Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | Matematika Aktuaria |
| Periode Ujian | : | November 2018 |
| Nomor Soal | : | 1 |
SOAL
Diketahui
| \(k\) | \({\ddot a_{\left. {\overline {\, k \,}}\! \right| }}\) | \({}_{\left. {k – 1} \right|}{q_x}\) |
| 1 | 1,00 | 0,33 |
| 2 | 1,93 | 0,24 |
| 3 | 2,80 | 0,16 |
| 4 | 3,62 | 0,11 |
Tentukan nilai \({\ddot a_{x:\left. {\overline {\, 4 \,}}\! \right| }}\) !
- 2,22
- 4,44
- 6,67
- 8,89
- 11,11
| Diketahui |
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| Rumus yang digunakan |
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| Proses pengerjaan | \({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \ddot a_{x:\left. {\overline {\, 4 \,}}\! \right| }^1 + {{\ddot a}_{x:\mathop {\left. {\overline {\, 4 \,}}\! \right| }\limits^1 }}\) \({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \sum\limits_{k = 0}^3 {{{\ddot a}_{\left. {\overline {\, {k + 1} \,}}\! \right| }} \cdot {}_{\left. k \right|}{q_x}} + {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot {}_4{p_x} = \sum\limits_{k = 0}^3 {{{\ddot a}_{\left. {\overline {\, {k + 1} \,}}\! \right| }} \cdot {}_{\left. k \right|}{q_x}} + {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot \left( {1 – \sum\limits_{k = 0}^{n – 1} {{}_{\left. k \right|}{q_x}} } \right)\) \({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \left[ {{{\ddot a}_{\left. {\overline {\, 1 \,}}\! \right| }} \cdot {}_{\left. 0 \right|}{q_x} + {{\ddot a}_{\left. {\overline {\, 2 \,}}\! \right| }} \cdot {}_{\left. 1 \right|}{q_x} + {{\ddot a}_{\left. {\overline {\, 3 \,}}\! \right| }} \cdot {}_{\left. 2 \right|}{q_x} + {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot {}_{\left. 3 \right|}{q_x}} \right]\) \(+ {{\ddot a}_{\left. {\overline {\, 4 \,}}\! \right| }} \cdot \left( {1 – \left[{{}_{\left. 0 \right|}{q_x} + {}_{\left. 1 \right|}{q_x} + {}_{\left. 2 \right|}{q_x} + {}_{\left. 3\right|}{q_x}} \right]} \right)\) \({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = \left[ {1\left( {0.33} \right) + 1.93\left( {0.24} \right) + 2.8\left( {0.16} \right) + 3.62\left( {0.11} \right)}\right]\) \(+ 3.62\left( {1 – \left[ {0.33 + 0.24 + 0.16 + 0.11} \right]} \right)\) \({{\ddot a}_{x:\left. {\overline {\, 4 \,}}\! \right| }} = 2.2186\) | |||||||||||||||
| Jawaban | A. 2,22 |