Diketahui	\({}_t{p_{50}} = {e^{0.5\left( {1 – {{1.05}^t}} \right)}}\)
Rumus yang digunakan	\({\mu _x} = \frac{{f\left( x \right)}}{{S\left( x \right)}}\) \(f\left( x \right) = F\left( x \right) – F\left( {x – 1} \right)\) \({}_t{p_x} = \frac{{S\left( {x + t} \right)}}{{S\left( x \right)}}\)
Proses Pengerjaan	\({\mu _{80}} = \frac{{f\left( {80} \right)}}{{S\left( {80} \right)}}\) \(= \frac{{F\left( {80} \right) – F\left( {79} \right)}}{{S\left( {80} \right)}}\) \(= \frac{{F\left( {80} \right) – F\left( {79} \right)}}{{S\left( {79} \right)}} \cdot \frac{{S\left( {79} \right)}}{{S\left( {80} \right)}}\) \(= \frac{{S\left( {79} \right) – S\left( {80} \right)}}{{S\left( {79} \right)}} \cdot \frac{{S\left( {79} \right)}}{{S\left( {80} \right)}}\) \(= \left( {1 – \frac{{S\left( {80} \right)}}{{S\left( {79} \right)}}} \right) \cdot \left( {\frac{{S\left( {79} \right)}}{{S\left( {80} \right)}}} \right)\) \(= \frac{{S\left( {79} \right)}}{{S\left( {80} \right)}} – 1\) \(= \left( {\frac{{S\left( {50 + 29} \right)}}{{S\left( {50} \right)}} \cdot \frac{{S\left( {50} \right)}}{{S\left( {50 + 30} \right)}}} \right) – 1\) \(= \frac{{\frac{{S\left( {50 + 29} \right)}}{{S\left( {50} \right)}}}}{{\frac{{S\left( {50 + 30} \right)}}{{S\left( {50} \right)}}}} – 1\) \(= \frac{{{}_{29}{p_{50}}}}{{{}_{30}{p_{50}}}} – 1\) \(= \frac{{\exp \left( {0.5\left( {1 – {{1.05}^{29}}} \right)} \right)}}{{\exp \left( {0.5\left( {1 – {{1.05}^{30}}} \right)} \right)}} – 1\) \(= 0,108384\)
Jawaban	d. 0,1054