Revision as of 02:56, 20 January 2024 by Admin (Created page with "'''Answer: A''' Let <math>P=0.00253</math> be the monthly net premium per 1 of insurance. <math <math display="block"> \begin{aligned} { }_{10} V & =100,000\left[\frac{i}{\delta} A_{55: 10 \mid}^{1}+A_{55: 10 \mid}-12 P \ddot{a}_{55: 10 \mid}^{(12)}\right] \\ & =100,000[1.02480(0.02471)+0.59342-(12)(0.00253)(7.8311)] \\ & \approx 38,100 \end{aligned} </math> Where <math <math display="block"> \begin{aligned} A_{55: \overline{10}}^{1} & =A_{55: \overline{10}}-{ }...")
Exercise
ABy Admin
Jan 20'24
Answer
Answer: A
Let [math]P=0.00253[/math] be the monthly net premium per 1 of insurance.
[[math]]
\begin{aligned}
{ }_{10} V & =100,000\left[\frac{i}{\delta} A_{55: 10 \mid}^{1}+A_{55: 10 \mid}-12 P \ddot{a}_{55: 10 \mid}^{(12)}\right] \\
& =100,000[1.02480(0.02471)+0.59342-(12)(0.00253)(7.8311)] \\
& \approx 38,100
\end{aligned}
[[/math]]
Where
[[math]]
\begin{aligned}
A_{55: \overline{10}}^{1} & =A_{55: \overline{10}}-{ }_{10} E_{55}=0.61813-0.59342=0.02471 \\
A_{55: 10 \mid} & ={ }_{10} E_{55}=0.59342 \\
\ddot{a}_{55: \overline{10}} & =8.0192 \\
\ddot{a}_{55: 10 \mid}^{(12)} & =\alpha(12) \ddot{a}_{55: \overline{10}}-\beta(12)\left[1-{ }_{10} E_{55}\right] \\
& =1.00020(8.0192)-0.46651(1-0.59342)=7.8311
\end{aligned}
[[/math]]
Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.