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An insurance policy offers coverage for three years. You are given the following about the annual loss:

  • Annual losses are independent
  • The annual loss has an exponential distribution with mean 500

Determine the mode of the loss distribution for the three year coverage.

  1. 500
  2. 750
  3. 1,000
  4. 2,000
  5. 2,500
  • Created by Admin, Jun 02'22

The random variables [math]X,Y[/math] have the joint density function

[[math]] f_{X,Y}(x,y) = \begin{cases} c xy \, e^{-xy}, 1 \lt x \lt y \\ 0, \, \textrm{Otherwise} \end{cases} [[/math]]

for a constant [math]c[/math]. Determine the joint density function for the random variables [math]W = Y^{-1}, Z = X^{-1}.[/math]

  • [[math]] f_{W,Z}(w,z) = \begin{cases} c wz \, e^{-1/wz}, 0 \lt w \lt z \lt 1 \\ 0, \, \textrm{Otherwise} \end{cases} [[/math]]
  • [[math]] f_{W,Z}(w,z) = \begin{cases} c wz \, e^{-w/z}, 0 \lt w \lt z \lt 1 \\ 0, \, \textrm{Otherwise} \end{cases} [[/math]]
  • [[math]]f_{W,Z}(w,z) = \begin{cases} c wz \, e^{-wz}, 1 \lt z \lt w \\ 0, \, \textrm{Otherwise} \end{cases}[[/math]]
  • [[math]]f_{W,Z}(w,z) = \begin{cases} c w^2z \, e^{-1/(wz)}, 0\lt z \lt w \lt 1 \\ 0, \, \textrm{Otherwise} \end{cases}[[/math]]
  • [[math]]f_{W,Z}(w,z) = \begin{cases} c wz^2 \, e^{-1/(wz)}, 0\lt z \lt w \lt 1 \\ 0, \, \textrm{Otherwise} \end{cases}[[/math]]
  • Created by Admin, Jun 02'22

A portfolio of four insurance policies have independent losses and identical ground-up (no deductible) loss distributions that are exponential with mean 1,000. Two of the policies have no deductible while the other two have a deductible equal to $400. Determine the density function of the sum of the losses for the policies with no deductible plus the sum of the payments for the policies with a deductible.

  • [[math]] \frac{xe^{-x/1000}}{10^6}[[/math]]
  • [[math]] \frac{xe^{-x/4000}}{16 \cdot 10^6}[[/math]]
  • [[math]] \frac{x^3e^{-x/1000}}{6 \cdot 10^{12}}[[/math]]
  • [[math]] \frac{x^2e^{-x/1000}}{ 24 \cdot 10^{9}}[[/math]]
  • [[math]] \frac{x^3e^{-x/4000}}{96 \cdot 10^{12}}[[/math]]
  • Created by Admin, Jun 02'22

A device containing two key components fails when, and only when, both components fail. The lifetimes, [math]T_1 [/math] and [math]T_2[/math] of these components are independent with common density function

[[math]] f(t) = \begin{cases} e^{-t}, \, 0\lt t \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

The cost, [math]X[/math], of operating the device until failure is [math]2T_1 + T_2[/math] . Let [math]g[/math] be the density function for [math]X[/math].

Determine [math]g(x)[/math], for [math]x \gt 0 [/math].

  • [math]e^{−x/2} − e^{−x}[/math]
  • [math]2(e^{-x/2} - e^{-x} )[/math]
  • [math]\frac{x^2e^{-x}}{2}[/math]
  • [math]\frac{e^{-x/2}}{2}[/math]
  • [math]\frac{e^{-x/3}}{3}[/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

  • Created by Admin, May 06'23

Let [math]X[/math] denote the loss amount sustained by an insurance company’s policyholder in an auto collision. Let [math]Z[/math] denote the portion of [math]X[/math] that the insurance company will have to pay. An actuary determines that [math]X[/math] and [math]Z[/math] are independent with respective density and probability functions

[[math]] f(x) = \begin{cases} (1/8)e^{-x/8}, \, x \gt0 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

and

[[math]] \operatorname{P}[Z = z] = \begin{cases} 0.45, \, z= 1 \\ 0.55, \, z=0 \end{cases} [[/math]]

Calculate the variance of the insurance company’s claim payment [math]ZX[/math].

  • 13.0
  • 15.8
  • 28.8
  • 35.2
  • 44.6

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

  • Created by Admin, May 06'23

For Company A there is a 60% chance that no claim is made during the coming year. If one or more claims are made, the total claim amount is normally distributed with mean 10,000 and standard deviation 2,000.

For Company B there is a 70% chance that no claim is made during the coming year. If one or more claims are made, the total claim amount is normally distributed with mean 9,000 and standard deviation 2,000.

The total claim amounts of the two companies are independent.

Calculate the probability that, in the coming year, Company B’s total claim amount will exceed Company A’s total claim amount.

  • 0.180
  • 0.185
  • 0.217
  • 0.223
  • 0.240

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

  • Created by Admin, May 07'23

A company offers earthquake insurance. Annual premiums are modeled by an exponential random variable with mean 2. Annual claims are modeled by an exponential random variable with mean 1. Premiums and claims are independent. Let [math]X[/math] denote the ratio of claims to premiums, and let [math]f[/math] be the density function of [math]X[/math].

Determine [math]f(x)[/math], where it is positive.

  • [math]\frac{1}{2x+1}[/math]
  • [math]\frac{2}{(2x+1)^2}[/math]
  • [math]e^{-x}[/math]
  • [math]2e^{-2x}[/math]
  • [math]xe^{-x}[/math]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

  • Created by Admin, May 07'23

In each of the months June, July, and August, the number of accidents occurring in that month is modeled by a Poisson random variable with mean 1. In each of the other 9 months of the year, the number of accidents occurring is modeled by a Poisson random variable with mean 0.5. Assume that these 12 random variables are mutually independent.

Calculate the probability that exactly two accidents occur in July through November.

  • 0.084
  • 0.185
  • 0.251
  • 0.257
  • 0.271

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

  • Created by Admin, May 08'23

In a given region, the number of tornadoes in a one-week period is modeled by a Poisson distribution with mean 2. The numbers of tornadoes in different weeks are mutually independent.

Calculate the probability that fewer than four tornadoes occur in a three-week period.

  • 0.13
  • 0.15
  • 0.29
  • 0.43
  • 0.86

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

  • Created by Admin, May 08'23

A certain brand of refrigerator has a useful life that is normally distributed with mean 10 years and standard deviation 3 years. The useful lives of these refrigerators are independent.

Calculate the probability that the total useful life of two randomly selected refrigerators will exceed 1.9 times the useful life of a third randomly selected refrigerator.

  • 0.407
  • 0.444
  • 0.556
  • 0.593
  • 0.604

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

  • Created by Admin, May 08'23