of Teachers of Mathematics. BeAnActuary.org is sponsored by the Joint Career Encouragement

### Poisson processes (and mixture distributions)

Poisson processes (and mixture distributions) James W. Daniel Austin Actuarial Seminars www.actuarialseminars.com June 26, 2008 c Copyright 2007 by James W. Daniel; reproduction in whole or in part without

### 39.2. The Normal Approximation to the Binomial Distribution. Introduction. Prerequisites. Learning Outcomes

The Normal Approximation to the Binomial Distribution 39.2 Introduction We have already seen that the Poisson distribution can be used to approximate the binomial distribution for large values of n and

### Lahore University of Management Sciences

Lahore University of Management Sciences CMPE 501: Applied Probability (Fall 2010) Homework 3: Solution 1. A candy factory has an endless supply of red, orange, yellow, green, blue and violet jelly beans.

### Manual for SOA Exam MLC.

Chapter 4. Life Insurance. c 29. Miguel A. Arcones. All rights reserved. Extract from: Arcones Manual for the SOA Exam MLC. Fall 29 Edition. available at http://www.actexmadriver.com/ c 29. Miguel A. Arcones.

### Section 1.5 Linear Models

Section 1.5 Linear Models Some real-life problems can be modeled using linear equations. Now that we know how to find the slope of a line, the equation of a line, and the point of intersection of two lines,

### 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

### Section 5 Part 2. Probability Distributions for Discrete Random Variables

Section 5 Part 2 Probability Distributions for Discrete Random Variables Review and Overview So far we ve covered the following probability and probability distribution topics Probability rules Probability

### Question: What is the probability that a five-card poker hand contains a flush, that is, five cards of the same suit?

ECS20 Discrete Mathematics Quarter: Spring 2007 Instructor: John Steinberger Assistant: Sophie Engle (prepared by Sophie Engle) Homework 8 Hints Due Wednesday June 6 th 2007 Section 6.1 #16 What is the

### Some special discrete probability distributions

University of California, Los Angeles Department of Statistics Statistics 100A Instructor: Nicolas Christou Some special discrete probability distributions Bernoulli random variable: It is a variable that

### Topic 8 The Expected Value

Topic 8 The Expected Value Functions of Random Variables 1 / 12 Outline Names for Eg(X ) Variance and Standard Deviation Independence Covariance and Correlation 2 / 12 Names for Eg(X ) If g(x) = x, then

### 3 Multiple Discrete Random Variables

3 Multiple Discrete Random Variables 3.1 Joint densities Suppose we have a probability space (Ω, F,P) and now we have two discrete random variables X and Y on it. They have probability mass functions f

### Ch5: Discrete Probability Distributions Section 5-1: Probability Distribution

Recall: Ch5: Discrete Probability Distributions Section 5-1: Probability Distribution A variable is a characteristic or attribute that can assume different values. o Various letters of the alphabet (e.g.

### MAT 155. Key Concept. September 27, 2010. 155S5.5_3 Poisson Probability Distributions. Chapter 5 Probability Distributions

MAT 155 Dr. Claude Moore Cape Fear Community College Chapter 5 Probability Distributions 5 1 Review and Preview 5 2 Random Variables 5 3 Binomial Probability Distributions 5 4 Mean, Variance and Standard

### Manual for SOA Exam MLC.

Chapter 4. Life insurance. Extract from: Arcones Fall 2009 Edition, available at http://www.actexmadriver.com/ (#1, Exam M, Fall 2005) For a special whole life insurance on (x), you are given: (i) Z is

### 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

### 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

### Section 6.1 Discrete Random variables Probability Distribution

Section 6.1 Discrete Random variables Probability Distribution Definitions a) Random variable is a variable whose values are determined by chance. b) Discrete Probability distribution consists of the values

### Manual for SOA Exam MLC.

Chapter 4. Life Insurance. Extract from: Arcones Manual for the SOA Exam MLC. Fall 2009 Edition. available at http://www.actexmadriver.com/ 1/14 Level benefit insurance in the continuous case In this chapter,

### Lecture Notes on Actuarial Mathematics

Lecture Notes on Actuarial Mathematics Jerry Alan Veeh May 1, 2006 Copyright 2006 Jerry Alan Veeh. All rights reserved. 0. Introduction The objective of these notes is to present the basic aspects of the

### CHAPTER 7 SECTION 5: RANDOM VARIABLES AND DISCRETE PROBABILITY DISTRIBUTIONS

CHAPTER 7 SECTION 5: RANDOM VARIABLES AND DISCRETE PROBABILITY DISTRIBUTIONS TRUE/FALSE 235. The Poisson probability distribution is a continuous probability distribution. F 236. In a Poisson distribution,

### Practice Exam 1. x l x d x 50 1000 20 51 52 35 53 37

Practice Eam. You are given: (i) The following life table. (ii) 2q 52.758. l d 5 2 5 52 35 53 37 Determine d 5. (A) 2 (B) 2 (C) 22 (D) 24 (E) 26 2. For a Continuing Care Retirement Community, you are given

### Practice Problems #4

Practice Problems #4 PRACTICE PROBLEMS FOR HOMEWORK 4 (1) Read section 2.5 of the text. (2) Solve the practice problems below. (3) Open Homework Assignment #4, solve the problems, and submit multiple-choice

### STAT 315: HOW TO CHOOSE A DISTRIBUTION FOR A RANDOM VARIABLE

STAT 315: HOW TO CHOOSE A DISTRIBUTION FOR A RANDOM VARIABLE TROY BUTLER 1. Random variables and distributions We are often presented with descriptions of problems involving some level of uncertainty about

### Normal distribution Approximating binomial distribution by normal 2.10 Central Limit Theorem

1.1.2 Normal distribution 1.1.3 Approimating binomial distribution by normal 2.1 Central Limit Theorem Prof. Tesler Math 283 October 22, 214 Prof. Tesler 1.1.2-3, 2.1 Normal distribution Math 283 / October

### Selecting an Insurance Provider

Insurance Selecting an Insurance Provider A responsible agent or broker will: Take time to clearly explain your policies and coverage. Review your insurance needs annually to see that your policies are

### Review the following from Chapter 5

Bluman, Chapter 6 1 Review the following from Chapter 5 A surgical procedure has an 85% chance of success and a doctor performs the procedure on 10 patients, find the following: a) The probability that

### INSTITUTE OF ACTUARIES OF INDIA

INSTITUTE OF ACTUARIES OF INDIA EXAMINATIONS 17 th November 2011 Subject CT5 General Insurance, Life and Health Contingencies Time allowed: Three Hours (10.00 13.00 Hrs) Total Marks: 100 INSTRUCTIONS TO

### 4. Life Insurance Payments

4. Life Insurance Payments A life insurance premium must take into account the following factors 1. The amount to be paid upon the death of the insured person and its present value. 2. The distribution

### Notes on Continuous Random Variables

Notes on Continuous Random Variables Continuous random variables are random quantities that are measured on a continuous scale. They can usually take on any value over some interval, which distinguishes

### Developing a New Freshman Mathematics Course for Biology and Environmental Sciences Majors. Department of Mathematics December 2015

Developing a New Freshman Mathematics Course for Biology and Environmental Sciences Majors. Department of Mathematics December 2015 Programs changing to allow Math 1044 as an alternate to Math 1042: GSTC