Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Pemodelan dan Teori Risiko |
| Periode Ujian |
: |
April 2019 |
| Nomor Soal |
: |
16 |
SOAL
Terdapat 30 klaim tercatat dalam suatu sampel acak. Terdapat 2 klaim dengan jumlah masing- masing 2.000, 6 klaim dengan jumlah masing-masing 4.000, dan 10 klaim dengan jumlah masing-masing 8.000. Tentukan koefisien skewness untuk distribusi empirik tersebut
- -0,559
- -0,449
- -0,515
- -0,959
- -0,459
| Diketahui |
Terdapat 30 klaim tercatat dalam suatu sampel acak.
- 2 klaim dengan jumlah masing- masing 2.000.
- 6 klaim dengan jumlah masing-masing 4.000
- 10 klaim dengan jumlah masing-masing 8.000
|
| Rumus yang digunakan |
Skewness: \({\gamma _1} = \frac{{{\mu _3}}}{{{\sigma ^3}}}\)
Distribusi Empirik: \(p\left( x \right) = \frac{1}{n}\)
\({\mu ‘_k} = E\left[ {{X^k}} \right] = \sum\nolimits_j {x_j^kp\left( {{x_j}} \right)} \) dengan \({{\mu ‘}_1} = \mu \)
\({\mu _k} = E\left[ {{{\left( {X – \mu } \right)}^k}} \right] = \sum\nolimits_j {\left( {{x_j} – \mu } \right)p\left( {{x_j}} \right)} \)
\({\sigma ^2} = {\mu _2} = \sum\nolimits_j {{{\left( {{x_j} – \mu } \right)}^2}p\left( {{x_j}} \right)} = \sum\nolimits_j {\left( {{x_j}^2 – 2{x_j}{\mu ^2} + {\mu ^2}} \right)p\left( {{x_j}} \right)} = {{\mu ‘}_2} – 2{\mu ^2} + {\mu ^2}\)
\({\mu _3} = \sum\nolimits_j {{{\left( {{x_j} – \mu } \right)}^3}p\left( {{x_j}} \right)} = \sum\nolimits_j {\left( {{x_j}^3 – 3{x_j}^2\mu + 3{x_j}{\mu ^2} – {\mu ^3}} \right)p\left( {{x_j}} \right)} \)
\({\mu _3} = {{\mu ‘}_3} – 3{{\mu ‘}_2}\mu + 3{\mu ^3} – {\mu ^3}\) |
| Proses pengerjaan |
| Besaran Klaim |
Peluang |
| 0 |
\(\frac{2}{{30}} = 0.4\) |
| 2000 |
\(\frac{2}{{30}} = 0.067\) |
| 4000 |
\(\frac{6}{{30}} = 0.2\) |
| 8000 |
\(\frac{{10}}{{30}} = 0.333\) |
|
|
\({{\mu ‘}_1} = \mu = 0.4\left( 0 \right) + 0.067\left( {2000} \right) + 0.2\left( {4000} \right) + 0.333\left( {8000} \right) = 3600\)
\({{\mu ‘}_2} = 0.4{\left( 0 \right)^2} + 0.067{\left( {2000} \right)^2} + 0.2{\left( {4000} \right)^2} + 0.333{\left( {8000} \right)^2} = 24.8 \times {10^6}\)
\({{\mu ‘}_3} = 0.4{\left( 0 \right)^3} + 0.067{\left( {2000} \right)^3} + 0.2{\left( {4000} \right)^3} + 0.333{\left( {8000} \right)^3} = 1.84 \times {10^{11}}\) |
|
\({\gamma _1} = \frac{{{\mu _3}}}{{{\sigma ^3}}} = \frac{{{\mu _3}}}{{{{\left( {{\sigma ^2}} \right)}^{\frac{3}{2}}}}} = \frac{{{{\mu ‘}_3} – 3{{\mu ‘}_2}\mu + 2{\mu ^3}}}{{{{\left( {{{\mu ‘}_2} – {\mu ^2}} \right)}^{\frac{3}{2}}}}}\)
\({\gamma _1} = \frac{{1.84 \times {{10}^{11}} – 3\left( {24.8 \times {{10}^6}} \right)\left( {3600} \right) + 2{{\left( {3600} \right)}^3}}}{{{{\left( {24.8 \times {{10}^6} – {{3600}^2}} \right)}^{\frac{3}{2}}}}} = 0.232495\) |
| Jawaban |
0,232495 (Tetapi dari PAI jawabannya A?) |
