Pembahasan Soal Ujian Profesi Aktuaris
Institusi |
: |
Persatuan Aktuaris Indonesia (PAI) |
Mata Ujian |
: |
A60 – Matematika Aktuaria |
Periode Ujian |
: |
November 2017 |
Nomor Soal |
: |
30 |
SOAL
Suatu “age-at-failure” variabel acak mempunyai distribusi sebagai berikut:
\({F_X}(x) = 1 – 0,1{(100 – x)^{1/2}},\,0 \le x \le 100\)
Tentukan nilai dari E[X] dan median dari distribusi tersebut
- 100/3 ; 75
- 100/3 ; 100
- 200/3 ; 100
- 200/3 ; 75
- 200/3 ; 50
PEMBAHASAN
Rumus |
\({F_X}(x) = 1 – 0,1{(100 – x)^{1/2}},\,0 \le x \le 100\)
\({S_X}(x) = 1 – {F_X}(x)\)
\({S_X}(x) = 1 – \left( {1 – 0,1{{(100 – x)}^{1/2}}} \right)\)
\({S_X}(x) = 0,1{(100 – x)^{1/2}},0 \le x \le 100\) |
Step 1 |
\(E[X] = \int\limits_0^{100} {{S_X}(x)} \,\,dx\)
\(E[X] = \int\limits_0^{100} {0,1{{(100 – x)}^{1/2}}} \,dx\)
\(E[X] = \frac{{0,1}}{{ – 1,5}}\left( {{{(100 – 100)}^{1,5}} – {{(100 – 0)}^{1,5}}} \right)\)
\(E[X] = \frac{{0,1}}{{ – 1,5}}\left( {0 – {{(100)}^{1,5}}} \right)\)
\(E[X] = \frac{{0,1}}{{1,5}}\left( {{{(100)}^{1,5}}} \right)\)
\(E[X] = \frac{{200}}{3}\) |
Step 2 |
\({F_X}({x_{med}}) = {S_X}({x_{med}}) = 0,5\)
\({S_X}({x_{med}}) = 0,1{(100 – {x_{med}})^{1/2}}\)
\(0,5 = 0,1{(100 – {x_{med}})^{1/2}}\)
\(5 = {(100 – {x_{med}})^{1/2}}\)
\({x_{med}} = 100 – {5^2}\)
\({x_{med}} = 75\) |
Jawaban |
d. 200/3 ; 75 |