Revision as of 21:41, 19 January 2024 by Admin (Created page with "'''Answer: A''' Premium at issue for (40): <math>\frac{1000 A_{40}}{\ddot{a}_{40}}=\frac{121.06}{18.4578}=6.5587</math> Premium at issue for (80): <math>\frac{1000 A_{80}}{\ddot{a}_{80}}=\frac{592.93}{8.5484}=69.3615</math> Lives in force after ten years: Issued at age <math>40: 10,000_{10} p_{40}=10,000 \times \frac{98,576.4}{99,338.3}=9923.30</math> Issued at age <math>80: 10,000_{10} p_{80}=10,000 \times \frac{41,841.1}{75,657.2}=5530.35</math> The total number...")
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)

Exercise


ABy Admin
Jan 19'24

Answer

Answer: A

Premium at issue for (40): [math]\frac{1000 A_{40}}{\ddot{a}_{40}}=\frac{121.06}{18.4578}=6.5587[/math]

Premium at issue for (80): [math]\frac{1000 A_{80}}{\ddot{a}_{80}}=\frac{592.93}{8.5484}=69.3615[/math]

Lives in force after ten years:

Issued at age [math]40: 10,000_{10} p_{40}=10,000 \times \frac{98,576.4}{99,338.3}=9923.30[/math]

Issued at age [math]80: 10,000_{10} p_{80}=10,000 \times \frac{41,841.1}{75,657.2}=5530.35[/math]

The total number of lives after ten years is therefore: [math]9923.30+5530.35=15,453.65[/math]

The average premium after ten years is therefore:

[math]\frac{(6.5587 \times 9923.30)+(69.3615 \times 5530.35)}{15,453.65}=29.03[/math]

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

00