Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Untuk sebuah table double decrement, diberikan:
| Usia \(\left( x \right)\) | \(l_x^{\left( \tau \right)}\) | \(d_x^{\left( 1 \right)}\) | \(d_x^{\left( 2 \right)}\) |
| 40 | 1000 | 60 | 55 |
| 41 | – | – | 70 |
| 42 | 750 | – | – |
Setiap decrement menyebar secara uniform, hitunglah nilai \(q_{41}^{‘\left( 1 \right)}\)
- 0,077
- 0,078
- 0,079
- 0,080
- 0,081
| Diketahui |
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| Rumus yang digunakan | \({}_tp_x^{‘\left( j \right)} = 1 – {}_tq_x^{‘\left( j \right)}\), \(d_x^{\left( \tau \right)} = l_x^{\left( \tau \right)} – l_{x + 1}^{\left( \tau \right)}\), \(d_x^{\left( \tau \right)} = d_x^{\left( 1 \right)} + d_x^{\left( 2 \right)}\), \(d_x^{\left( j \right)} = l_x^{\left( \tau \right)}q_x^{\left( j \right)}\), \(d_x^{\left( \tau \right)} = l_x^{\left( \tau \right)}q_x^{\left( \tau \right)}\), \(p_x^{‘\left( j \right)} = {\left( {p_x^{\left( \tau \right)}} \right)^{\frac{{q_x^{\left( j \right)}}}{{q_x^{\left( \tau \right)}}}}}\) | ||||||||||||||||
| Proses pengerjaan | \(l_{41}^{\left( \tau \right)} = l_{40}^{\left( \tau \right)} – d_{40}^{\left( 1 \right)} – d_{40}^{\left( 2 \right)}\) \(= 1000 – 60 – 55\) \(= 885\) | ||||||||||||||||
| \(l_{42}^{\left( \tau \right)} = l_{41}^{\left( \tau \right)} – d_{41}^{\left( 1 \right)} – d_{41}^{\left( 2 \right)}\) \(750 = 885 – d_{41}^{\left( 1 \right)} – 70\) \(d_{41}^{\left( 1 \right)} = 65\) | |||||||||||||||||
| \(q_{41}^{\left( 1 \right)} = \frac{{d_{41}^{\left( 1 \right)}}}{{l_{41}^{\left( \tau \right)}}} = \frac{{65}}{{885}} = 0,073446\) dan \(q_{41}^{\left( \tau \right)} = \frac{{d_{41}^{\left( \tau \right)}}}{{l_{41}^{\left( \tau \right)}}} = \frac{{65 + 70}}{{885}} = 0,152542\) \(q_{41}^{‘\left( 1 \right)} = 1 – p_{41}^{‘\left( 1 \right)}\) \(= 1 – {\left( {p_{41}^{\left( \tau \right)}} \right)^{\frac{{q_{41}^{\left( 1 \right)}}}{{q_{41}^{\left( \tau \right)}}}}}\) \(= 1 – {\left( {1 – 0,152542} \right)^{\frac{{0,073446}}{{0,152542}}}}\) \(= 0,076599\) | |||||||||||||||||
| Jawaban | a. 0,077 |