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 |
