Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | April 2019 |
| Nomor Soal | : | 29 |
SOAL
Diberikan informasi \({\mu _x} = 0,002x + 0,005\)
Hitunglah \({}_{\left. 5 \right|}{q_{20}}\) (gunakan pembulatan terdekat)
- 0,042
- 0,050
- 0,064
- 0,075
- 0,082
| Diketahui | \({\mu _x} = 0,002x + 0,005\) |
| Rumus yang digunakan | \({}_{\left. t \right|u}{q_x} = {}_t{p_x} – {}_{t + u}{p_x} = \frac{{S\left( {x + t} \right)}}{{S\left( x \right)}} – \frac{{S\left( {x + t + u} \right)}}{{S\left( x \right)}}\) \({}_t{p_0} = S\left( t \right) = \exp \left[ { – \int\limits_0^t {{\mu _t}dt} } \right]\) |
| Proses pengerjaan | \({}_{\left. 5 \right|}{q_{20}} = \frac{{S\left( {25} \right) – S\left( {26} \right)}}{{S\left( {20} \right)}}\) dengan \(\int\limits_0^t {{\mu _x}dx} = \int\limits_0^t {0.002x + 0.005dx} = 0.001{x^2} + 0.005x\) |
| \(S\left( {20} \right) = \exp \left[ { – \left( {0.001\left( {{{20}^2}} \right) + 0.005\left( {20} \right)} \right)} \right] = \exp \left[ { – 0.5} \right] = 0.606531\) \(S\left( {25} \right) = \exp \left[ { – \left( {0.001\left( {{{25}^2}} \right) + 0.005\left( {25} \right)} \right)} \right] = \exp \left[ { – 0.75} \right] = 0.472367\) \(S\left( {26} \right) = \exp \left[ { – \left( {0.001\left( {{{26}^2}} \right) + 0.005\left( {26} \right)} \right)} \right] = \exp \left[ { – 0.806} \right] = 0.446641\) | |
| \({}_{\left. 5 \right|}{q_{20}} = \frac{{S\left( {25} \right) – S\left( {26} \right)}}{{S\left( {20} \right)}} = \frac{{0.472367 – 0.446641}}{{0.606531}} = 0.042415\) | |
| Jawaban | a. 0,042 |