Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
Juni 2015 |
| Nomor Soal |
: |
18 |
SOAL
Untuk suatu tabel double decrement, diberikan:
- \(q_x^{‘\left( 2 \right)} = 2q_x^{‘\left( 1 \right)}\)
- \(q_x^{‘\left( 1 \right)} + q_x^{‘\left( 2 \right)} = q_x^{\left( \tau \right)} + 0,18\)
Hitunglah \(q_x^{‘\left( 2 \right)}\)
- 0,2
- 0,3
- 0,4
- 0,6
- 0,7
| Diketahui |
- \(q_x^{‘\left( 2 \right)} = 2q_x^{‘\left( 1 \right)}\)
- \(q_x^{‘\left( 1 \right)} + q_x^{‘\left( 2 \right)} = q_x^{\left( \tau \right)} + 0,18\)
|
| Rumus yang digunakan |
\({}_tp_x^{\left( \tau \right)} = \prod\limits_{j = 1}^n {{}_tp_x^{‘\left( j \right)}} = \prod\limits_{j = 1}^n {\left( {1 – {}_tq_x^{‘\left( j \right)}} \right)} \)
\(q_x^{\left( \tau \right)} = 1 – p_x^{\left( \tau \right)} = 1 – \left( {1 – q_x^{‘\left( 1 \right)}} \right)\left( {1 – q_x^{‘\left( 2 \right)}} \right)\) |
| Proses pengerjaan |
\(q_x^{‘\left( 1 \right)} + q_x^{‘\left( 2 \right)} = q_x^{\left( \tau \right)} + 0.18\)
\(0.5q_x^{‘\left( 2 \right)} + q_x^{‘\left( 2 \right)} = 1 – \left[ {\left( {1 – 0.5q_x^{‘\left( 2 \right)}} \right)\left( {1 – q_x^{‘\left( 2 \right)}} \right)} \right] + 0.18\)
\(1.5q_x^{‘\left( 2 \right)} = 1 – 1 + 1.5q_x^{‘\left( 2 \right)} – 0.5{\left( {q_x^{‘\left( 2 \right)}} \right)^2} + 0.18\)
\(0.5{\left( {q_x^{‘\left( 2 \right)}} \right)^2} = 0.18\)
\(q_x^{‘\left( 2 \right)} = \sqrt {\frac{{0.18}}{{0.5}}} = 0.6\) |
| Jawaban |
d. 0,6 |
