problemset5_a

problemset5_a - 1. The exponential distribution with mean 1...

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Unformatted text preview: 1. The exponential distribution with mean 1 is being used as the model for a loss distribution. An actuary attempts to "discretize" the distribution by assigning a probability to k + % for k = 0, 1, 2, . The probability assigned to h +é- is PU: < X g k + 1] ,where X is the exponential random variable with mean I. Find the mean of the discretized distribution. A) 1.00 B) 1.02 C) 1.04 D) 1.06 B) 1.08 2. The frequency distribution is Poisson with a mean of 20 and the severity distribution is exponential with a mean of 100. A deductible cf 20 is imposed on each individual loss. Find the mean and variance of the aggregate payment made by the insurer. 3. The frequency distribution N has a negative binomial distribution with r = 3 and ,6 m 2 . The severity distribution X is Pareto with parameters a = 3 and t? = 200 . A deductible of 100 is applied to every individual loss. Find the mean of the aggregate payment. 4. A company has 1500 employees. The insurer for the company's health plan has noticed that the employees can be divided into three groups with regard to likelihood and amount of claim. Ciaim Amount Distribution Probability Group of Claim Mean Variance 1 .02 100 100 2 .01 200 200 3 .01 500 1000 The insurer charges EiS] + %10§1 , where S is the aggregate claim random variable, and calculates the premium to be 4,454.40 . The insurer later notices that the mean claim amount for group 3 should actually have been 400, and recalcuiates the premium to be 4,076.20 . How many employees are in group 1'? A) 50 B) 100 C) 200 D) 400 E) 800 5. An insurer has a portfolio of 500 policies described below: Number in Pmbabifity _ Claim. Category Category “9M1. 13mm 1 m .01 a=1c,aE=9 2 SOD—m .04 ,u=3,o‘2=l The insurer wiil charge a premium of E [S] + t/VarLS‘] , where S is the aggregate ciaim random variable. What is the maximum possible premium that will be charged? A) 74.25 13) 74.5 C) 74.75 D) 75.0 a) 75.25 6. (SOA) The number of claims, N , made on an insurance portfolio follows the following distribution: If a claim occurs, the benefit is 0 or 10 with probability 0.8 and 0.2, respectively. The number of claims and the benefit for each claim are independent. Calculate the probability that aggregate benefits will exceed expected benefits by more than 2 standard deviations. A) .02 B) .05 C) .07 D) .09 E) .12 7. (SOA) You own a fancy Eight bulb factory. Your workforce is a bit clumsy w» they keep dropping boxes of light bulbs. The boxes have varying numbers of light bulbs in them, and when dropped, the entire box is destroyed. You are given: Expected number of boxes dropped per month: 50 Variance of the number of boxes dropped per month: 100 Expected value per box: 200 Variance of the value per box: 400 You pay your employees a bonus if the vaiue of light bulbs destroyed in a month is less than 8000. Assuming independence and using the normal approximation, calculate the probability that you will pay your employees a bonus next month. A) 0.16 B) 0.19 C) 0.23 D) 0.27 E) 0.31 8. (BOA) A dam is prepcsed for a river which is currently used for salmon breeding. You hate modeled: (i) For each hour the dam is Opened the number of salmon that will pass through and reach the breeding grounds has a distribution with mean 100 and variance 900. (ii) The number of eggs released by each saimon has a distribution with mean of 5 and variance of 5. (iii) The number of salmon going through the dam each hour it is open and the numbers of eggs released by the salmon are independent. Using the normal approximation for the aggregate number of eggs released, determine the least number of whole hours the darn should be left open so the probability that 10,000 eggs will be released is greater than 95%. A) 20 B) 23 C) 26 D) 29 E) 32 9. (SOA) In a clinic, physicians volunteer their time on a daily basis to provide care to those who are not eligible to obtain care otherwise. The number of physicians who volunteer in any day is uniformly distributed on the integers 1 through 5. The number of patients that can be served by a given physician has a Poisson distribution with mean 30. Determine the probability that 120 or more patients can be served in a day at the ciinic, using the normal approximation with continuity correction. A) l — @(068) B) l — c13(072) C) 1 — @(093) D) 1— (H313) E) 1— (36.16) 10. (SOA) For an individual over 65: (i) The number of pharmacy claims is a i’oisson random variable with mean 25. (ii) The amount of each pharmacy claim is uniformly distributed between 5 and 95. (iii) The amounts of the claims and the number of claims are mutually independent. Determine the probability that aggregate claims for this individual will exceed 2000 using the normal approximation. A) 1— @(133) B) 1— @(166) C) 1— (13(233) D) 1 — @(266) E) 1— (56.33) ll.