Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Metoda Statistika |
| Periode Ujian |
: |
Mei 2018 |
| Nomor Soal |
: |
17 |
SOAL
Untuk sebuah tabel mortalita dengan dua tahun seleksi, diberikan:
- \({q_{[x]}} = (1 – 2k){q_x},\) untuk semua x
- \({q_{[x] + 1}} = (1 – k){q_{x + 1}},\) untuk semua x
- \({l_{[32]}} = 90\)
- \({l_{32}} = 100\)
- \({l_{33}} = 90\)
- \({l_{34}} = 63\)
Hitunglah \({q_{[32]}}\)
- 1/12
- 2/25
- 1/15
- 2/31
- 2/35
| Diketahui |
- \({q_{[x]}} = (1 – 2k){q_x},\) untuk semua x
- \({q_{[x] + 1}} = (1 – k){q_{x + 1}},\) untuk semua x
- \({l_{[32]}} = 90\)
- \({l_{32}} = 100\)
- \({l_{33}} = 90\)
- \({l_{34}} = 63\)
|
| Rumus yang digunakan |
\({q_x} = \frac{{{l_x} – {l_{x + 1}}}}{{{l_x}}}\)
\({q_{[x]}} = \frac{{{l_{[x]}} – {l_{[x] + 1}}}}{{{l_{[x]}}}}\)
\({q_{[x] + 1}} = \frac{{{l_{[x] + 1}} – {l_{x + 2}}}}{{{l_{[x] + 1}}}}\) |
| Proses pengerjaan |
\({q_{32}} = \frac{{{l_{32}} – {l_{33}}}}{{{l_{32}}}} = \frac{{100 – 90}}{{100}} = \frac{1}{{10}}\)
\({q_{33}} = \frac{{{l_{33}} – {l_{34}}}}{{{l_{33}}}} = \frac{{90 – 63}}{{90}} = \frac{{27}}{{90}} = \frac{3}{{10}}\)
\({q_{32}} = \frac{{{q_{[31] + 1}}}}{{1 – k}} = \frac{1}{{10}}\)
\({q_{32}} = \frac{{{q_{[32]}}}}{{1 – 2k}} = \frac{1}{{10}}\)
\({q_{33}} = \frac{{{q_{[32] + 1}}}}{{1 – k}} = \frac{3}{{10}}\)
\({q_{33}} = \frac{{{q_{[33]}}}}{{1 – 2k}} = \frac{3}{{10}}\)
\({q_{[32] + 1}} = \frac{{{l_{[32] + 1}} – {l_{34}}}}{{{l_{[32] + 1}}}} = \frac{{{l_{[32] + 1}} – 63}}{{{l_{[32] + 1}}}} = \frac{3}{{10}}(1 – k){\rm{ (*)}}\)
\({q_{[32]}} = \frac{{{l_{[32]}} – {l_{[32] + 1}}}}{{{l_{[32]}}}} = \frac{{90 – {l_{[32] + 1}}}}{{90}} = (1 – 2k)\frac{1}{{10}}{\rm{ (**)}}\)
berdasarkan (**), misalkan \(x = {l_{[32] + 1}},\) maka
\(90 – x = 9(1 – 2k)\)
\(\Leftrightarrow x = 90 – 9 + 18k\)
\(\Leftrightarrow x = 81 + 18k\)
substitusikan nilai x kedalam (*)
\(\frac{{81 + 18k – 63}}{{81 + 18k}} = \frac{{3 – 3k}}{{10}}\)
\(\Leftrightarrow \frac{{18k + 18}}{{81 + 18k}} = \frac{{3 – 3k}}{{10}}\)
\(\Leftrightarrow 180k + 180 = 243 – 243k + 54k – 54{k^2}\)
\(\Leftrightarrow 180k + 180 = 243 – 189k – 54{k^2}\)
\(\Leftrightarrow 54{k^2} + 369k – 63 = 0\)
\(\Leftrightarrow 6{k^2} + 41k – 7 = 0\)
\(\Leftrightarrow (6k – 1)(k + 7) = 0\)
artinya \(k = – 7\) dan \(k = \frac{1}{6}\)
untuk \(k = – 7{\rm{ }}\) diperoleh \({q_{[32]}} = (1 – 2( – 7))\frac{1}{{10}} = \frac{{15}}{{10}}\) (tidak memenuhi)
untuk \(k = \frac{1}{6}\) diperoleh \({q_{[32]}} = (1 – 2(\frac{1}{6}))\frac{1}{{10}} = \frac{1}{{15}}\) |
| Jawaban |
c. 1/15 |
