Pembahasan Soal Ujian Profesi Aktuaris
SOAL
Tentukan nilai \(Var\left( {{Y_{95}}} \right)\), bila menggunakan tingkat bunga tahunan 5% dan nilai sebagai brikut: \({l_{95}} = 100\), \({l_{96}} = 70\), \({l_{97}} = 40\), \({l_{98}} = 20\), \({l_{99}} = 4\), \({l_{100}} = 0\), \({a_{95}} = 1,2352\), dan \({}^2{a_{95}} = 1,1403\)
- 1,0933
- 1,0399
- 2,0933
- 2,2352
- 2,2532
| Diketahui | Tingkat bunga tahunan 5% dan nilai sebagai brikut: \({l_{95}} = 100\), \({l_{96}} = 70\), \({l_{97}} = 40\), \({l_{98}} = 20\), \({l_{99}} = 4\), \({l_{100}} = 0\), \({a_{95}} = 1,2352\), dan \({}^2{a_{95}} = 1,1403\) |
| Rumus yang digunakan | \(Var\left( Y \right) = \frac{{{}^2{A_x} – {{\left( {{A_x}} \right)}^2}}}{{{d^2}}}\) dengan \(d = 1 – v = 1 – \frac{1}{{1 + i}}\) \({A_x} = 1 – d{\ddot a_x}\) dan \({\ddot a_x} = 1 + {a_x}\) |
| Proses pengerjaan | \(d = 1 – \frac{1}{{1.05}} = \frac{1}{{21}}\) dan \({d^*} = 1 – \frac{1}{{{{1.05}^2}}} = \frac{{41}}{{441}}\) \({\ddot a_{95}} = 1 + {a_{95}} = 1 + 1.2352 = 2.2352\) \({A_{95}} = 1 – d{\ddot a_{95}} = 1 – \frac{1}{{21}}\left( {2.2352} \right) = 0.893562\) |
| \({}^2{\ddot a_{95}} = 1 + {}^2{a_{95}} = 1 + 1.1403 = 2.1403\) \({}^2{A_{95}} = 1 – {d^*}{}^2{\ddot a_{95}} = 1 – \frac{{41}}{{441}}\left( {2.1403} \right) = 0.801015\) | |
| \(Var\left( Y \right) = \frac{{{}^2{A_x} – {{\left( {{A_x}} \right)}^2}}}{{{d^2}}} = \frac{{0.8010 – {{0.8936}^2}}}{{{{\left( {\frac{1}{{21}}} \right)}^2}}} = 1.093257\) | |
| Jawaban | a. 1,0933 |