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28 exercise(s) shown, 18 hidden

Suppose the claim frequency [math]N[/math] has a probability function defined recursively by

[[math]] \operatorname{P}(N=k) = a \operatorname{P}(N=k-1) [[/math]]

where [math]a\gt0[/math] and [math]k \geq 1 [/math]. Which of the following expressions equals the expected value of [math]N[/math]?

  1. [math]a/(1-a)[/math]
  2. [math]a [/math]
  3. [math]a/(1-a^2)[/math]
  4. [math]a^2/(1-a^2) [/math]
  5. [math]1-a [/math]
  • Created by Admin, Jun 01'22

Suppose [math]F(x)[/math] is a continuous cumulative probability distribution function with [math]\lim_{x\rightarrow 1}F(x)=1[/math] and [math]F(x)\gt0[/math] for all [math]x[/math]. For which of the following [math]g(x)[/math] is [math]F(g(x))[/math] also a cumulative probability distribution function?

  • [math]x^2[/math]
  • [math]\sqrt{|x| + 1} [/math]
  • [math]e^{-x}[/math]
  • [math](1 + e^{-x})^{-1}[/math]
  • [math]1-\ln(1 + e^{-x})[/math]
  • Created by Admin, Jun 01'22

Suppose the loss has a continuous cumulative distribution function [math]F(x)[/math] with the following values:

F(0) F(250) F(500) F(800) F(1000) F(1500) F(2000)
0.25 0.4375 0.5 0.75 0.8125 0.9 1

If [math]P[/math] denotes the payment, determine the 25th percentile of [math]P^{-1}[/math].

  • 800
  • 1/800
  • 250
  • 1/250
  • 1/1000
  • Created by Admin, Jun 01'22

The number of injury claims per month is modeled by a random variable [math]N[/math] with

[[math]] \operatorname{P}[N=n] = \frac{1}{(n+1)(n+2)} [[/math]]

, for nonnegative integers, [math]n[/math]. Calculate the probability of at least one claim during a particular month, given that there have been at most four claims during that month.

  • 1/3
  • 2/5
  • 1/2
  • 3/5
  • 5/6

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

  • Created by Admin, May 01'23

The loss due to a fire in a commercial building is modeled by a random variable [math]X[/math] with density function

[[math]] f(x) = \begin{cases} 0.005(20 − x), \, 0\lt x \lt 20 \\ 0, \, \textrm{otherwise.} \end{cases} [[/math]]

Given that a fire loss exceeds 8, calculate the probability that it exceeds 16.

  • 1/25
  • 1/9
  • 1/8
  • 1/3
  • 3/7

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

  • Created by Admin, May 01'23

The lifetime of a machine part has a continuous distribution on the interval (0, 40) with probability density function [math]f(x)[/math], where [math]f(x)[/math] is proportional to [math](10+x)^{-2}[/math] on the interval.

Calculate the probability that the lifetime of the machine part is less than 6.

  • 0.04
  • 0.15
  • 0.47
  • 0.53
  • 0.94

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

  • Created by Admin, May 01'23

A group insurance policy covers the medical claims of the employees of a small company. The value, [math]V[/math], of the claims made in one year is described by [math]V = 100,000Y[/math] where [math]Y[/math] is a random variable with density function

[[math]] f(y) = \begin{cases} k(1-y)^4, \, 0\lt y \lt 1 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

where [math]k[/math] is a constant. Calculate the conditional probability that [math]V[/math] exceeds 40,000, given that [math]V[/math] exceeds 10,000.

  • 0.08
  • 0.13
  • 0.17
  • 0.20
  • 0.51

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

  • Created by Admin, May 01'23

An insurance policy pays for a random loss [math]X[/math] subject to a deductible of [math]C[/math], where [math]0 \lt C \lt 1[/math] . The loss amount is modeled as a continuous random variable with density function

[[math]] f(x) = \begin{cases} 2x, \, 0 \lt x \lt 1 \\ 0, \, \textrm{otherwise.} \end{cases} [[/math]]

Given a random loss [math]X[/math], the probability that the insurance payment is less than 0.5 is equal to 0.64. Calculate [math]C[/math].

  • 0.1
  • 0.3
  • 0.4
  • 0.6
  • 0.8

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

  • Created by Admin, May 01'23

An insurance policy pays 100 per day for up to three days of hospitalization and 50 per day for each day of hospitalization thereafter. The number of days of hospitalization, [math]X[/math], is a discrete random variable with probability function

[[math]] \operatorname{P}[X=k] = \begin{cases} \frac{6-k}{15}, \, k =1,2,3,4,5 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Determine the expected payment for hospitalization under this policy.

  • 123
  • 210
  • 220
  • 270
  • 367

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

  • Created by Admin, May 01'23

Let [math]X[/math] be a continuous random variable with density function

[[math]] f(x) = \begin{cases} \frac{|x|}{10}, \, -2 \leq x \leq 4 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the expected value of [math]X[/math].

  • 1/5
  • 3/5
  • 1
  • 28/15
  • 12/5

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

  • Created by Admin, May 01'23