Pembahasan Ujian PAI: A50 – No. 14 – November 2014

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

1. 2.061,81
2. 2.081,61
3. 2.161,81
4. 2.181,16
5. 2.186,11

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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

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