Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Pemodelan dan Teori Risiko |
| Periode Ujian |
: |
Mei 2017 |
| Nomor Soal |
: |
11 |
SOAL
Sebuah perusahaan asuransi mempunyai 2(dua) jenis klaim asuransi. Untuk setiap jenis klaim tersebut, banyaknya klaim mengikuti distribusi Poisson dan besarnya klaim berdistribusi secara seragam (uniform) sebagai berikut:
| Jenis Klaim |
Parameter Poisson \(\lambda \) untuk jumlah klaim |
Range dari masing-masing besaran klaim |
| I |
12 |
(0, 1) |
| II |
4 |
(0, 5) |
Banyaknya klaim dari dua jenis pertanggungan saling bebas (independent) serta besarnya klaim dan banyaknya klaim saling bebas.
Hitunglah probabilitas nilai total klaim melebihi 18 dengan menggunakan aproksimasi normal.
- 0,37
- 0,39
- 0,41
- 0,43
- 0,45
| Diketahui |
Sebuah perusahaan asuransi mempunyai 2(dua) jenis klaim asuransi. Untuk setiap jenis klaim tersebut, banyaknya klaim mengikuti distribusi Poisson dan besarnya klaim berdistribusi secara seragam (uniform) sebagai berikut:
| Jenis Klaim |
Parameter Poisson \(\lambda \) untuk jumlah klaim |
Range dari masing-masing besaran klaim |
| I |
12 |
(0, 1) |
| II |
4 |
(0, 5) |
Banyaknya klaim dari dua jenis pertanggungan saling bebas (independent) serta besarnya klaim dan banyaknya klaim saling bebas. |
| Rumus yang digunakan |
Uniform: \(E\left[ X \right] = \frac{{b – a}}{2}\) ; \(Var\left( X \right) = \frac{{{{\left( {b – a} \right)}^2}}}{{12}}\)
Agregat: \(E\left[ S \right] = E\left[ N \right]E\left[ X \right]\) dan \(Var\left[ S \right] = E\left[ N \right]Var\left[ X \right] + Var\left[ N \right]E{\left[ X \right]^2}\)
\(P\left( {S \le s} \right) = \Phi \left( {\frac{{S – E\left[ S \right]}}{{\sqrt {Var\left( S \right)} }}} \right)\) |
| Proses pengerjaan |
Untuk masing-masing jenis klaim (berdistribusi uniform)
- \(E\left[ {{X_I}} \right] = \frac{1}{2}\) dan \(Var\left( {{X_I}} \right) = \frac{1}{{12}}\)
- \(E\left[ {{X_{II}}} \right] = \frac{5}{2}\) dan \(Var\left( {{X_{II}}} \right) = \frac{{25}}{{12}}\)
|
|
Aggregat masing-masing jenis klaim
- \(E\left[ {{S_I}} \right] = 12\left( {\frac{1}{2}} \right) = 6\) dan \(Var\left( {{S_I}} \right) = 12\left( {\frac{1}{{12}}} \right) + 12{\left( {\frac{1}{2}} \right)^2} = 4\)
- \(E\left[ {{S_{II}}} \right] = 4\left( {\frac{5}{2}} \right) = 10\) dan \(Var\left( {{S_{II}}} \right) = 4\left( {\frac{{25}}{{12}}} \right) + 4{\left( {\frac{5}{2}} \right)^2} = \frac{{100}}{3}\)
|
|
DIketahui total klaim \(S = {S_I} + {S_{II}}\) diperoleh
\(E\left[ S \right] = 6 + 10 = 16\) dan \(Var\left( S \right) = 4 + \frac{{100}}{3} = \frac{{112}}{3}\) |
|
Peluang menggunakan aproksimasi normal
\(P\left( {S > 18} \right) = 1 – \Phi \left( {\frac{{18 – 16}}{{\sqrt {\frac{{112}}{3}} }}} \right) = 1 – \Phi \left( {0.33} \right) = 1 – 0.06293 = 0.3707\) |
| Jawaban |
a. 0,37 |
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