Pembahasan-Soal-Ujian-Profesi-Aktuaris-1024x481 (3) pai

Pembahasan Ujian PAI: A60 – No. 12 – November 2018

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 \(\mathop {q’}\nolimits_{41}^{(1)} \) !

  1. 0,077
  2. 0,079
  3. 0,081
  4. 0,083
  5. 0,085
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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
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