Pembahasan Soal Ujian Profesi Aktuaris
Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian |
: |
A20 – Probabilitas dan Statistika |
Periode Ujian |
: |
Juni 2016 |
Nomor Soal |
: |
20 |
SOAL
Seorang aktuaris menentukan besar klaim untuk suatu jenis kecelakaan diri ialah suatu peubah acak X dengan fungsi pembangkit peluang :
\({M_X}(t) = \frac{1}{{{{(1 – 2.500t)}^4}}}\)
Berapa simpangan baku untuk peubah acak X ?
- 1.000
- 5.000
- 2.000
- 8.660
- 11.000
PEMBAHASAN
Step 1 |
\({M_X}(t) = \frac{1}{{{{(1 – 2.500t)}^4}}}\)
\(X \sim Gamma\,(\alpha = 4\,,\,\theta = 2.500)\)
\(Var[X] = \,\alpha \,{\theta ^2}\,\, = \,\,4{(2.500)^2}\, = \,25.000.000\)
\({\sigma _X} = \sqrt {Var[X]} = 5.000\) |
Step 2 |
\(E[X] = {M_X}'(0)\)
\(E[X] = \frac{{\partial \left( {\frac{1}{{{{(1 – 2.500t)}^4}}}} \right)}}{{\partial t}}|t = 0\)
\(E[X] = ( – 4)( – 2.500){(1 – 2.500(0))^{ – 5}}\)
\(E[X] = 10.000\)\(E[{X^2}] = {M_X}”(0)\)
\(E[{X^2}] = \frac{{{\partial ^2}\left( {\frac{1}{{{{(1 – 2.500t)}^4}}}} \right)}}{{\partial {t^2}}}|t = 0\)
\(E[{X^2}] = \frac{{\partial \left( {10.000{{(1 – 2.500t)}^{ – 5}}} \right)}}{{\partial t}}\left| {_{t = 0}} \right.\)
\(E[{X^2}] = (10.000)( – 5)( – 2.500){(1 – 2.500(0))^{ – 6}}\)
\(E[{X^2}] = 125.000.000\) |
Sterp 3 |
\(Var[X] = 125.000.000 – {(10.000)^2}\)
\(Var[X] = 25.000.000\)
\({\sigma _X} = 5.000\) |
Jawaban |
b. 5.000 |