Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | November 2018 |
| Nomor Soal | : | 12 |
SOAL
Apabila setiap decrement berdistribusi seragam untuk setiap tahun usia dalam tabel double decrement berikut:
| Umur | \(l_x^{\left( \tau \right)}\) | \(d_x^{\left( 1 \right)}\) | \(d_x^{\left( 2 \right)}\) |
| 40 | 1000 | 60 | 55 |
| 41 | | | 70 |
| 42 | 750 | | |
Tetukanlah \(\mathop {q’}\nolimits_{41}^{(1)} \) !
- 0,077
- 0,079
- 0,081
- 0,083
- 0,085
| Diketahui | Apabila setiap decrement berdistribusi seragam untuk setiap tahun usia dalam tabel double decrement berikut:| Umur | \(l_x^{\left( \tau \right)}\) | \(d_x^{\left( 1 \right)}\) | \(d_x^{\left( 2 \right)}\) | | 40 | 1000 | 60 | 55 | | 41 | | | 70 | | 42 | 750 | | |
|
| Rumus yang digunakan | - \(l_{x + 1}^{\left( \tau \right)} = l_x^{\left( \tau \right)} – d_x^{\left( 1 \right)} – d_x^{\left( 2 \right)}\)
- \(q_x^{\left( \tau \right)} = \frac{{d_x^{\left( \tau \right)}}}{{l_x^{\left( \tau \right)}}} = \frac{{d_x^{\left( 1 \right)} + d_x^{\left( 2 \right)}}}{{l_x^{\left( \tau \right)}}}\)
- \(q_x^{\left( 1 \right)} = \frac{{d_x^{\left( 1 \right)}}}{{l_x^{\left( \tau \right)}}}\)
- \(\mathop {P’}\nolimits_x^{(1)} = {\left( {p_x^{\left( \tau \right)}} \right)^{\frac{{q_x^{\left( 1 \right)}}}{{q_x^{\left( \tau \right)}}}}}\)
|
| Proses pengerjaan | \(l_{41}^{\left( \tau \right)} = l_{40}^{\left( \tau \right)} – d_{40}^{\left( 1 \right)} – d_{40}^{\left( 2 \right)} = 1000 – 60 – 55 = 885\)
\(d_{41}^{\left( 1 \right)} = l_{41}^{\left( \tau \right)} – l_{42}^{\left( \tau \right)} – d_{41}^{\left( 2 \right)} = 885 – 750 – 70 = 65\)| Umur | \(l_x^{\left( \tau \right)}\) | \(d_x^{\left( 1 \right)}\) | \(d_x^{\left( 2 \right)}\) | | 40 | 1000 | 60 | 55 | | 41 | 885 | 65 | 70 | | 42 | 750 | | |
\(q_{41}^{\left( \tau \right)} = \frac{{d_{41}^{\left( 1 \right)} + d_{41}^{\left( 2 \right)}}}{{l_{41}^{\left( \tau \right)}}} = \frac{{65 + 70}}{{885}} = 0.152542\)
\(q_{41}^{\left( 1 \right)} = \frac{{d_{41}^{\left( 1 \right)}}}{{l_{41}^{\left( \tau \right)}}} = \frac{{65}}{{885}} = 0.073446\)
\(\mathop {P’}\nolimits_{41}^{(1)} = {\left( {p_{41}^{\left( \tau \right)}} \right)^{\frac{{\mathop q\nolimits_{41}^{(1)} }}{{\mathop q\nolimits_{41}^{(\tau )} }}}} = {\left( {1 -0.152542}\right)^{\frac{{0.073446}}{{0.152542}}}} = 0.923401\)
\(q_{41}^{\left( 1 \right)} = 1 – \mathop {P’}\nolimits_{41}^{(1)} = 1 – 0.923401 = 0.076599\) |
| Jawaban | A. 0,077 |
