Chapter 2 1. Model 1 in the table handed out in class is a uniform distribution from 0 to 100. Determine what the table entries would be for a generalized uniform distribution covering the range from a to b where a < b. 2. Let X be a discrete random variable with probability function p(x) = 2(1/3) x for x = 1, 2, 3, What is the probability that X is odd? 3. * For a distribution where x > 2, you are given: Calculate z. The hazard rate function: h(x) = z 2 /2x, for x > 2 F(5) = 0.84. 4. FX(t) = (t 2-1)/9999 for 1<t<100. Calculate fx(50). 5. You are given that the random variable X is distributed as a Weibull distribution with parameters θ = 3 and τ = 0.5. Calculate: a. Pr[X < 5] b. Pr[3 < X < 5] 6. (Spreadsheet Problem) You are given that the random variable X is distributed as a Geometric distribution with parameters β = 3. Calculate: a. Pr[X < 5] b. Pr[3 < X < 5] 7. A Weibull Distribution with parameter τ = 1 becomes what distribution? 8. A random variable X has a density function f(x) = 4x(1+x 2 ) -3, for x > 0. Determine the mode of X. Chapter 3 9. Determine the following for a generalized uniform distribution covering the range from a to b where a < d < b: a. E[X k ] b. E[X] c. Var(X) d. e(d) e. VaRp f. TVaRp February 12, 2016
10. For the Pareto distribution, determine E[X], Var(X), and the coefficient of variation in terms of α and θ. 11. For the Gamma distribution, determine E[X], Var(X), the coefficient of variation, and skewness in terms of α and θ. 12. For the Exponential distribution, determine E[X], Var(X), and e(d) in terms of θ. 13. You are given: F(x) = x 3 /27, for 0 < x < 3 1, for x >3 Calculate: a. E[X] b. Var(X) c. e(1) d. E[(X-1)+] e. E[X Λ 2] f. The Median g. The standard deviation principle with k = 1 h. VaR.80 i. TVaR.80 14. (Spreadsheet) If you roll two fair die, X is the sum of the dice. Calculate: a. E[X] b. Var(X) c. e(4) d. E[(X-4)+] e. E[X Λ 10] f. The Mode g. π20 15. You are given a sample of 2, 2, 3, 5, 8. For this empirical distribution, determine: a. The mean b. The variance c. The standard deviation d. The coefficient of variation e. The skewness f. The kurtosis g. VaR.80 h. TVaR.80 February 12, 2016
16. * Losses follow a Pareto distribution has parameters of α = 7 and θ = 10,000. Calculate e(5,000) 17. The amount of an individual claim has a Pareto distribution with θ = 8000 and α = 9. Use the central limit theorem to approximate the probability that the sum of 500 independent claims will exceed 550,000. 18. Lifetimes of an ipod follows a Single Parameter Pareto distribution with 1 and 4. The expected lifetime of an ipod is 8 years. Calculate the probability that the lifetime of an ipod is at least 6 years. 19. You are given that F x x x. 4 X ( ) 1 (100 / ) for 100 You are also given that f ( y ) is: Y Calculate Var(Y) Var(X). y f ( y ) 100 0.4 200 0.3 300 0.2 400 0.1 20. A company has 50 employees whose dental expenses are mutual independent. For each employee, the company reimburses 100% of the dental expenses. The dental expense for each employee is distributed as follows: Y Expense Probability 0 0.5 100 0.3 400 0.1 900 0.1 Using the normal approximation, calculate the 95 th percentile of the cost to the company. February 12, 2016
Chapter 4 Homework Problems 21. The distribution function for losses from your renter s insurance is the following: F(x) = 1 0.8[1000/(1000+x)] 5 0.2[12000/(12000+x)] 3 Calculate: a. E[X] b. Var(X) c. Use the normal approximation to determine the probability that the sum of 100 independent claims will not exceed 200,000. 22. * X has a Burr distribution with parameters α = 1, γ = 2, and θ = 1000 0.5. Y has a Pareto distribution with parameters α = 1 and θ = 1000. Z is a mixture of X and Y with equal weights on each component. Determine the median of Z. 23. The random variable X is distributed as a Pareto distribution with parameters α and θ. E(X) = 1 and Var(X) = 3. The random variable Y is equal to 2X. Calculate the Var(Y). 24. * Claim severities are modeled using a continuous distribution and inflation impacts claims uniformly at an annual rate of s. Which of the following are true statements regarding the distribution of claim severities after the effect of inflation? i. An exponential distribution will have a scale parameter of (1+s)θ. ii. A Pareto distribution will have scale parameters (1+s)α and (1+s)θ. iii. An Inverse Gaussian distribution will have a scale parameter of (1+s)θ. 25. * The aggregate losses of Eiffel Auto Insurance are denoted in euro currency and follow a Lognormal distribution with μ = 8 and σ = 2. Given that 1 euro = 1.3 dollars, determine the lognormal parameters for the distribution of Eiffel s losses in dollars. February 12, 2016
Chapter 5 Homework Problems 26. (Spreadsheet) Calculate Γ(1.2). 27. The random variable X is the number of dental claims in a year and is distributed as a gamma distribution given parameter θ and with parameter α = 1. θ is distributed uniformly between 1 and 3. Calculate E(X) and Var(X). 28. A dental insurance company has 1000 insureds. Assume the number of claims from each insured is independent. Using the information in Problem 27 and the normal approximation, calculate the probability that the Company will incur more than 2100 claims. 29. * Let N have a Poisson distribution with mean Λ. Let Λ have a uniform distribution on the interval (0,5). Determine the unconditional probability that N > 2. Chapter 6 30. The number of hospitalization claims in a year follows a Poisson distribution with a mean of λ. The probability of exactly three claims during a year is 60% of the probability that there will be 2 claims. Determine the probability that there will be 5 claims. 31. If the number of claims is distributed as a Poison distribution with λ = 3, calculate: a. Pr(N = 0) b. Pr(N = 1) c. Pr(N = 2) d. E(N) e. Var(N) 32. If the number of claims is distributed as a zero truncated Poison distribution with λ = 3, calculate: a. Pr(N = 0) b. Pr(N = 1) c. Pr(N = 2) d. E(N) e. Var(N) February 12, 2016
