Revision as of 00:27, 16 January 2024 by Admin (Created page with "For a mortality table with a select period of two years, you are given: {| class="table" ! <math>x</math> !! <math>q_{[x]}</math> !! <math>q_{[x]+1}</math> !! <math>q_{x+2}</math> !! <math>x+2</math> |- | 50 || 0.0050 || 0.0063 || 0.0080 || 52 |- | 51 || 0.0060 || 0.0073 || 0.0090 || 53 |- | 52 || 0.0070 || 0.0083 || 0.0100 || 54 |- | 53 || 0.0080 || 0.0093 || 0.0110 || 55 |} The force of mortality is constant between integral ages. Calculate <math>1000_{2.5} q_{[50]...")
Jan 16'24
Exercise
For a mortality table with a select period of two years, you are given:
| [math]x[/math] | [math]q_{[x]}[/math] | [math]q_{[x]+1}[/math] | [math]q_{x+2}[/math] | [math]x+2[/math] |
|---|---|---|---|---|
| 50 | 0.0050 | 0.0063 | 0.0080 | 52 |
| 51 | 0.0060 | 0.0073 | 0.0090 | 53 |
| 52 | 0.0070 | 0.0083 | 0.0100 | 54 |
| 53 | 0.0080 | 0.0093 | 0.0110 | 55 |
The force of mortality is constant between integral ages.
Calculate [math]1000_{2.5} q_{[50]+0.4}[/math].
- 15.2
- 16.4
- 17.7
- 19.0
- 20.2
Jan 16'24
Answer: B
[[math]]
\begin{aligned}
{ }_{2.5} q_{[50]+0.4} & =1-{ }_{2.5} p_{[50]+0.4}=1-{ }_{2.9} p_{[50]} /\left(p_{[50]}\right)^{0.4} \\
& =1-\left\{p_{[50]} p_{[50]+1}\left(p_{52}\right)^{0.9}\right\} /\left(1-q_{[50]}\right)^{0.4} \\
& =1-\left\{\left(1-q_{[50]}\right)\left(1-q_{[50]+1}\right)\left(1-q_{52}\right)^{0.9}\right\} /\left(1-q_{[50]}\right)^{0.4} \\
& =1-\left\{(1-0.0050)(1-0.0063)(1-0.0080)^{0.9}\right\} /(1-0.0050)^{0.4} \\
& =0.01642
\end{aligned}
[[/math]]
[math]1000_{2.5} q_{[50]+0.4}=16.42[/math]
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.