Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Metoda Statistika |
| Periode Ujian | : | Mei 2018 |
| Nomor Soal | : | 4 |
SOAL
Diberikan
- \(\mu _{x + t}^{} = \left\{ \begin{array}{l} 0.03,0 \le t < 1\\ 0.08,1 \le t < 2 \end{array} \right.\)
- \(Y = \min ({T_x},2)\)
Hitunglah nilai \(E(Y)\)
- 1,52
- 1,61
- 1,73
- 1,8
- 1,92
| Diketahui |
|
| Rumus yang digunakan | \(E[\min ({T_x},n)] = \int\limits_0^n t {{\rm{ }}_t}{P_x}{\rm{ }}{\mu _{x + t}}{\rm{ }}dt{\rm{ + }}n{{\rm{ }}_t}{P_x}\) \(_t{p_x} = {e^{ – \int\limits_0^2 {{\mu _t}dt} }}\) |
| Proses Pengerjaan | Dengan mengaplikasikan rumus di atas diperoleh \(E(Y) = E[\min ({T_x},2)]\) \(= \int\limits_0^2 t {{\rm{ }}_t}{P_x}{\rm{ }}{\mu _{x + t}}{\rm{ }}dt{\rm{ + 2}}{{\rm{ }}_t}{P_x}\) \(= \int\limits_0^1 t {{\rm{ }}_t}{P_x}{\rm{ }}{\mu _{x + t}}{\rm{ }}dt + \int\limits_0^2 t {{\rm{ }}_t}{P_x}{\rm{ }}{\mu _{x + t}}{\rm{ }}dt{\rm{ + 2}}{{\rm{ }}_t}{P_x}\) \(= \int\limits_0^1 t {\rm{ }}{e^{ – 0.03t}}{\rm{ }}0.03{\rm{ }}dt + \int\limits_1^2 t {\rm{ }}{e^{ – 0.08t}}{\rm{ }}0.08{\rm{ }}dt + 2{e^{ – \int\limits_0^2 {{\mu _t}dt} }}\) \(= \int\limits_0^1 t {\rm{ }}{e^{ – 0.03t}}{\rm{ }}0.03{\rm{ }}dt + \int\limits_1^2 t {\rm{ }}{e^{ – 0.08t}}{\rm{ }}0.08{\rm{ }}dt + 2{e^{ – (\int\limits_0^1 {{\mu _t}dt} + \int\limits_1^2 {{\mu _t}dt} )}}\) \(= 0.03\int\limits_0^1 t {\rm{ }}{e^{ – 0.03t}}{\rm{ }}dt + \int\limits_1^2 t {\rm{ }}{e^{ – 0.08t}}{\rm{ }}0.08{\rm{ }}dt + 2{e^{ – (\int\limits_0^1 {{\rm{0}}{\rm{.03t }}dt} + \int\limits_1^2 {0.08dt} )}}\) \(= 0.03(0.490112) + 0.08(1.32482) + 2{e^{ – (0.03 + 0.08)}}\) \(= 1.912\) \(= 1.92\) |
| Jawaban | b. 1,92 |