For an [math]n[/math]-year endowment insurance of 1000 on ( [math]x[/math] ), you are given:

(i) Death benefits are payable at the moment of death

(ii) Premiums are payable annually at the beginning of each year

(iii) Deaths are uniformly distributed over each year of age

(iv) [math]\quad i=0.05[/math]

(v) [math]{ }_{n} E_{x}=0.172[/math]

(vi) [math]\quad \bar{A}_{x: n}=0.192[/math]

Calculate the annual net premium for this insurance.

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• 11.3
• 12.5
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• 14.9

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

Answer: B

Let [math]P[/math] be the annual net premium

[math]P=\frac{1000 \bar{A}_{x: n}}{\ddot{a}_{x: n}}=\frac{1000(0.192)}{\ddot{a}_{x: n}}[/math]

where

[math]\ddot{a}_{x: n]}=\frac{1-A_{x: n}}{d}=\frac{(1.05)}{(0.05)}\left(1-A_{x: n]}^{1}-A_{x: n]}^{1}\right)[/math]

[math]A_{x: n}=\frac{i}{\delta}\left(A_{x: n}^{1}\right)+{ }_{n} E_{x}[/math]

[math]\Rightarrow 0.192=\frac{0.05}{0.04879}\left(A_{x: n}^{1}\right)+0.172[/math]

[math]\Rightarrow A_{x: n}^{1}=0.019516[/math]

[math]\Rightarrow \ddot{a}_{x: n \mid}=\frac{1.05}{0.05}(1-0.019516-0.172)=16.978[/math]

Therefore, we have

[math]P=\frac{1000(0.192)}{16.978}=11.31[/math]

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