(SOA) A company has 50 employees whose dental expenses are mutually independent. For each employee, the company reimburses 100% of dental expenses in excess of a $100 deductible. The dental expense for each employee is distributed as follows: Expense Probability 0 0.20 50 0.30 200 0.30 500 0.10 1,000 0.10 Determine, by normal approximation, the 95—th percentile of the cost to the company. A) 8,000 B) 9,000 C) 10,000 1)) 11,000 E) 12,000 12. (SOA) An insurer provides life insurance for the following group of independentilives: NumberDeath Probability of Lives Benefit of Death 100 1 0.01 200 2 0.02 300 3 0.03 The insurer purchases reinsurance with a retention of 2 on each life. The reinsurer charges a premium of H equal to its expected claims plus the standard deviation of its claims. The insurer charges a premium G equal to expected retained claims plus the standard deviation of retained claims plus H . Determine G. A) 44 B) 46 C) 70 D) 94 E) 96 13. (SOA) Two portfolios of independent insurance policies have the following characteristics: Portfolio A: Class NumberProb. of Claim Claim Amount in Class per Policy 1 2000 0.05 1 2 500 0.10 2 Portfolio B: Class NumberProb. of Claim Claim Amount Distribution in Class per Policy Mean Variance 1 2000 0.05 1 1 2 500 0.10 2 4 The aggregate claims in the portfolios are denoted by 8,; and 85, respectively. Var SB] Determine VarlSA] . A) 1.0 B) 1.1 C) 1.2 D) 2.0 E) 2.1 14. (SOA) A group medical insurance policy cover the medical expenses incurred by 100,000 mutually independent lives. The annual loss amount X incurred by each life is distributed as follows: 0 50 200 500 1000 10, 000 x . P(X = x): .3 .1 .1 .2 .2 .1 The policy pays 80% of the annual losses for each life. The premium is equal to the 95-th percentile of the normal distribution which approximates the distribution of total claims. Determine the security loading. A) 1,213,000 B) 1,356,000 C) 1,446,000 D) 1,516,000 E) 1,624,000 15. (SOA) The policies of a building insurance company are classified according to the location of the building insured: Number of Claim Region Claim Amount Policies in Region Probability: A 20 300 0.01 B 10 500 0.02 C 5 600 0.03 D 15 500 0.02 E 1 8 100 0.01 Using the normal approximation, relative security loadings are computed for each region such that the probability that the total claims for the region do not exceed premium collected from policies in that region is 0.95. Which region pays the iargest relative security loading? A) A B) B C) C D) D E) E 16. (SOA) An insurer has the following portfolio of policies: Benefit Number Probability Class Amount Covered of a Claim 1 1 400 0.02 2 10 100 0.02 The insurer reinsures the amount in excess of R (R > 1) per policy at a cost of 0.025 per unit of coverage. The insurer wants to minimize the probability, as determined by the normal approximation, that retained claims plus cost of reinsurance exceeds 34. Determine R. A) 1.5 B) 2.0 C) 2.5 D) 3.0 B) 3.5 17. (SOA) An insurance company is selling policies to individuals with independent future iifetimes and identical mortality profiles. For each individual, the probability of death by all causes is 0.10 and the probability of death due to accident is .01. Each insurance policy pays the following benefits: 10 for accidental death, 1 for non-accidental death . The company wishes to have at least 95% confidence that premiums with a relative security loading of 0.20 are adequate to cover claims. Using the normal approximation, determine the minimum number of policies that must be sold. A) 1793 B) 1975 C) 2043 D) 2545 E) 2804 18. (CAS May 06) Between 9am and 3pm Big National Bank employs 2 tellers to service customer transactions. The time it takes Teller X to complete a transaction follows an exponential distribution with a mean of 10 minutes. Transaction times for Teller Y follow an exponential distribution with a mean of 15 minutes. Both Teller X and Teller Y are continuously busy while the bank is open. On