Pembahasan Soal Ujian Profesi Aktuaris
| Institusi | : | Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian | : | A20 – Probabilitas dan Statistika |
| Periode Ujian | : | Juni 2016 |
| Nomor Soal | : | 25 |
SOAL
Suatu peubah acak besar klaim kendaraan bermotor mempunyai informasi sebagai berikut :
| Besar Klaim | 20 | 30 | 40 | 50 | 60 | 70 | 80 |
| Peluang | 0,15 | 0,10 | 0,05 | 0,20 | 0,10 | 0,10 | 0,30 |
Berapa persentase bahwa klaim yang terjadi akan berada dalam 1 simpangan baku dari rataan besar klaim?
- 45%
- 55%
- 68%
- 85%
- 92%
| Misalkan | X ialah besar klaim kendaraan bermotor |
| Step 1 | \(E[X] = \sum {x\,P(X = x)} \)
\(E[X] = 20(0,15) + 30(0,10) + 40(0,05) + 50(0,20) + 60(0,10) + 70(0,10) + 80(0,30)\)
\(E[X] = 55\)\(E[{X^2}] = \sum {{x^2}\,P(X = x)} \)
\(\begin{array}{l} E[{X^2}] = {20^2}(0,15) + {30^2}(0,10) + {40^2}(0,05) + {50^2}(0,20) +{60^2}(0,10) + {70^2}(0,10)\\ & + {80^2}(0,30) \end{array}\)
\(E[{X^2}] = 3.500\)\(Var[X] = 3.500 – {55^2}\)
\(Var[X] = 475\)
\({\sigma _X} \cong 21,7945\) |
| Step 2 | \(P({\mu _X} – {\sigma _X} < x < {\mu _X} + {\sigma _X}) = P(55 – 21,7945 < x < 55 + 21,7945)\)
\(P({\mu _X} – {\sigma _X} < x < {\mu _X} + {\sigma _X}) = P(33,2055 < x < 76,7945)\)
\(P({\mu _X} – {\sigma _X} < x < {\mu _X} + {\sigma _X}) = P(X = 40) + P(X = 50) + P(X = 60) + P(X = 70)\)
\(P({\mu _X} – {\sigma _X} < x < {\mu _X} + {\sigma _X}) = 0,05 + 0,20 + 0,10 + 0,10\)
\(P({\mu _X} – {\sigma _X} < x < {\mu _X} + {\sigma _X}) = 0,45\)
\(P({\mu _X} – {\sigma _X} < x < {\mu _X} + {\sigma _X}) = 45\% \) |
| Jawaban | a. 45% |