33. If the number of claims is distributed as a zero modified Poison distribution with λ = 3 and p0 M = 0.5, calculate: a. Pr(N = 0) b. Pr(N = 1) c. Pr(N = 2) d. E(N) e. Var(N) 34. If the number of claims is distributed as a Geometric distribution with β = 3, calculate: a. Pr(N = 0) b. Pr(N = 1) c. Pr(N = 2) d. E(N) e. Var(N) 35. (Spreadsheet) If the number of claims is distributed as a Binomial distribution with m = 6 and q = 0.4, calculate: a. Pr(N = 0) b. Pr(N = 1) c. Pr(N = 2) d. E(N) e. Var(N) 36. (Spreadsheet) If the number of claims is distributed as a Negative Binomial distribution with γ = 3 and β = 2, calculate: a. Pr(N = 0) b. Pr(N = 1) c. Pr(N = 2) d. E(N) e. Var(N) 37. * The Independent Insurance Company insures 25 risks, each with a 4% probability of loss. The probabilities of loss are independent. What is the probability of 4 or more losses in the same year? (Hint: Use the binomial distribution.) 38. * You are given a negative binomial distribution with γ = 2.5 and β = 5. For what value of k does pk take on its largest value? 39. * N is a discrete random variable from the (a, b, 0) class of distributions. The following information is known about the distribution: P(N = 0) = 0.327680 P(N = 1) = 0.327680 P(N = 2) = 0.196608 E[N] = 1.25 February 12, 2016
Based on this information, which of the following are true statements? I. P(N = 3) = 0.107965 II. N is from a binomial distribution. III. N is from a Negative Binomial Distribution. 40. * You are given: Claims are reported at a Poisson rate of 5 per year. The probability that a claim will settle for less than 100,000 is 0.9. What is the probability that no claim of 100,000 or more will be reported in the next three years? 41. N is distributed as a zero modified geometric distribution with β = 3 and p0 M = 0.5. Calculate: a. e(2) b. Var[ N 3] c. E[(N-2)+] 42. N is the random variable representing the number of claims under homeowners insurance. N is distributed as a zero modified geometric distribution with β = 2. M M p 4 = 2/27. Calculate p 2. 43. Under an unmodified geometric distribution, Var[N] 20. Chapter 8 Under a zero-modified geometric distribution, Var[N] 20.25. The parameter is the same for both distributions. M Calculate p 0. 44. Losses are distributed exponentially with θ = 1000. Losses are subject to an ordinary deductible of 500. Calculate: a. f L ( y ) Y b. F L ( y ) Y L c. EY [ ] d. The loss elimination ratio 45. Losses are distributed exponentially with parameter θ = 1000. Losses are subject to an ordinary deductible of 500. Calculate: a. f P ( y ) Y b. F P ( y ) Y February 12, 2016
P c. EY [ ] Homework Problems 46. Losses are distributed exponentially with θ = 1000. Losses are subject to a franchise deductible of 500. Calculate: a. f L ( y ) Y b. F L ( y ) Y L c. EY [ ] d. The loss elimination ratio 47. Losses are distributed exponentially with parameter θ = 1000. Losses are subject to a franchise deductible of 500. Calculate: a. f P ( y ) Y b. F P ( y ) Y P c. EY [ ] 48. Last year, losses were distributed exponentially with θ = 1000. This year losses are subject to 10% inflation. Losses in both years are subject to an ordinary deductible of 500. Calculate the following for this year: a. E(Y L ) b. E(Y P ) February 12, 2016
49. * Losses follow an exponential distribution with parameter θ. For a deductible of 100, the expected payment per loss is 2000. What is the expected payment per loss in terms of θ for a deductible of 500? (Note: Assume it is an ordinary deductible.) 50. * In year 2007, claim amounts have the following Pareto distribution F(x) = 1 (800/[x+800]) 3 The annual inflation rate is 8%. A franchise deductible of 300 will be implemented in 2008. Calculate the loss elimination ratio of the franchise deductible. 51. (Spreadsheet) If you roll two fair die, X is the sum of the dice. Let X represent the distribution of losses from a particular event. If an ordinary deductible of 3 is applied, calculate: a. E(Y L ) b. E(Y P ) c. Var(Y L ) d. Var(Y P ) 52. Losses are distributed uniformly between 0 and 100,000. An insurance policy which covers the losses has a 10,000 deductible and an 80,000 upper limit. The upper limit is applied prior to applying the deductible. Calculate: a. E(Y L ) b. E(Y P ) 53. Last year, losses were distributed exponentially with θ = 1000. This year losses are subject to 25% inflation. Losses in both years are subject to an ordinary deductible of 500, an upper limit of 4000, and coinsurance where the company pays 80% of the claim. Calculate the following for this year: a. E(Y L ) b. E(Y P ) 54. * An insurance company offers two types of policies: Type Q and Type R. Type Q has no deductible, but has a policy limit of 3000. Type R has no limit, but has an ordinary deductible of d. Losses follow a Pareto distribution with θ = 2000 and α = 3. Calculate the deductible d such that both policies have the same expected cost per loss. February 12, 2016
55. * Well Traveled Insurance Company sells a travel insurance policy that reimburses travelers for any expense incurred for a planned vacation that is cancelled because of airline bankruptcies. Individual claims follow a Pareto distribution with α = 2 and θ = 500. Well Traveled imposes a limit of 1000 on each claim. If a planned policyholder s planned vacation is cancelled due to airline bankruptcy and he or she has incurred more than 1000 of expenses, what is the expected non-reimbursable amount of the claim? 56. If X is uniformly distributed on from 0 to b. An ordinary deductible of d is applied. Calculate: a. E[Y L ] b. Var[Y L ] c. E[Y P ] d. Var[Y P ] 57. Losses follow a Pareto distribution with α = 3 and θ = 200. A policy covers the loss except for a franchise deductible of 50. X is the random variable representing the amount paid by the insurance company per payment. Calculate the expected value of the X. 58. Losses in 2007 are uniformly distributed between 0 and 100,000. An insurance policy pays for all claims in excess of an ordinary deductible of $20,000. For 2008, claims will be subject to uniform inflation of 20%. The Company implements an upper limit u (without changing the deductible) so that the expected cost per loss in 2008 is the same as the expected cost per loss in 2007. Calculate u. 59. Losses follow an exponential distribution with a standard deviation of 100. An insurance company applies an ordinary policy deductible d which results in a Loss Elimination Ratio of 0.1813. Calculate d. 60. Losses are distributed following the single parameter Pareto with α = 4 and θ = 90. An insurance policy is issued with an ordinary deductible of 100. Calculate the Var(Y L ). February 12, 2016
