Pembahasan Soal Ujian Profesi Aktuaris
| Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
| Mata Ujian |
: |
Probabilita dan Statistika |
| Periode Ujian |
: |
Juni 2014 |
| Nomor Soal |
: |
21 |
SOAL
Jika X berdistribusi uniform kontinu pada interval [0,10], maka nilai probabilitas \(\Pr \left( {X + \frac{{10}}{X} > 7} \right)\) sama dengan …
- 3/10
- 31/70
- 1/2
- 39/70
- 7/10
| Diketahui |
X berdistribusi uniform kontinu pada interval [0,10] |
| Rumus yang digunakan |
\(\Pr (X > x) = \int {{f_X}(x)dx} \) |
| Proses pengerjaan |
\({f_X}(x) = \frac{1}{{10}},x \in \left[ {0,10} \right]\)
\(\Pr \left( {X + \frac{{10}}{X} > 7} \right) = \Pr \left( {\frac{{{X^2} – 7X + 10}}{X} > 0} \right)\)
\(\Pr \left( {X + \frac{{10}}{X} > 7} \right) = \Pr \left( {\frac{{(X – 2)(X – 5)}}{X} > 0} \right)\)
\(\Pr \left( {X + \frac{{10}}{X} > 7} \right) = \Pr \left( {0 < X < 2{\rm{ }}atau{\rm{ }}5 < X \le 10} \right)\)
\(\Pr \left( {X + \frac{{10}}{X} > 7} \right) = \Pr \left( {0 < X < 2} \right) + \Pr \left( {5 < X \le 10} \right)\)
\(\Pr \left( {X + \frac{{10}}{X} > 7} \right) = \int\limits_0^2 {\frac{1}{{10}}dx + } \int\limits_5^{10} {\frac{1}{{10}}dx = \frac{7}{{10}}} \) |
| Jawaban |
e. 7/10 |
