Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Diketahui variabel acak X berdistribusi Poisson dengan mean \(\lambda \) . Jika Pr( X = 1 | X \(\le \) 1) = 80%, maka nilai dari \(\lambda \) sama dengan …
- 4
- – ln (2)
- 0,80
- 0,25
- – ln (0,8)
Diketahui | Variabel acak X berdistribusi Poisson dengan mean \(\lambda \) Jika Pr( X = 1 | X \(\le \) 1) = 80% |
Rumus yang digunakan | \(\Pr \left( {X = 1\left| {X \le 1)} \right.} \right) = \frac{{\Pr (X = 1)}}{{\Pr (X \le 1)}}\) |
Proses pengerjaan | \({f_X}(x) = {e^{ – \lambda }}\frac{{{\lambda ^x}}}{{x!}},x = 0,1,2,…\) \(Pr(X = 1|X \le 1){\rm{ }} = {\rm{ }}80\% = 0,8\) \(\Pr \left( {X = 1\left| {X \le 1)} \right.} \right) = \frac{{\Pr (X = 1)}}{{\Pr (X \le 1)}} = 0,8\) \(\Leftrightarrow \frac{{{e^{ – \lambda }}\frac{{{\lambda ^1}}}{{1!}}}}{{{e^{ – \lambda }}\frac{{{\lambda ^0}}}{{0!}} + {e^{ – \lambda }}\frac{{{\lambda ^1}}}{{1!}}}} = 0,8\) \(\Leftrightarrow \frac{\lambda }{{1 + \lambda }} = 0,8\) \(\Leftrightarrow \lambda = 4\) |
Jawaban | a. 4 |