Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Sebuah asuransi dwiguna (endowment insurance) n-tahun sebesar 1.000 untuk (x), diketahui:
- Manfaat kematian dibayarkan pada saat kematian
- Premium dibayarkan secara tahunan setiap awal tahun
- Kematian berdistribusi seragam pada seluruh usia
- \(i = 0,05\)
- \({}_n{E_x} = 0,172\)
- \({\bar A_{x:\left. {\overline {\, n \,}}\! \right| }} = 0,192\)
Tentukan premi netto tahunan untuk asuransi di atas.
- 10,1
- 11,3
- 12,5
- 13,7
- 14,9
| Misalkan | P menyatakan premi netto tahunan |
| Step 1 | \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{1 – {A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{d}\) \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{1 – \left( {{A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} + {A_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }}} \right)}}{{\left( {iv} \right)}}\) \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{1 – \left( {{A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} + {A_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }}} \right)}}{{\left( {iv} \right)}}\) |
| Step 2 | \({\bar A_{x:\left. {\overline {\, n \,}}\! \right| }} = \left( {{{\bar A}_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} + {{\bar A}_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^| }}} \right)\) \({\bar A_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{i}{\delta }\left( {{A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}} \right) + {}_n{E_x}\) \(0,192 = \frac{{0,05}}{{\ln (1,05)}}\left( {{A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }}} \right) + 0,172\) \({A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} = 0,01951616567\) \({A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} \cong 0,0195\) |
| \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{1 – \left( {{A_{\mathop x\limits^| :\left. {\overline {\, n \,}}\! \right| }} + {}_n{E_x}} \right)}}{{\left( {iv} \right)}}\) \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{1 – \left( {0,0195 + 0,172} \right)}}{{\left( {\frac{{0,05}}{{1,05}}} \right)}}\) \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = 16,9785\) | |
| Maka | \(P = \frac{{B{{\bar A}_{x:\left. {\overline {\, n \,}}\! \right| }}}}{{{{\ddot a}_{x:\left. {\overline {\, n \,}}\! \right| }}}}\) \(P = \frac{{1.000(0,192)}}{{16,9785}}\) \(P = 11,30841947\) \(P \cong 11,3\) |
| Jawaban | b. 11,3 |