⧼exchistory⧽
To view all exercises, please subscribe to guide
11 exercise(s) shown, 9 hidden
Jun 01'22

You are given the following:

  • Year 1 loss has an exponential distribution with mean $1,250.
  • Claim costs rise 5% from year 1 to year 2
  • The policy has a deductible of $400 in effect during years 1 and 2.
  • x1 denotes the probability that payment exceeds $500 in year 1, given that payment is positive.
  • x2 denotes the probability that payment exceeds $500 in year 2, given that payment is positive.

Which interval contains the ratio x2/x1?

  • [0.5, 0.7]
  • [0.98, 1.03]
  • [1.1, 1.2]
  • [1.3, 1.4]
  • 1.5+
May 04'23

The time, [math]T[/math], that a manufacturing system is out of operation has cumulative distribution function

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

The resulting cost to the company is [math]Y = T^2[/math]. Let [math]g[/math] be the density function for [math]Y[/math]. Determine [math]g(y)[/math], for [math]y \gt 4 [/math].

  • [math]\frac{4}{y^2}[/math]
  • [math]\frac{8}{y^{3/2}}[/math]
  • [math]\frac{8}{y^3}[/math]
  • [math]\frac{16}{y}[/math]
  • [math]\frac{1024}{y^5}[/math]

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

May 04'23

An investment account earns an annual interest rate [math]R[/math] that follows a uniform distribution on the interval (0.04, 0.08). The value of a 10,000 initial investment in this account after one year is given by [math]V = 10 000e^R. [/math] Let [math]F[/math] be the cumulative distribution function of [math]V[/math]. Determine [math]F(v)[/math] for values of [math]v[/math] that satisfy [math]0 \lt F(v) \lt 1 [/math].

  • [math]\frac{10000e^{v /10000} − 10408}{425}[/math]
  • [math]25e^{v/10000} - 0.04[/math]
  • [math]\frac{v-10408}{10833-10408}[/math]
  • [math]\frac{25}{v}[/math]
  • [math]25\left[ \ln(v/10000) - 0.04 \right][/math]

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

May 04'23

An actuary models the lifetime of a device using the random variable [math]Y = 10X^{0.8}[/math], where [math]X[/math] is an exponential random variable with mean 1. Let [math]f(y)[/math] be the density function for [math]Y[/math].

Determine [math]f(y)[/math] for [math]y\gt0[/math].

  • [math]10 y^{0.8} \exp(−8 y^{−0.2} )[/math]
  • [math]8 y^{−0.2} \exp(−10 y^{0.8} )[/math]
  • [math]8 y^{−0.2} \exp[−(0.1y)^{1.25} ][/math]
  • [math](0.1y )^{1.25} \exp[−0.125(0.1y)^{0.25} ][/math]
  • [math]0.125(0.1y )^{0.25} \exp[−(0.1y )^{1.25} ][/math]

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

May 04'23

Let [math]T[/math] denote the time in minutes for a customer service representative to respond to 10 telephone inquiries. [math]T[/math] is uniformly distributed on the interval with endpoints 8 minutes and 12 minutes. Let [math]R[/math] denote the average rate, in customers per minute, at which the representative responds to inquiries, and let [math]f(r)[/math] be the density function for [math]R[/math].

Determine [math]f(r)[/math], for [math]\frac{10}{12} \lt r \lt \frac{10}{8}[/math].

  • 12/5
  • 3-5/2r
  • [math]3r-5\ln(r)/2[/math]
  • [math]\frac{10}{r^2}[/math]
  • [math]\frac{5}{2r^2}[/math]

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

Jul 07'24

A number [math]U[/math] is chosen at random in the interval [math][0,1][/math]. Find the probability that [math]T = U/(1 - U) \lt 1/4[/math].

  • 1/6
  • 1/5
  • 1/4
  • 1/2
  • 2/3

References

Doyle, Peter G. (2006). "Grinstead and Snell's Introduction to Probability" (PDF). Retrieved June 6, 2024.

Jun 01'22

An investor has invested $50,000 in a stock. Denote by [math]S_t[/math] the stock price at time [math]t[/math] expressed in years with [math]S_0[/math] the current stock price. Suppose [math]\ln(S_t/S_0) [/math] is normally distributed with mean [math]\mu = 0.05t [/math] and standard deviation [math]\sigma = 0.1\sqrt{t} [/math]. Determine the interval containing the 99th percentile for the investor's $50,000 investment for a 30 day period. Assume 365 days in a year

  • [46000, 50000]
  • [52000, 55000]
  • [58000, 60000]
  • [75000, 77000]
  • [77000, [math]\infty[/math]]
Jun 01'22

Losses for year 1 equal

[[math]]\frac{1500(1-X^{1/3})}{1 + X^{1/3}}[[/math]]

with [math]X[/math] a non-negative random variable bounded by 1 with cumulative distribution function

[[math]] F(u) = u^2. [[/math]]

Determine the probability that losses exceed $300.

  • 0.064
  • 0.0878
  • 0.148
  • 0.296
  • 0.444
Jun 01'22

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

[[math]] f(x) = 6x(1 - x), \quad \text{for } 0 \le x \le 1 [[/math]]

Suppose that [math]Y = \exp(\exp(X)) [/math], determine the mode of the distribution of [math]Y[/math].

  • 2.7
  • 3.9
  • 5.2
  • 6.4
  • 7.8
May 04'23

The monthly profit of Company I can be modeled by a continuous random variable with density function [math]f[/math]. Company II has a monthly profit that is twice that of Company I. Let [math]g[/math] be the density function for the distribution of the monthly profit of Company II.

Determine [math]g(y)[/math] where it is not zero.

  • [math]\frac{1}{2}f(\frac{y}{2})[/math]
  • [math]f(\frac{y}{2})[/math]
  • [math]2f(\frac{y}{2})[/math]
  • [math]2f(y)[/math]
  • [math]2f(2y)[/math]

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