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28 exercise(s) shown, 16 hidden
May 01'23

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.

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.

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.

May 01'23

A manufacturer’s annual losses follow a distribution with density function

[[math]] f(x) = \begin{cases} \frac{2.5(0.6)^{2.5}}{x^{3.5}}, \, x \gt 0.6 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

To cover its losses, the manufacturer purchases an insurance policy with an annual deductible of 2. Calculate the mean of the manufacturer’s annual losses not paid by the insurance policy.

  • 0.84
  • 0.88
  • 0.93
  • 0.95
  • 1.00

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

May 01'23

An insurance policy reimburses a loss up to a benefit limit of 10. The policyholder’s loss, [math]Y[/math], follows a distribution with density function:

[[math]] f(y) = \begin{cases} 2y^{-3}, \, y \gt 1 \\ 0, \, \textrm{Otherwise.} \end{cases} [[/math]]

Calculate the expected value of the benefit paid under the insurance policy.

  • 1.0
  • 1.3
  • 1.8
  • 1.9
  • 2.0

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

May 01'23

An insurance company’s monthly claims are modeled by a continuous, positive random variable [math]X[/math], whose probability density function is proportional to [math](1 + x)^{- 4}[/math] for [math] x \gt 0 [/math] .

Calculate the company’s expected monthly claims.

  • 1/6
  • 1/3
  • 1/2
  • 1
  • 3

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

May 01'23

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

[[math]] f(x) = \begin{cases} \frac{p-1}{x^p}, \, x \gt 1 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Calculate the value of [math]p[/math] such that [math]\operatorname{E}(X) = 2 [/math].

  • 1
  • 2.5
  • 3
  • 5
  • There is no such [math]p[/math]

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

May 01'23

The lifetime of a light bulb has density function, [math]f[/math], where [math]f(x)[/math] is proportional to

[[math]] \frac{x^2}{1+x^3}, \, 0 \lt x \lt5, \, \textrm{and}, \, 0 \, \textrm{otherwise}. [[/math]]

Calculate the mode of this distribution.

  • 0.00
  • 0.79
  • 1.26
  • 4.42
  • 5.00

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

May 08'23

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

[[math]] f(x) = \begin{cases} 2e^{-2x}, \, x \gt0 \\ 0, \, \textrm{otherwise} \end{cases} [[/math]]

Calculate [math]\operatorname{P}[ X \leq 0.5 | X \leq 1.0].[/math]

  • 0.433
  • 0.547
  • 0.632
  • 0.731
  • 0.865

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

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 a non-zero loss, determine the 25th percentile of [math]P^{-1}[/math].

  • 800
  • 1/800
  • 250
  • 1/250
  • 1/1000