Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | November 2015 |
| Nomor Soal | : | 19 |
SOAL
Diberikan sebagai berikut:
- \({A_x} = 0,28\)
- \({A_{x + 20}} = 0,40\)
- \({A_{x:\mathop {\left. {\overline {\, {20} \,}}\! \right| }\limits^1 }} = 0,25\)
- \(i = 0,05\)
Hitunglah \({a_{x:\left. {\overline {\, {20} \,}}\! \right| }}\)
- 11,0
- 11,2
- 11,7
- 12,0
- 12,3
| Diketahui |
|
| Rumus yang digunakan | \({A_{x:\left. {\overline {\, n \,}}\! \right| }} = A_{x:\left. {\overline {\, n \,}}\! \right| }^1 + {A_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^1 }} = {A_x} – {A_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^1 }} \cdot {A_{x + 20}} + {A_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^1 }}\) \({\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} = \frac{{1 – {A_{x:\left. {\overline {\, n \,}}\! \right| }}}}{d}\) dengan \(d = \frac{i}{{1 + i}}\) \({a_{x:\left. {\overline {\, n \,}}\! \right| }} = {\ddot a_{x:\left. {\overline {\, n \,}}\! \right| }} – 1 + {A_{x:\mathop {\left. {\overline {\, n \,}}\! \right| }\limits^1 }}\) |
| Proses pengerjaan | \({A_{x:\left. {\overline {\, {20} \,}}\! \right| }} = {A_x} – {A_{x:\mathop {\left. {\overline {\, {20} \,}}\! \right| }\limits^1 }} \cdot {A_{x + 20}} + {A_{x:\mathop {\left. {\overline {\, {20} \,}}\! \right| }\limits^1 }}\) \({A_{x:\left. {\overline {\, {20} \,}}\! \right| }} = 0.28 – 0.25\left( {0.40} \right) + 0.25 = 0.43\) |
| \({{\ddot a}_{x:\left. {\overline {\, {20} \,}}\! \right| }} = \frac{{1 – {A_{x:\left. {\overline {\, {20} \,}}\! \right| }}}}{d}\) \({{\ddot a}_{x:\left. {\overline {\, {20} \,}}\! \right| }} = \frac{{1 – {A_{x:\left. {\overline {\, {20} \,}}\! \right| }}}}{{\frac{i}{{1 + i}}}}\) \({{\ddot a}_{x:\left. {\overline {\, {20} \,}}\! \right| }} = \frac{{1 – 0.43}}{{\frac{{0.05}}{{1.05}}}} = 11.97\) | |
| \({a_{x:\left. {\overline {\, {20} \,}}\! \right| }} = {{\ddot a}_{x:\left. {\overline {\, {20} \,}}\! \right| }} – 1 + {A_{x:\mathop {\left. {\overline {\, {20} \,}}\! \right| }\limits^1 }}\) \({a_{x:\left. {\overline {\, {20} \,}}\! \right| }} = 11.97 – 1 + 0.25 = 11.22\) | |
| Jawaban | b. 11,2 |
