Pembahasan Ujian PAI: A50 – No. 28 – November 2018

\(\mu _{10 + t}^{\left( \tau \right)} = \frac{{50 – 2t}}{{600 – 50t + {t^2}}},0 \le t < 20\)

\({}_tp_x^{\left( \tau \right)} = \exp \left( { – \int\limits_0^t {\mu _x^{\left( \tau \right)}\left( s \right)ds} } \right)\)

\({f_T}\left( t \right) = {}_tp_x^{\left( \tau \right)}\mu _x^{\left( \tau \right)}\left( t \right)\)

\({f_{T,J}}\left( {t,j} \right) = {}_tp_x^{\left( \tau \right)}\mu _x^{\left( j \right)}\left( t \right)\)

\(\frac{{50 – 2t}}{{600 – 50t + {t^2}}} = \frac{1}{{30 – t}} + \mu _{10 + t}^{\left( 2 \right)}\)

\(\mu _{10 + t}^{\left( 2 \right)} = \frac{{50 – 2t}}{{600 – 50t + {t^2}}} – \frac{1}{{30 – t}}\)

\(= \frac{{50 – 2t – 20 + t}}{{\left( {20 – t} \right)\left( {30 – t} \right)}}\)

\(= \frac{{30 – t}}{{\left( {20 – t} \right)\left( {30 – t} \right)}}\)

\(= \frac{1}{{20 – t}},{\rm{ }}0 \le t < 20\)

\(_tp_{10}^{\left( \tau \right)} = \exp \left( { – \int\limits_0^t {\frac{{50 – 2s}}{{600 – 50s + {s^2}}}ds} } \right),{\rm{ Misal\_}}u = 600 – 50s + {s^2} \to du = – \left( {50 – 2s} \right)ds\)

\(= \exp \left( {\int\limits_{600}^{600 – 50t + {t^2}} {\frac{1}{u}du} } \right)\)

\(= \exp \left[ {\ln \left( {600 – 50t + {t^2}} \right) – \ln \left( {600} \right)} \right]\)

\(= \frac{{600 – 50t + {t^2}}}{{600}}\)

\(f\left( {t,j = 2} \right) = \frac{{600 – 50t + {t^2}}}{{600\left( {20 – t} \right)}} = \frac{{\left( {20 – t} \right)\left( {30 – t} \right)}}{{600\left( {20 – t} \right)}}\)
\(= \frac{{30 – t}}{{600}}\)
\(f\left( {t = 6,j = 2} \right) = \frac{{30 – 6}}{{600}}\)
\(= 0,04\)