61. Losses follow a Pareto distribution with α = 5 and θ = 2000. A policy pays 100% of losses from 500 to 1000. In other words, if a loss occurs for less than 500, no payment is made. If a loss occurs which exceeds 1000, no payment is made. However, if a loss occurs which is between 500 and 1000, then the entire amount of the loss is paid. Calculate E(Y P ). 62. * You are given the following: Losses follow a lognormal distribution with parameters μ =10 and σ = 1. For each loss less than or equal to 50,000, the insurer makes no payment. For each loss greater than 50,000, the insurer pays the entire amount of the loss up to the maximum covered loss of 100,000. Determine the expected amount of the loss. 63. You are given the following: Losses follow a distribution prior to the application of any deductible with a mean of 2000. The loss elimination ratio at a deductible of 1000 is 0.3. 60% of the losses in number are less than the deductible of 1000. Determine the average size of the loss that is less than the deductible of 1000. 64. Losses follow a Pareto distribution with α = 5 and θ = 2000. An insurance policy covering these losses has a deductible of 100 and makes payments directly to the physician. Additionally, the physician is entitled to a bonus if the loss is less than 500. The bonus is 10% of the difference between 500 and the amount of the loss. The following table should help clarify the arrangement: Amount of Loss Loss Payment Bonus 50 0 45 100 0 40 250 150 25 400 300 10 500 400 0 1000 900 0 Calculate the expected total payment (loss payment plus bonus) from the insurance policy to the physician per loss. February 12, 2016
Chapter 9 Homework Problems 65. * For an insured, Y is the random variable representing the total amount of time spent in a hospital each year. The distribution of the number of hospital admissions in a year is: Number of Admissions Probability 0 0.60 1 0.30 2 0.10 The distribution of the length of each stay for an admission follows a Gamma distribution with α = 1 and θ = 5. Calculate E[Y] and Var[Y]. 66. An automobile insurer has 1000 cars covered during 2007. The number of automobile claims for each car follows a negative binomial distribution with β = 1 and γ = 1.5. Each claim is distributed exponentially with a mean of 5000. Assume that the number of claims and the amount of the loss are independent and identically distributed. Using the normal distribution as an approximating distribution of aggregate losses, calculate the probability that losses will exceed 8 million. 67. For an insurance company, each loss has a mean of 100 and a variance of 100. The number of losses follows a Poisson distribution with a mean of 500. Each loss and the number of losses are mutually independent. The loss ratio for the insurance company is defined as the ratio of aggregate losses to the total premium collected. The premium collected is 110% of the expected aggregate losses. Using the normal approximation, calculate the probability that the loss ratio will exceed 95%. 68. The number of claims follows a Poisson distribution with a mean of 3. The distribution of claims is fx(1) = 1/3 and fx(2) = 2/3. Calculate fs(4). February 12, 2016
69. (Spreadsheet) The number of claims and the amount of each claim is mutually independent. The frequency distribution is: The severity distribution is: n pn 1 0.05 2 0.20 3 0.30 4 0.25 5 0.15 6 0.05 x fx(x) 100 0.25 200 0.20 300 0.15 400 0.12 500 0.10 600 0.08 700 0.06 800 0.04 Calculate fs(x). (See Example 9.5 in the book for an example.) February 12, 2016
70. You are given the following table for aggregate claims: s FS(s) E[(S-d)+] 0 100 200 0.50 300 0.65 60 400 0.75 500 600 Losses can only occur in multiples of 100. Calculate the net stop loss premium for stop loss insurance covering losses in excess of 325. 71. * Losses follow a Poisson frequency distribution with a mean of 2 per year. The amount of a loss is 1, 2, or 3 with each having a probability of 1/3. Loss amounts are independent of the number of losses and from each other. An insurance policy covers all losses in a year subject to an annual aggregate deductible of 2. Calculate the expected claim payments for this insurance policy. 72. * An insurance portfolio produces N claims with the following distribution: n P(N=n) 0 0.1 1 0.5 2 0.4 Individual claim amounts have the following distribution: x fx(x) 0 0.7 10 0.2 20 0.1 Individual claim amounts and counts are independent. Stop Loss insurance is purchased with an aggregate deductible of 400% of expected claims. Calculate the net stop loss premium. 73. Losses are modeled assuming that the amount of all losses is 40 and that the number of losses follows a geometric distribution with a mean of 4. Calculate the net stop loss premium for coverage with an aggregate deductible of 100. February 12, 2016
74. * An aggregate claim distribution has the following characteristics: P[S = i] = 1/6 for i = 1, 2, 3, 4, 5, or 6. A stop loss insurance with a deductible amount of d has an expected insurance payment of 1.5 Determine d. 75. For an automobile policy, the severity distribution is Gamma with α = 3 and θ = 1000. The number of claims in a year is distributed as follows: Calculate: a. ES [ ] b. Var[ S ] c. F S (10,000) Number of Claims Probability 1 0.65 2 0.20 3 0.10 4 0.05 76. (Spreadsheet) You are given that the number of losses follow a Poisson with λ = 6. Losses are distributed as follows: f X (1) f X (2) f X (4) 1/ 3. Find fs () s for total losses of 100 or less. 77. On a given day, a doctor provides medical care for NA adults and NC children. Assume that NA and NC have Poisson distributions with parameters 3 and 2 respectively. The distributions for the length of care per patient are as follows: Time Adult Child 1 hour 0.4 0.9 2 hours 0.6 0.1 Let NA, NC, and the lengths of stay for all patients be independent. The doctor charges 200 per hour. Determine the probability that the office income on a given day will be less than or equal to 800. 78. (Spreadsheet) Losses follow a Pareto distribution with α = 3 and θ = 1000. a. Using the Method of Rounding with a span of 100, calculate fj for 0 < j <1000. Calculate the mean of the discretized distribution. Compare this mean to the actual mean. b. Using the Method of Local Moment Matching and matching the mean (with k = 1) and a span of 100, calculate fj for 0 < j <1000. Calculate the mean of the discretized distribution. Why does this mean not equal the actual mean? February 12, 2016
