Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Metoda Statistika |
| Periode Ujian |
: |
November 2014 |
| Nomor Soal |
: |
14 |
SOAL
Jika force of mortality didefinisikan sebagai:
\({\mu _x} = \frac{2}{{x + 1}} + \frac{2}{{100 – x}},0 \le x < 100\)
Tentukan jumlah kematian yang terjadi di antara usia 1 dan 4 dalam life table dengan radix 10.000
- 2.061,81
- 2.081,61
- 2.161,81
- 2.181,16
- 2.186,11
| Diketahui |
\({\mu _x} = \frac{2}{{x + 1}} + \frac{2}{{100 – x}},0 \le x < 100\) |
| Rumus yang digunakan |
- \({}_t{p_0} = S\left( t \right) = \exp \left[ { – \int\limits_0^t {\mu \left( s \right)ds} } \right]\)
- \({l_x} = {l_0} \cdot S\left( x \right)\)
- \({}_n{d_x} = {l_x} – {l_{x + n}}\)
|
| Proses pengerjaan |
\(S\left( t \right) = \exp \left[ { – \int\limits_0^t {\left( {\frac{2}{{s + 1}} + \frac{2}{{100 – s}}} \right)ds} } \right]\)
\(S\left( t \right) = \exp \left[ { – 2\int\limits_0^t {\left( {\frac{1}{{s + 1}}} \right)ds} – 2\int\limits_0^t {\left( {\frac{1}{{100 – s}}} \right)ds} } \right]\)
\(S\left( t \right) = \exp \left[ { – 2\ln \left( {t + 1} \right) – 2\int\limits_0^t {\left( {\frac{1}{{100 – s}}} \right)ds} } \right]\)
\({S\left( t \right) = \exp \left[ { – 2\ln \left( {t + 1} \right) + 2\int\limits_{100}^{100 – t} {\left( {\frac{1}{u}} \right)du} } \right]}\) \({{\rm{misal\_}}u = 100 – s \Rightarrow du = – ds}\)
\(S\left( t \right) = \exp \left[ { – 2\ln \left( {t + 1} \right) + 2\ln \left( {100 – t} \right) – 2\ln (100)} \right]\)
\(S\left( t \right) = \frac{{{{\left( {100 – t} \right)}^2}}}{{10,000{{\left( {t + 1} \right)}^2}}}\)
- \({l_1} = {l_0} \cdot S\left( 1 \right) = 10,000\frac{{{{\left( {100 – 1} \right)}^2}}}{{10,000{{\left( {1 + 1} \right)}^2}}} = 2,450.25\)
- \({l_4} = {l_0} \cdot S\left( 4 \right) = 10,000\frac{{{{\left( {100 – 4} \right)}^2}}}{{10,000{{\left( {4 + 1} \right)}^2}}} = 368.64\)
- \({}_3{d_1} = {l_1} – {l_4} = 2,450.25 – 368.64 = 2,081.61\)
|
| Jawaban |
B. 2.081,61 |
