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 |
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| 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 |