79. Using the example that was handed out in class, find fs(6) and fs(7). 80. * You are given: a. S has a compound Poisson distribution with λ = 2; and b. Individual claim amounts x are distributed as follows: fx(1) = 0.4 and fx(2) = 0.6 Determine fs(4). 81. * Aggregate claims S has a compound Poisson distribution with discrete individual claim amount distributions of fx(1) = 1/3 and fx(3) = 2/3. Also, fs(4) = fs(3) + 6fS(1). Calculate the Var(S). 82. * For aggregate claims S, you are given: a. fs(x) = n 0 f * n X ( x) e 50 (50) n! b. Losses are distributed as follows: Determine Var(S) x fx(x) 1 0.4 2 0.5 3 0.1 n 83. With no deductible, the number of payments for losses under warranty coverage for an Iphone follows a negative binomial distribution with a mean of 0.25 and a variance of 0.375. The company imposes a deductible of d such that the number of expected payments for losses is reduced to 50% of the prior expected payments. Calculate the Var (N P ) after the imposition of the deductible. February 12, 2016
84. * Prior to the application of any deductible, aggregate claim counts during 2005 followed a Poisson with λ = 14. Similarly, individual claim sizes followed a Pareto with α = 3 and θ = 1000. Annual inflation for the claim sizes is 10%. All policies in 2005 and 2006 are subject to a 250 ordinary deductible. Calculate the increase in the number of claims that exceed the deductible in 2006 when compared to 2005. 85. Purdue University has decided to provide a new benefit to each class at the university. Each class will be provided with group life insurance. Students in each class will have 10,000 of coverage while the professor for the class will have 20,000. STAT 479 has 27 students all age 22 split 12 males and 15 females. The class also has an aging professor. The probability of death for each is listed below: Age Gender Probability of Death 22 Male 0.005 22 Female 0.003 58 Male 0.020 The insurance company providing the coverage charges a premium equal to the expected claims plus one standard deviation. Calculate the premium. 86. Purdue University has decided to provide a new benefit to each class at the university. Each class will be provided with group life insurance. Students in each class will have 10,000 of coverage while the professor for the class will have 20,000. STAT 479 has 27 students all age 22 split 12 males and 15 females. The class also has an aging professor. The probability of death for each is listed below: Age Gender Probability of Death 22 Male 0.005 22 Female 0.003 58 Male 0.020 Purdue purchases a Stop Loss Policy such with an aggregate deductible of 20,000. Calculate the net premium that Purdue will pay for the stop loss coverage. February 12, 2016
87. * An insurer provides life insurance for the following group of independent lives: Number of Lives Death Benefit Probability 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 the its expected claims plus the standard deviation of its claims. The insurer charges a premium of G which is equal to its expected retained claims plus the standard deviation of the retained claims plus H Calculate G. 88. X1, X2, and X3 are mutually independent random variables. These random variables have the following probability functions: Calculate fs(x). x f1(x) f2(x) f3(x) 0 0.6 0.4 0.1 1 0.4 0.3 0.4 2 0.0 0.2 0.5 3 0.0 0.1 0.0 February 12, 2016
89. * Two portfolios of independent insurance policies have the following characteristics: Portfolio A Class Number in Class Probability of Claim Claim Amount Per Policy 1 2000 0.05 1 2 500 0.10 2 Class Number in Class Portfolio B Probability of Claim Claim Amount Distribution Mean Variance 1 2000 0.05 1 1 2 500 0.10 2 4 The aggregate claims in the portfolios are denoted by SA and SB, respectively. Calculate Var[SB]/Var[SA]. 90. * An insurance company is selling policies to individuals with independent future lifetimes and identical mortality profiles. For each individual, the probability of death by all causes is 0.10 and the probability of death due to an accident is 0.01. Each insurance policy pays a benefit of 10 for an accidental death and 1 for nonaccidental 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. (In other words, the premium is 1.20E(S).) Using the normal approximation, determine the minimum number of policies that must be sold. February 12, 2016
1. Not Provided 2. ¾ 3. 2 4. 100/9999 5. a. 0.725 b. 0.093 6. a..822 b. 0.244 7. No Answer Given 8. 1/ 5 9. a. (b k+1 a k+1 )/(b-a)(k+1) b. (b+a)/2 c. (b-a) 2 /12 d. (b-d)/2 e. (1-p)a+pb f. [b+(1-p)a+pb]/2 10. a. θ/(α-1) b. αθ 2 /[(α-1) 2 (α-2)] c. [α/( α-2)] 0.5 11. a. θα b. θ 2 α c. 1/ α d. 2/ α 12. a. θ b. θ 2 c. θ 13. a. 2.25 b. 27/80 c. 17/13 d. 34/27 e. 50/27 f. 2.3811 g. 2.83 h. 2.785 i. 2.895 Homework Problems Answers February 12, 2016
Answers 14. a. 7 b. 5.8333 c. 3.7333 d. 3.1111 e. 6.8888 f. 7 g. 5 15. a. 4 b. 5.2 c. 2.28 d. 0.57 e. 0.8096 f. 2.145 g. 5 h. 8 16. 2500 17. 0.0244 18. 4/9 19. 7777.78 20. 11,173 21. a. 1400 b. 26,973,333 c. 87.7% 22. 100 23. 12 24. i only. Be sure to indicate how to make the others true. 25. μ = 8.26 and σ = 2.00 26. 0.918169 27. 2 and 14/3 28. 7.2% 29. 0.6094 30. 2.6% 31. a. 0.0498 b. 0.1494 c. 0.2240 d. 3 e. 3 32. a. 0 b. 0.1572 February 12, 2016
c. 0.2358 d. 3.1572 e. 2.6609 Homework Problems Answers 33. 34. 35. a. 0.5000 b. 0.0786 c. 0.1179 d. 1.5786 e. 3.8224 a. 0.2500 b. 0.1875 c. 0.1406 d. 3 e. 12 a. 0.0467 b. 0.1866 c. 0.3110 d. 2.4 e. 1.44 36. a. 0.0370 b. 0.0741 c. 0.0988 d. 6 e. 18 37. 0.0165 38. 7 39. Statement III is only true statement 40. 0.2231 41. a. 4 b. 1.6943 c. 1.125 42. 1/6 43. 0.1 44. a. fyl(y) = 1 e -0.5, y = 0 fyl(y) = 0.001e -(0.001y + 0.5), y > 0 b. FYL(y) = 1 e -0.5, y = 0 FYL(y) = 1 e -(.001 y + 0.5), y > 0 February 12, 2016
