Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
November 2014 |
| Nomor Soal |
: |
26 |
SOAL
Diketahui \({\mu _x} = 0,04\) untuk \(0 < x \le 40\) dan \({\mu _x} = 0,05\) untuk \(x > 40\). Manakah dari pilhan nilai di bawah ini yang paling mendekati untuk \(e_{25:\left. {\overline {\, {25} \,}}\! \right| }^0\)
- 12,6
- 15,6
- 10,6
- 8,6
- 6,6
| Diketahui |
\({\mu _x} = 0,04\) untuk \(0 < x \le 40\) dan \({\mu _x} = 0,05\) untuk \(x > 40\) |
| Rumus yang digunakan |
\({}_t{p_x} = \exp \left[ { – \int\limits_0^t {{\mu _{x + s}}ds} } \right]\)
\(e_{x:\left. {\overline {\, n \,}}\! \right| }^0 = \int\limits_0^n {{}_t{p_x}dt} \) |
| Proses pengerjaan |
Untuk seseorang yang berusia \(x = 25\)
- \(t \le 15\)
\({}_t{p_{25}} = \exp \left[ { – \int\limits_0^t {0.04ds} } \right] = \exp \left[ { – 0.04t} \right]\)
- \(t > 15\)
\({}_t{p_{25}} = {}_{15}{p_{25}} \cdot \exp \left[ { – \int\limits_{15}^t {0.05ds} } \right]\)
=\(\exp \left[ { – 0.04\left( {15} \right)} \right] \cdot \exp \left[ { – 0.05\left( {t – 15} \right)} \right]\)
|
|
\(e_{25:\left. {\overline {\, {25} \,}}\! \right| }^0 = \int\limits_0^{40} {{}_t{p_{25}}dt} = \int\limits_0^{15} {{}_t{p_{25}}dt} + \int\limits_{15}^{25} {{}_t{p_{25}}dt} \)
\(e_{25:\left. {\overline {\, {25} \,}}\! \right| }^0 = \int\limits_0^{15} {\exp \left[ { – 0.04t} \right]dt} + \exp \left[ { – 0.04\left( {15} \right)} \right]\int\limits_{15}^{25} {\exp \left[ { – 0.05\left( {t – 15} \right)} \right]dt} \)
\(e_{25:\left. {\overline {\, {25} \,}}\! \right| }^0 = – \frac{1}{{0.04}}\left[ {\exp \left( { – 0.04\left( {15} \right)} \right) – 1} \right] – \frac{{\exp \left( { – 0.04\left( {15} \right)} \right)}}{{0.05}}\left[ {\exp \left[ { – 0.05\left( {25 – 15} \right)} \right] – 1} \right]\)
\(e_{25:\left. {\overline {\, {25} \,}}\! \right| }^0 = \frac{1}{{0.04}}\left[ {1 – \exp \left( { – 0.04\left( {15} \right)} \right)} \right] + \frac{{\exp \left( { – 0.04\left( {15} \right)} \right)}}{{0.05}}\left[ {1 – \exp \left[ { – 0.05\left( {25 – 15} \right)} \right]} \right]\)
\(e_{25:\left. {\overline {\, {25} \,}}\! \right| }^0 = 25 – 25 \cdot \exp \left( { – 0.6} \right) + 20 \cdot \exp \left( { – 0.6} \right) – 20 \cdot \exp \left( { – 1.1} \right)\)
\(e_{25:\left. {\overline {\, {25} \,}}\! \right| }^0 = 15.598520\) |
| Jawaban |
b. 15,6 |
