Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Jika variabel acak “age-at-failure” berdistribusi “exponential” dengan “mean” \(\frac{1}{\lambda }\) , manakah dari pernyataan berikut yang benar untuk \(\bar P\left( {{{\bar A}_x}} \right)\)?
- \(\bar P\left( {{{\bar A}_x}} \right) = \lambda \)
- \(\bar P\left( {{{\bar A}_x}} \right) = \frac{1}{\lambda }\)
- \(\bar P\left( {{{\bar A}_x}} \right) = \frac{\lambda }{{\lambda + \delta }}\)
- \(\bar P\left( {{{\bar A}_x}} \right) = \frac{\delta }{{\lambda + \delta }}\)
- \(\bar P\left( {{{\bar A}_x}} \right) = \frac{1}{{\lambda + \delta }}\)
| Diketahui | variabel acak “age-at-failure” berdistribusi “exponential” dengan “mean” \(\frac{1}{\lambda }\) |
| Rumus yang digunakan | \({\bar a_x} = \frac{1}{{\mu + \delta }}\)
\(\bar P\left( {{A_x}} \right) = \frac{{{{\bar A}_x}}}{{{{\bar a}_x}}}\)
Untuk \(\mu \) dan \(\delta \) konstan \({\bar A_x} = \frac{\mu }{{\mu + \delta }}\) |
| Proses pengerjaan | \(\bar P\left( {{A_x}} \right) = \frac{{{{\bar A}_x}}}{{{{\bar a}_x}}} = \frac{{\frac{\lambda }{{\lambda + \delta }}}}{{\frac{1}{{\lambda + \delta }}}} = \lambda \) |
| Jawaban | A. \(\bar P\left( {{{\bar A}_x}} \right) = \lambda \) |