c. 1000e -0.5 Homework Problems d. 1 e -0.5 45. a. fyp(y) = 0.001e -0.001y b. FYP(y) = 1 e -.001 y c. 1000 46. a. fyl(y) = 1 e -0.5, y = 0 fyl(y) = 0.001e -0.001y, y > 500 b. FYL(y) = 1 e -0.5, y = 0 FYL(y) = 1 e -.001 y, y > 500 c. 1500e -0.5 d. 1 1.5e -0.5 47. a. fyp(y) = 0.001e -0.001y+0.5 -.001 y+0.5 b. FYP(y) = 1 e c. 1500 February 12, 2016
Answers 48. a. 1100e -(0.5/1.1) b. 1100 49. 2000e -400/θ 50. 0.165 51. a. 4.0278 b. 4.39 c. 5.58 d. 4.48 52. a. 38,500 b. 42,777.78 53. a. 629.56 b. 939.19 54. 182.18 55. 1500 56. a. (b-d) 2 /2b b. [(b-d) 3 /3b] [(b-d) 2 /2b] 2 c. (b-d)/2 d. (b-d) 2 /12 57. 175 58. 71,833.62 59. 20 60. 1708.70 61. 705.06 62. 16,224 63. 333.33 64. 431.83 65. 2.5 and 23.75 66. 0.068 67. 0.1587 February 12, 2016
68. 0.151436 69. None provided 70. 51.25 71. 2.360894 72. 0.224 73. 92.16 74. 2.25 75. a. 4,650 b. 11,377,500 c. 76. No Answer Provided 77. 0.2384 78. a. Mean = 498.61 b. Mean = 499.85 79. 0.106978 and 0.060129 80. 0.1517 81. 76 82. 165 83. 0.15625 84. 0.406 85. 5,727 86. 22.60 87. 46.13 88. x fs(x) Homework Problems 89. 2.1 90. 1975 91. 0 0.024 1 0.130 2 0.280 3 0.280 4 0.180 5 0.086 6 0.020 February 12, 2016
Chapter 10 91. * A random sample, X1, X2,, Xn, is drawn from a distribution with a mean of 2/3 and a variance of 1/18. ˆ = (X1 + X2 + + Xn)/(n-1) is the estimator of the distribution mean θ. Find MSE( ˆ ). 92. * Claim sizes are uniformly distributed over the interval [0, θ]. A sample of 10 claims, denoted by X1, X2,,X10 was observed and an estimate of θ was obtained using: ˆ = Y = max(x1, X2,,X10) Recall that the probability density function for Y is: fy(y) = 10y 9 /θ 10 Calculate the mean square error for ˆ for θ =100. 93. * You are given two independent estimates of an unknown quantity θ: a. Estimator A: E( ˆ A) = 1000 and σ( ˆ A) = 400 b. Estimator B: E( ˆ B) = 1200 and σ( ˆ B) = 200 Estimator C is a weighted average of Estimator A and Estimator B such that: ˆ C = (w) ˆ A + (1-w) ˆ B Determine the value of w that minimizes σ( ˆ C). April 15, 2014
94. * You are given: Homework Problems x Pr(X = x) 0 0.5 1 0.3 2 0.1 3 0.1 Using a sample size of n, the population mean is estimated by the sample mean X. The variance is estimated by: 2 S n = ( X i n X ) 2 Calculate the bias of S 2 n when n = 4. 95. * For the random variable X, you are given: a. E(X) = θ, θ > 0 b. Var(θ) = θ 2 /25 c. ˆ = kx/(k+1) d. MSEθ(θ) = 2[bias θ(θ)] 2 Determine k. 96. You are given the following sample of claims: X: 12, 13, 16, 16, 22, 24, 26, 26, 28, 30 The sum of X is 213 and the sum of X 2 is 4921. H0 is that μx = 17 and H1 is that μx > 17. Calculate the z statistic, the critical value(s) assuming a significance level of 1%, and the p value. State your conclusion with regard to the Hypothesis Testing. April 15, 2014
Wang Warranty Corporation is testing ipods. Wang starts with 100 ipods and tests them by dropping them on the ground. Wang records the number of drops before each ipod will no longer play. The following data is collected from this test: Drops to Failure Number Drops to Failure Number 1 6 9 7 2 2 10 6 4 3 11 7 6 4 12 7 7 4 13 48 8 5 22 1 Ledbetter Life Insurance Company is completing a mortality study on a 3 year term insurance policy. The following data is available: Life Date of Entry Date of Exit Reason for Exit 1 0 0.2 Lapse 2 0 0.3 Death 3 0 0.4 Lapse 4 0 0.5 Death 5 0 0.5 Death 6 0 0.5 Lapse 7 0 1.0 Death 8 0 3.0 Expiry of Policy 9 0 3.0 Expiry of Policy 10 0 3.0 Expiry of Policy 11 0 3.0 Expiry of Policy 12 0 3.0 Expiry of Policy 13 0 3.0 Expiry of Policy 14 0 3.0 Expiry of Policy 15 0 3.0 Expiry of Policy 16 0.5 2.0 Lapse 17 0.5 1.0 Death 18 1.0 3.0 Expiry of Policy 19 1.0 3.0 Expiry of Policy 20 2.0 2.5 Death Schneider Trucking Company had the following losses during 2013: Amount of Claim Number of Payments Total Amount of Losses 0 10 8 60 10-20 5 70 20-30 4 110 30 + 3 200 Total 20 440 April 15, 2014
Chapter 11 97. Using the data from Wang Warranty Corporation, calculate: a. p ( x ) 100 b. F ( x ) 100 c. The empirical mean d. The empirical variance e. Ĥ(x) where Ĥ(x) is the cumulative hazard function from the Nelson Åalen estimate f. Ŝ(x) where Ŝ(x) is the survival function from the Nelson Åalen estimate 98. Using the data from Schneider Trucking Company, calculate: a. The ogive, F20(x) b. The histogram, f20(x) c. E(XΛ30) minus E[(X-20)+] 99. * You are given: Claim Size (X) Number of Claims (0,25] 30 (25,50] 32 (50,100] 20 (100,200] 8 Assume a uniform distribution of claim sizes within each interval. Estimate the mean of the claim size distribution. Estimate the second raw moment of the claim size distribution. April 15, 2014
Chapter 12 Homework Problems 100. Using the data for Ledbetter Life Insurance Company, calculate the following where death is the decrement of interest: a. S () 20 t using the Kaplan Meier Product Limit Estimator b. Ĥ(t) where Ĥ(t) is the cumulative hazard function from the Nelson Åalen estimate c. Ŝ(t) where Ŝ(t) is the survival function from the Nelson Åalen estimate 101. Using the data for Ledbetter Life Insurance Company, and treating all expiries as lapses, calculate the following where lapse is the decrement of interest: a. S () 20 t using the Kaplan Meier Product Limit Estimator b. Ĥ(t) where Ĥ(t) is the cumulative hazard function from the Nelson Åalen estimate c. Ŝ(t) where Ŝ(t) is the survival function from the Nelson Åalen estimate 102. * Three hundred mice were observed at birth. An additional 20 mice were first observed at age 2 (days) and 30 more were first observed at age 4. There were 6 deaths at age 1, 10 at age 3, 10 at age 4, a at age 5, b at age 9, and 6 at age 12. In addition, 45 mice escaped and were lost to observation at age 7, 35 at age 10, and 15 at age 13. The following product-limit estimates were obtained: S (7) 0.892 and S (13) 0.856. 350 350 Determine a and b. 103. * There are n lives observed from birth. None are censored and no two lives die at the same age. At the time of the ninth death, the Nelson Åalen estimate of the cumulative hazard rate is 0.511 and at the time of the tenth death it is 0.588. Estimate the value of the survival function at the time of the third death. April 15, 2014
