Pembahasan Ujian PAI: A50 – No. 2 – Mei 2018

dan setiap decrement  menyebar seragam (UDD)

\(_tp_x^{‘(j)} = 1{ – _t}q_x^{‘(j)}\),  \(d_x^{(\tau )} = l_x^{(\tau )} – l_{x + 1}^{(\tau )}\),  \(l_x^{(\tau )} = \sum\limits_{j = 1}^n {l_x^{(j)}} \)latex ,  \(d_x^{(\tau )} = \sum\limits_{j = 1}^m {d_x^{(j)}} \)
\(d_x^{(j)} = l_x^{(\tau )}q_x^{(j)},d_x^{(\tau )} = l_x^{(\tau )}q_x^{(\tau )},\)
\(_{{\rm{ }}s}P{‘_x}^{(j)} = {{(_s}{P_x}^{(\tau )})^{q_x^{(j)}/q_x^{(\tau )}}}\)

\(d_{20}^{(\tau )} = \sum\limits_{j = 1}^2 {d_x^{(j)} = 170} \)
\(d_{20}^{(\tau )} = l_{20}^{(\tau )} – l_{21}^{(\tau )}\)

maka
\(l_{21}^{(\tau )} = l_{20}^{(\tau )} – d_{20}^{(\tau )} = 1000 – 170 = 830\)

selanjutnya
\(d_{21}^{(\tau )} = l_{21}^{(\tau )} – l_{22}^{(\tau )} = 830 – 650 = 180\)
\(d_{21}^{(\tau )} = d_{21}^{(1)} + d_{21}^{(2)}\)

maka
\(d_{21}^{(2)} = 180 – 66 = 114\)
\(q_{21}^{(2)} = \frac{{d_{21}^{(2)}}}{{l_{21}^{(\tau )}}} = \frac{{114}}{{830}}\)
\(q_{21}^{(\tau )} = \frac{{d_{21}^{(\tau )}}}{{l_{21}^{(\tau )}}} = \frac{{180}}{{830}}\)

Selanjutnya dipunyai \(q’\)\({}_{21}^{\left( 2 \right)}\)
artinya
\(q’\)\({}_{21}^{\left( 2 \right)}\)\(= 1 – \) \({p’}\)\({}_{21}^{\left( 2 \right)}\)
\(= 1 – {({P_{21}}^{(\tau )})^{q_{21}^{(2)}/q_{21}^{(\tau )}}}\)

\(= 1 – {(\frac{{650}}{{830}})^{\frac{{(\frac{{114}}{{830}})}}{{(\frac{{180}}{{830}})}}}}\)

\(= 0.1434\)