Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
April 2019 |
| Nomor Soal |
: |
1 |
SOAL
Diberikan informasi ssebagai berikut :
- \({S_0}\left( t \right) = {\left( {1 – \frac{t}{\omega }} \right)^{\frac{1}{4}}}\), untuk \(0 \le t \le \omega \)
- \({\mu _{65}} = \frac{1}{{180}}\)
Tentukan nilai \({e_{106}}\), the curtate expectation of life untuk usia 106
(Gunakan pembulatan terdekat).
- 2,2
- 2,5
- 2,7
- 3,0
- 3,2
| Diketahui |
- \({S_0}\left( t \right) = {\left( {1 – \frac{t}{\omega }} \right)^{\frac{1}{4}}}\), untuk \(0 \le t \le \omega \)
- \({\mu _{65}} = \frac{1}{{180}}\)
|
| Rumus yang digunakan |
Hukum De Moivre
\({\begin{array}{*{20}{c}} {{S_0}\left( t \right) = {{\left( {1 – \frac{t}{\omega }} \right)}^\alpha },}&{\mu \left( x \right) = \frac{\alpha }{{\omega – x}},}&{{}_t{p_x} = \left( {\frac{{\omega – x – t}}{{\omega – x}}} \right)} \end{array}^\alpha }\) |
| Proses pengerjaan |
\(\mu \left( x \right) = \frac{\alpha }{{\omega – x}}\)
\(\frac{1}{{180}} = \frac{{0.25}}{{\omega – 65}}\)
\(\omega – 65 = 45\)
\(\omega = 110\) |
| \({}_t{p_x} = {\left( {\frac{{\omega – x – t}}{{\omega – x}}} \right)^\alpha }\)
\(= {\left( {\frac{{110 – x – t}}{{110 – x}}} \right)^{\frac{1}{4}}}\) |
| \({e_{106}} = \sum\limits_{k = 1}^4 {_k{p_{106}}} \)
\(= \sum\limits_{k = 1}^4 {{{\left( {\frac{{4 – k}}{4}} \right)}^{\frac{1}{4}}}} \)
\(= 2.4786\) |
| Jawaban |
b. 2,5 |