104. Astleford Ant Farm is studying the life expectancy of ants. The farm is owned by two brothers who are both actuaries. They isolate 100 ants and record the following data: Number of Days till Death Number of Ants Dying 1 4 2 8 3 12 4 20 5 40 6 8 7 4 8 2 9 1 10 1 a. One of the brothers, Robert, uses the Nelson-Åalen estimator to determine H ˆ (5). Determine the 90% linear confidence interval for H ˆ (5). b. The other brother, Daniel, decides that since he has complete data for these 100 ants, he will just use the unbiased estimator of S ˆ(5). Using this approach, determine the 90% confidence interval for S ˆ(5). April 15, 2014
105. The following information on students in the actuarial program at Purdue is used to complete an analysis of students leaving the program because they are switching majors. Student Time of Time of Exit Reason for Exit Entry 1 0.5 Switching Major 2-5 0 1 Switching Major 6 0 2 Switching Major 7 0 3 Graduation 8 0 3 Switching Major 9-12 0 3.5 Graduation 13-23 0 4 Graduation 24 0.5 2 Switching Major 25 0.5 3 Switching Major 26 1 3.5 Graduation 27 1 4 Switching Major 28 1.5 4 Graduation 29 2 5 Graduation 30 3 5 Graduation Sx ˆ( ) is estimated using the product limit estimator. Estimate Var[ S 30(2)] using the Greenwood approximation. 106. A mortality study is conducted on 50 lives, all from age 0. At age 15, there were two deaths; at age 17, there were three censored observations; at age 25 there were four deaths; at age 30, there were c censored observations; at age 32 there were eight deaths; and at age 40 there were two deaths. Let S be the product limit estimate of S (35) and let V be the Greenwood estimate of this estimator s variance. You are given Determine c. 2 V / S 0.011467. April 15, 2014
107. * Fifteen cancer patients were observed from the time of diagnosis until the earlier of death of 36 months from diagnosis. Deaths occurred as follows: at 15 months, there were two deaths; at 20 months there were three deaths; at 24 months there were 2 deaths; at 30 months there were d deaths; at 34 months there were two deaths; and at 36 months there were one death. The Nelson Åalen estimate of H (35) is 1.5641. Determine the variance of this estimator. 108. A mortality study is completed on 30 people. The following deaths occur during the five years: 3 deaths at time 1.0 4 deaths at time 2.0 5 deaths at time 3.0 8 deaths at time 3.8 10 deaths at time 4.5 There were no other terminations and no lives entered the study after the start of the study. The data was smoothed using a uniform kernel with a bandwidth of 1. Calculate fˆ (x) and Fˆ (x) for all x > 0. 109. * From a population having a distribution function F, you are given the following sample: 2.0, 3.3, 3.3, 4.0, 4.0, 4.7, 4.7, 4.7 Calculate the kernel density estimate of F(4), using a uniform kernel with bandwidth of 1.4. 110. * You study five lives to estimate the time from the onset of a disease until death. The times to death are: 2 3 3 3 7 Using a triangular kernel with a bandwidth of 2, estimate the density function at 2.5. April 15, 2014
111. You are given the following random sample: 12 15 27 42 The data is smoothed using a uniform kernel with a bandwidth of 6. Calculate the mean and variance of the smoothed distribution. 112. You are given the following random sample: 12 15 27 42 The data is smoothed using a triangular kernel with a bandwidth of 12. Calculate the mean and variance of the smoothed distribution. 113. You are given the following random sample: 12 15 27 42 The data is smoothed using a gamma kernel with a bandwidth of 3. Calculate the mean and variance of the smoothed distribution. April 15, 2014
Chapter 13 Homework Problems 114. You are given the following sample of claims obtained from an inverse gamma distribution: X: 12, 13, 16, 16, 22, 24, 26, 26, 28, 30 The sum of X is 213 and the sum of X 2 is 4921. Calculate α and θ using the method of moments. 115. * You are given the following sample of five claims: 4 5 21 99 421 Find the parameters of a Pareto distribution using the method of moments. 116. * A random sample of death records yields the follow exact ages at death: 30 50 60 60 70 90 The age at death from which the sample is drawn follows a gamma distribution. The parameters are estimated using the method of moments. Determine the estimate of α. 117. * You are given the following: i. The random variable X has the density function f(x) = αx -α-1, 1 < x <, α >1 ii. A random sample is taken of the random variable X. Calculate the estimate of α in terms of the sample mean using the method of moments. 118. * You are given the following: i. The random variable X has the density function f(x) = {2(θ x)}/θ 2, 0 < x < θ ii. A random sample of two observations of X yields values of 0.50 and 0.70. Determine θ using the method of moments. April 15, 2014
119. You are given the following sample of claims: X: 12, 13, 16, 16, 22, 24, 26, 26, 28, 30 Calculate the smoothed empirical estimate of the 40 th percentile of this distribution. 120. * For a complete study of five lives, you are given: a. Deaths occur at times t = 2, 3, 3, 5, 7. b. The underlying survival distribution S(t) = 4 -λt, t > 0 Using percentile matching at the median, calculate the estimate of λ. 121. You are given the following 9 claims: X: 10, 60, 80, 120, 150, 170, 190, 230, 250 The sum of X = 1260 and the sum of X 2 = 227,400. The data is modeled using an exponential distribution with parameters estimated using the percentile matching method. Calculate θ based on the empirical value of 120. 122. * For a sample of 10 claims, x1 < x2 < < x10 you are given: a. The smoothed empirical estimate of the 55 th percentile is 380. b. The smoothed empirical estimate of the 60 th percentile is 402. Determine x6. 123. You are given the following: a. Losses follow a Pareto distribution with parameters α and θ. b. The 10 th percentile of the distribution is θ k, where k is a constant. c. The 90 th percentile of the distribution is 5θ 3k. Determine α. April 15, 2014
