⧼exchistory⧽
12 exercise(s) shown, 0 hidden
ABy Admin
Jan 15'24

You are given:

(i) [math]\quad S_{0}(t)=\left(1-\frac{t}{\omega}\right)^{\frac{1}{4}}[/math], for [math]0 \leq t \leq \omega[/math]

(ii) [math]\quad \mu_{65}=\frac{1}{180}[/math]

Calculate [math]e_{106}[/math], the curtate expectation of life at age 106 .

  • 2.2
  • 2.5
  • 2.7
  • 3.0
  • 3.2

Copyright 2024 . The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

ABy Admin
Jan 15'24

Scientists are searching for a vaccine for a disease. You are given:

(i) 100,000 lives age [math]x[/math] are exposed to the disease

(ii) Future lifetimes are independent, except that the vaccine, if available, will be given to all at the end of year 1

(iii) The probability that the vaccine will be available is 0.2

(iv) For each life during year [math]1, q_{x}=0.02[/math]

(v) For each life during year 2, [math]q_{x+1}=0.01[/math] if the vaccine has been given, and [math]q_{x+1}=0.02[/math] if it has not been given

Calculate the standard deviation of the number of survivors at the end of year 2.

  • 100
  • 200
  • 300
  • 400
  • 500

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

ABy Admin
Jan 15'24

You are given that mortality follows Gompertz Law with [math]B=0.00027[/math] and [math]c=1.1[/math]. Calculate [math]f_{50}(10)[/math].

  • 0.048
  • 0.050
  • 0.052
  • 0.054
  • 0.056

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

ABy Admin
Jan 15'24

You are given the survival function:

[[math]]S_{0}(x)=\left(1-\frac{x}{60}\right)^{\frac{1}{3}}, \quad 0 \leq x \leq 60.[[/math]]

Calculate [math]1000 \mu_{35}[/math].

  • 5.6
  • 6.7
  • 13.3
  • 16.7
  • 20.1

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

ABy Admin
Jan 15'24

You are given the following survival function of a newborn:

[[math]] S_{0}(x)= \begin{cases}1-\frac{x}{250}, & 0 \leq x\lt40 \\ 1-\left(\frac{x}{100}\right)^{2}, & 40 \leq x \leq 100\end{cases} [[/math]]


Calculate the probability that (30) dies within the next 20 years.

  • 0.13
  • 0.15
  • 0.17
  • 0.19
  • 0.21

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

ABy Admin
Jan 15'24

In a population initially consisting of [math]75 \%[/math] females and [math]25 \%[/math] males, you are given:

(i) For a female, the force of mortality is constant and equals [math]\mu[/math]

(ii) For a male, the force of mortality is constant and equals [math]1.5 \mu[/math]

(iii) At the end of 20 years, the population is expected to consist of [math]85 \%[/math] females and [math]15 \%[/math] males

Calculate the probability that a female survives one year.

  • 0.89
  • 0.92
  • 0.94
  • 0.96
  • 0.99

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

ABy Admin
Jan 15'24

You are given that mortality follows Makeham's Law with the following parameters:

[[math]] \begin{array}{ll} \text { i) } & A=0.004 \\ \text { ii) } & B=0.00003 \\ \text { iii) } & c=1.1 \end{array} [[/math]]


Let [math]L_{15}[/math] be the random variable representing the number of lives alive at the end of 15 years if there are 10,000 lives age 50 at time 0 .

Calculate [math]\operatorname{Var}\left[L_{15}\right][/math].

  • 1,317
  • 1,328
  • 1,339
  • 1,350
  • 1,361

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

ABy Admin
Jan 15'24

You are given:

i)[math] q_{80}=0.04[/math]

ii)[math] q_{81}=0.06[/math]

iii)[math]q_{82}=0.08[/math]

iv) Deaths between ages 80 and 81 are uniformly distributed

v) Deaths between ages 81 and 82 are subject to a constant force of mortality

Calculate the probability that a person aged 80.6 will die between ages 81.1 and 81.6.

  • 0.0294
  • 0.0296
  • 0.0298
  • 0.0300
  • 0.0302

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

ABy Admin
Jan 15'24

For a new light bulb, you are given:

i) [math]{ }_{t} q_{0}=\frac{t^{2}+t}{72}[/math] for [math]0 \leq t \leq 8[/math]

ii) [math]T_{0}[/math] is the random variable representing the future lifetime

Calculate [math]\operatorname{Var}\left[T_{0}\right][/math].

  • 3.9
  • 4.1
  • 4.3
  • 4.5
  • 4.7

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

ABy Admin
Jan 16'24

You are given the following:

(i) [math] e_{40:20}=18[/math]

(ii) [math]e_{60}=25[/math]

(iii) [math]{ }_{20} q_{40}=0.2[/math]

(iv) [math]q_{40}=0.003[/math]

Calculate [math]e_{41}[/math].

  • 36.1
  • 37.1
  • 38.1
  • 39.1
  • 40.1

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.