Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Matematika Aktuaria |
| Periode Ujian |
: |
November 2014 |
| Nomor Soal |
: |
30 |
SOAL
Asuransi diskrit berjangka 2 tahun dijual untuk usia \(\left( x \right)\) dengan tingkat bunga \(i = 0\). Jika diketahui \({q_x} = 0,50\) dan \(Var\left( {Z_{x:\left. {\overline {\, 2 \,}}\! \right| }^1} \right) = 0,1771\). Hitunglah \({q_{x + 1}}\)
- 0,52
- 0,56
- 0,42
- 0,45
- 0,54
| Diketahui |
- Asuransi diskrit berjangka 2 tahun dijual untuk usia \(\left( x \right)\) dengan tingkat bunga \(i = 0\).
- Diketahui \({q_x} = 0,50\) dan \(Var\left( {Z_{x:\left. {\overline {\, 2 \,}}\! \right| }^1} \right) = 0,1771\).
|
| Rumus yang digunakan |
\(A_{x:\left. {\overline {\, n \,}}\! \right| }^1 = \sum\limits_{k = 0}^{n – 1} {{v^{k + 1}} \cdot {}_k{p_x} \cdot {q_{x + k}}} \)
\({}^2A_{x:\left. {\overline {\, n \,}}\! \right| }^1 = \sum\limits_{k = 0}^{n – 1} {{v^{2\left( {k + 1} \right)}} \cdot {}_k{p_x} \cdot {q_{x + k}}} \)
\(Var\left( {A_{x:\left. {\overline {\, n \,}}\! \right| }^1} \right) = {}^2A_{x:\left. {\overline {\, n \,}}\! \right| }^1 – {\left( {A_{x:\left. {\overline {\, n \,}}\! \right| }^1} \right)^2}\) |
| Proses pengerjaan |
Karena \(i = 0\) maka \(v = 1\)
\(A_{x:\left. {\overline {\, 2 \,}}\! \right| }^1 = \sum\limits_{k = 0}^1 {{v^{k + 1}} \cdot {}_k{p_x} \cdot {q_{x + k}}} \)
\(A_{x:\left. {\overline {\, 2 \,}}\! \right| }^1 = {v^1} \cdot {}_0{p_x} \cdot {q_x} + {v^2} \cdot {p_x} \cdot {q_{x + 1}}\)
\(A_{x:\left. {\overline {\, 2 \,}}\! \right| }^1 = {q_x} + {p_x} \cdot {q_{x + 1}}\) |
|
\({}^2A_{x:\left. {\overline {\, n \,}}\! \right| }^1 = \sum\limits_{k = 0}^{n – 1} {{v^{2\left( {k + 1} \right)}} \cdot {}_k{p_x} \cdot {q_{x + k}}} \)
\({}^2A_{x:\left. {\overline {\, n \,}}\! \right| }^1 = {v^2} \cdot {}_0{p_x} \cdot {q_x} + {v^4} \cdot {p_x} \cdot {q_{x + 1}}\)
\({}^2A_{x:\left. {\overline {\, n \,}}\! \right| }^1 = {q_x} + {p_x} \cdot {q_{x + 1}}\) |
|
\(Var\left( {A_{x:\left. {\overline {\, n \,}}\! \right| }^1} \right) = {}^2A_{x:\left. {\overline {\, n \,}}\! \right| }^1 – {\left( {A_{x:\left. {\overline {\, n \,}}\! \right| }^1} \right)^2}\)
\(Var\left( {A_{x:\left. {\overline {\, n \,}}\! \right| }^1} \right) = {q_x} + {p_x} \cdot {q_{x + 1}} – {\left( {{q_x} + {p_x} \cdot {q_{x + 1}}} \right)^2}\)
\(0.1771 = 0.5 + 0.5{q_{x + 1}} – {\left( {0.5 + 0.5{q_{x + 1}}} \right)^2}\)
\(0.1771 = 0.5 + 0.5{q_{x + 1}} – 0.25 – 0.5{q_{x + 1}} – 0.25{\left( {{q_{x + 1}}} \right)^2}\)
\({q_{x + 1}} = \sqrt {\frac{{0.5 – 0.25 – 0.1771}}{{0.25}}} \)
\({q_{x + 1}} = 0.54\) |
| Jawaban |
e. 0,54 |