124. You are given the following random sample of 3 data points from a population with a Pareto distribution with θ = 70: X: 15 27 43 Calculate the maximum likelihood estimate for α. 125. * You are given: a. Losses follow an exponential distribution with mean θ. b. A random sample of 20 losses is distributed as follows: Range Frequency [0,1000] 7 (1000, 2000) 6 (2000, ) 7 Calculate the maximum likelihood estimate of θ. 126. * You are given the following: i. The random variable X has the density function f(x) = {2(θ x)}/θ 2, 0 < x < θ ii. A random sample of two observations of X yields values of 0.50 and 0.90. Determine the maximum likelihood estimate for θ. 127. * You are given: a. Ten lives are subject to the survival function S(t) = (1-t/k) 0.5, 0 < t < k b. The first two deaths in the sample occured at time t = 10. c. The study ends at time t = 10. Calculate the maximum likelihood estimate of k. 128. * You are given the following: a. The random variable X follows the exponential distribution with parameter θ. b. A random sample of three observations of X yields values of 0.30, 0.55, and 0.80 Determine the maximum likelihood estimate of θ. April 15, 2014
129. * Ten laboratory mice are observed for a period of five days. Seven mice die during the observation period, with the following distribution of deaths: Time of Death in Days Number of Deaths 2 1 3 2 4 1 5 3 The lives in the study are subject to an exponential survival function with mean of θ. Calculate the maximum likelihood estimate of θ. 130. * A policy has an ordinary deductible of 100 and a policy limit of 1000. You observe the following 10 payments: 15 50 170 216 400 620 750 900 900 900 An exponential distribution is fitted to the ground up distribution function, using the maximum likelihood estimate. Determine the estimated parameter θ. 131. * Four lives are observed from time t = 0 until death. Deaths occur at t = 1, 2, 3, and 4. The lives are assumed to follow a Weibull distribution with τ = 2. Determine the maximum likelihood estimator for θ. 132. * The random variable X has a uniform distribution on the interval [0,θ]. A random sample of three observations of X are recorded and grouped as follows: Number of Interval Observations [0,k) 1 [k,5) 1 [5,θ] 1 Calculate the maximum likelihood estimate of θ April 15, 2014
133. * A random sample of three claims from a dental insurance plan is given below: 225 525 950 Claims are assumed to follow a Pareto distribution with parameters θ = 150 and α. Determine the maximum likelihood estimate of α. 134. * The following claim sizes are experienced on an insurance coverages: 100 500 1,000 5,000 10,000 You fit a lognormal distribution to this experience using maximum likelihood. Determine the resulting estimate of σ. Chapter 14 You are given the following data from a sample: k nk 0 20 1 25 2 30 3 15 4 8 5 2 Use this data for the next four problems. 135. Assuming a Binomial Distribution, estimate m and q using the Method of Moments. 136. Assuming a Binomial Distribution, find the MLE of q given that m = 6. 137. (Spreadsheet) Assuming a Binomial Distribution, find the MLE of m and q. 138. Assuming a Poisson Distribution, approximate the 90% confidence interval for the true value of λ. April 15, 2014
Chapter 16 Homework Problems 139. You are given the following 20 claims: X: 10, 40, 60, 65, 75, 80, 120, 150, 170, 190, 230, 340, 430, 440, 980, 600, 675, 950, 1250, 1700 The data is being modeled using an exponential distribution with θ = 427.5. Calculate D(200). 140. You are given the following 20 claims: X: 10, 40, 60, 65, 75, 80, 120, 150, 170, 190, 230, 340, 430, 440, 980, 600, 675, 950, 1250, 1700 The data is being modeled using an exponential distribution with θ = 427.5. You are developing a p-p plot for this data. What are the coordinates for x7 = 120. 141. Mark the following statements True or False with regard to the Kolmogorov- Smirnov test: The Kolmogorov-Smirnov test may be used on grouped data as well as individual data. If the parameters of the distribution being tested are estimated, the critical values do not need to be adjusted. If the upper limit is less than, the critical values need to be larger. April 15, 2014
142. Balog s Bakery has workers compensation claims during a month of: 100, 350, 550, 1000 Balog s owner, a retired actuary, believes that the claims are distributed exponentially with θ = 500. He decides to test his hypothesis at a 10% significance level. Calculate the Kolmogorov-Smirnov test statistic. State the critical value for his test and state his conclusion. He also tests his hypothesis using the Anderson-Darling test statistic. State the values of this test statistic under which Mr. Balog would reject his hypothesis. 143. * The observations of 1.7, 1.6, 1.6, and 1.9 are taken from a random sample. You wish to test the goodness of fit of a distribution with probability density function given by f(x) = 0.5x for 0 < x < 2. Using the Kolmogorov-Smirnov statistic, which of the following should you do? a. Accept at both levels b. Accept at the 0.01 level but reject at the 0.10 level c. Accept at the 0.10 level but reject at the 0.01 level d. Reject at both levels e. Cannot be determined. 144. * Two lives are observed beginning at time t=0. One dies at time 5 and the other dies at time 9. The survival function S(t) = 1 (t/10) is hypothesized. Calculate the Kolmogorov-Smirnov statistic. 145. * From a laboratory study of nine lives, you are given: a. The times of death are 1, 2, 4, 5, 5, 7, 8, 9, 9 b. It has been hypothesized that the underlying distribution is uniform with ω = 11. Calculate the Kolmogorov-Smirnov statistic for the hypothesis. April 15, 2014
146. You are given the following data: Homework Problems Claim Range Count 0-100 30 100-200 25 200-500 20 500-1000 15 1000+ 10 H0: The data is from a Pareto distribution. H1: The data is not from a Pareto distribution. Your boss has used the data to estimate the parameters as α = 4 and θ = 1200. Calculate the chi-square test statistic. Calculate the critical value at a 10% significance level. State whether you would reject the Pareto at a 10% significance level. 147. During a one-year period, the number of accidents per day in the parking lot of the Steenman Steel Factory is distributed: Number of Accidents Days 0 220 1 100 2 30 3 10 4+ 5 H0: The distribution of the number of accidents is distributed as Poison with a mean of 0.625. H1: The distribution of the number of accidents is not distributed as Poison with a mean of 0.625. Calculate the chi-square statistic. Calculate the critical value at a 10% significance level. State whether you would reject the H0 at a 10% significance level. April 15, 2014