average, every third customer transaction is a deposit and the amount of the deposit follows a Pareto distribution with parameters a = 3 and B = 5000. Each transaction that involves a deposit of at least $7500 is handled by the branch manager. Calculate the total deposits made through the tellers each day. A) Less than $31,000 B) At least $31,000, but less than 332,500 C) At least $32,500, but less than $35,000 D) At least $35,000, but less than $37,500 B) At least $37,500 19. (SOA) Each life within a group medical expense policy has loss amounts which follow a compound Poisson process with A = 0.16. Given a loss, the probability that it is for Disease 1 is 11-5 . Loss amount distributioris have the following parameters: Mean per loss Standard Deviation per loss Disease 1 5 50 Other diseases 10 20 Premiums for a group of 100 independent lives are set at a level such that the probability (using the normal approximation to the distribution for aggregate losses) that aggregate losses for the group will exceed aggregate premiums for the group is 0.24. A vaccine which will eliminate Disease 1 and costs 0.15 per person has been discovered. Define: A = the aggregate premium assuming that no one obtains the vaccine, and B = the aggregate premium assuming that everyone obtains the vaccine and the cost of the vaccine is a covered loss. Calculate A/ B . A) 0.94 B) 0.97 C) 1.00 D) 1.03 E) 1.06 20. (SOA) A company insures a fleet of vehicles. Aggregate losses have a compound Poisson distribution. The expected number of losses is 20. Loss amounts, regardless of vehicle type, have exponential distribution with 3 = 200. In order to reduce the cost of the insurance, two modifications are to be made: (i) A certain type of vehicle will not be insured. It is estimated that this will reduce loss frequency by 20%. (ii) A deductible of 100 per loss will be imposed. Calculate the expected aggregate amount paid by the insurer after the modifications. A) 1600 B) 1940 C) 2520 D) 3200 B) 3830 21. An insurer combines two independent lines of insurance, both of which have compound Poisson distributions. 31, 5'2 and S denote the aggregate claim random variables for line 1, line 2 and the two lines combined, respectively, and X‘1),X(2) and X denote the claim amount random variables for line 1, line 2 and the two lines combined. Given E[S] = 52, Var[S] = 900,135,] 2 20, E[X(2)] = 8, Vor[X(2}] = 61, and Var[X{1]] = 100, what is A1 ? A) l B) 2 C) 3 D) 4 E) 5 1. Y is the discretized distribution. P[Y = a + g] = Plk < X g k + 1] : e-k — 8+1 for a = 0, 1, 2, Em = f; (a + %)(e'k m 3+1): °° k(e"‘ — e‘k“1)+%§%(e‘k — 6"”1). 13:0 13:0 DO 2(ewk _ e-k—1)=(60 _ 6—1)+(e—1 W 3-2) + = 1, k=U 0° A wk _ 8‘1 0° I” —k—1 8‘2 F620 8 — (1_e.,_3)2 , Ag] 6 _' —-i.)2 —i 3—2 6—1 .— 00 "k. —}c—1 _ '3 _ flame ‘“ e )— (1—971)2 " (1—e-i)2 — 1—3-1 - E[Y] m 15:31.] + é— 2 1.08. Answer: E 2. S = Y1 + + YN , where N is Poisson with mean 10 and Y = (X — 20)+. ElS] 2 Elm - Em = (20)E[(X - 20m. Since X is exponential, E[(X — 20%} = fgg’n — Fx(t)] dt = [fie—“1”” at = 100i;-2 = 81.87. Then, E[S] = (20)(81.87) = 1637.4. E[(X — 20):] = E[(X — 2O)2|X > 20] - P(X > 20). For the exponential distribution, we know that the cost per payment X — 20|X > 20 also has an exponential distribution with the same mean of 100. Therefore, the second moment is 2 x 1002 = 20,000, so that E[(X — 20)?,_] m 20,000 - P{X > 20) m 20,0008—20/100 = 16,3746 Then, since N is Poisson, Var[S] 2 A v E[YE] = (20)(16, 374.6) = 327,482 . An alternative approach would be to let N * be the number of losses above 20, and let Yp be the cost per payment. Then S = Ypl + + YpN‘ . Since N is Poisson, N * is also Poisson with mean AP(X > 20) = wok-20000 = 16.3746. The cost per payment is exponential with mean 100, so = 2 x 1002 m 20, 000 and Vor[S] = (32.7492)(10,000) = 327, 492 . 3. The aggregate payment is S = Y1 + + YN ,where Y = (X -— 100%. . If we define N * to be the number of losses above the deductible, then S : Ypl + ng + + P)» ,where Y}: is the cost per payment. N "‘ is also negative binomial, with 7" = 3 and 200 16 1 16 [3' = 2 - P(X > 100} = 2- (10mm)3 = g ,so E[N] = (3)6?) = (—). SinceX is Pareto, Y}: is also Pareto with c2 = 3 and 6’ m 200 + 10 0 so E[Yp] = 332% = 150. Then, egg] = E[N*] - Em] = («133)(150) = i3? . ...
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problemset5_a - 1. The exponential distribution with mean 1...

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