Revision as of 21:40, 19 January 2024 by Admin (Created page with "'''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}...")
Exercise
ABy Admin
Jan 19'24
Answer
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]
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