148. * You are given the following random sample of automobile claims: 54 140 230 560 600 1,100 1,500 1,800 1,920 2,000 2,450 2,500 2,580 2,910 3,800 3,800 3,810 3,870 4,000 4,800 7,200 7,390 11,750 12,000 15,000 25,000 30,000 32,200 35,000 55,000 You test the hypothesis that automobile claims follow a continuous distribution F(x) with the following percentiles: x 310 500 2,498 4,876 7,498 12,930 F(x) 0.16 0.27 0.55 0.81 0.90 0.95 You group the data using the largest number of groups such that the expected number of claims in each group is at least 5. Calculate the Chi-Square goodness-of-fit statistic. 149. Based on a random sample, you are testing the following hypothesis: H0: The data is from a population distributed binomial with m = 6 and q = 0.3. H1: The data is from a population distributed binomial. You are also given: L(θ0) =.1 and L(θ1) =.3 Calculate the test statistic for the Likelihood Ratio Test State the critical value at the 10% significance level 150. State whether the following are true or false i. The principle of parsimony states that a more complex model is better because it will always match the data better. ii. In judgment-based approaches to determining a model, a modeler s experience is critical. iii. In most cases, judgment is required in using a score-based approach to selecting a model. April 15, 2014
Chapter 21 Homework Problems 151. A random number generated from a uniform distribution on (0, 1) is 0.6. Using the inverse transformation method, calculate the simulated value of X assuming: i. X is distributed Pareto with α = 3 and θ = 2000 0.5x 0 < x < 1.2 ii. F(x) = 0.6 1.2 < x < 2.4 0.5x -0.6 2.4 < x < 3.2 iii. F(x) = 0.1x -1 10 < x < 15 0.05x 15 < x < 20 152. * You are given that f(x) = (1/9)x 2 for 0 < x < 3. You are to simulate three observations from the distribution using the inversion method. The follow three random numbers were generated from the uniform distribution on [0,1]: 0.008 0.729 0.125 Using the three simulated observations, estimate the mean of the distribution. 153. * You are to simulate four observations from a binomial distribution with two trials and probability of success of 0.30. The following random numbers are generated from the uniform distribution on [0,1]: 0.91 0.21 0.72 0.48 Determine the number of simulated observations for which the number of successes equals zero. April 15, 2014
154. Kyle has an automobile insurance policy. The policy has a deductible of 500 for each claim. Kyle is responsible for payment of the deductible. The number of claims follows a Poison distribution with a mean of 2. Automobile claims are distributed exponentially with a mean of 1000. Kyle uses simulation to estimate the claims. A random number is first used to calculate the number of claims. Then each claim is estimated using random numbers using the inverse transformation method. The random numbers generated from a uniform distribution on (0, 1) are 0.7, 0.1, 0.5, 0.8, 0.3, 0.7, 0.2. Calculate the simulated amount that Kyle would have to pay in the first year. 155. * Insurance for a city s snow removal costs covers four winter months. You are given: i. There is a deductible of 10,000 per month. ii. The insurer assumes that the city s monthly costs are independent and normally distributed with mean of 15,000 and standard deviation of 2000. iii. To simulate four months of claim costs, the insurer uses the inversion method (where small random numbers correspond to low costs). iv. The four numbers drawn from the uniform distribution on [0,1] are: Calculate the insurer s simulated claim cost. 0.5398 0.1151 0.0013 0.7881 156. * Annual dental claims are modeled as a compound Poisson process where the number of claims has mean of 2 and the loss amounts have a two-parameter Pareto distribution with θ = 500 and α = 2. An insurance pays 80% of the first 750 and 100% of annual losses in excess of 750. You simulate the number of claims and loss amounts using the inversion method. The random number to simulate the number of claims is 0.80. The random numbers to simulate the amount of claims are 0.60, 0.25, 0.70, 0.10, and 0.80. Calculate the simulated insurance claims for one year. April 15, 2014
157. A sample of two selected from a uniform distribution over (1,U) produces the following values: 3 7 You estimate U as the Max(X1, X2). Estimate the Mean Square Error of your estimate of U using the bootstrap method. 158. * Three observed values from the random variable X are: Chapter 3 1 1 4 You estimate the third central moment of X using the estimator: g(x1, X2, X3) = 1/3 Σ(Xj - X ) 3 Determine the bootstrap estimate of the mean-squared error of g. 159. * Using the criterion of existence of moments, determine which of the following distributions have heavy tails. a. Normal distribution with mean μ and variance of σ 2. b. Lognormal distribution with parameters μ and σ 2. c. Single Parameter Pareto. April 15, 2014
Answers 91. (n+8)/[18(n-1) 2 ] 92. 151.52 93. 1/5 94. -0.24 95. 5 96. z = 2.0814; critical value = 2.33; Since 2.0814 is less than 2.33, we cannot reject the null hypothesis; p = 0.0188 97. x p100(x) x F100(x) Ĥ(x) Ŝ(x) x<1 0 0 1 1 0.06 1<x<2 0.06 0.060000 0.941765 2 0.02 2<x<4 0.08 0.081277 0.921939 4 0.03 4<x<6 0.11 0.113885 0.892360 6 0.04 6<x<7 0.15 0.158829 0.853142 7 0.04 7<x<8 0.19 0.205888 0.813924 8 0.05 8<x<9 0.24 0.267616 0.765201 9 0.07 9<x<10 0.31 0.359722 0.697871 10 0.06 10<x<11 0.37 0.446678 0.639750 11 0.07 11<x<12 0.44 0.557789 0.572473 12 0.07 12<x<13 0.51 0.682789 0.505206 13 0.48 13<x<22 0.99 1.662381 0.189687 22 0.01 x>22 1.00 2.662381 0.069782 Empirical Mean = 10.44 and Empirical Variance = 14.4064 98. a. 0.04x for 0 < x < 10 0.15 + 0.025x for 10 < x < 20 0.25 + 0.02x for 20 < x < 30 Undefined for x > 30 b. 0.04 for 0 < x < 10 0.025 for 10 < x < 20 0.02 for 20 < x < 30 Undefined for x > 30 c. 8 99. 47.50 and 3958.333333333 100. 20 Ht ˆ () St ˆ( ) 0 t 0.3 1 0 1 0.3 t 0.5 0.928571 0.071429 0.931063 0.5 t 1.0 0.773810 0.238095 0.788128 1.0 t 2.5 0.633117 0.419913 0.657104 t 2.5 0.575561 0.510823 0.600002 April 15, 2014