En una tabla de mortalidad \(lo = 10000\) , siendo \(l_{40}=9200\) y siendo \(l_{55}=8600\) Calcular \(_{15}p^{-}_{40}\) para un riesgo del \(0.05\)%
Estableciendo un nuevo \(lo =5000\)
\[ p_{0}=\frac{9200}{10000}=0.92 \phantom{1cm} q_{0}=1-0.92=0.08\phantom{rrr} lambda=1.6448536\\ \begin{aligned} l^{-}_{40} = l_{0}·p_{0} - \lambda·\sqrt{l_{0}·p_{0}·q_{0}}\\ l^{-}_{40}=5000· 0.92-1.6448536\sqrt{5000·0.92·0.08}=\\ l^{-}_{40} = 4568.446 \end{aligned}\\ p_{0}=\frac{8600}{10000}=0.86 \phantom{1cm} q_{0}=1-0.86=0.14\phantom{rrr} lambda=1.6448536\\ \begin{aligned} l^{-}_{55} = l_{0}·p_{0} - \lambda·\sqrt{l_{0}·p_{0}·q_{0}}\\ l^{-}_{55}=5000· 0.86-1.6448536\sqrt{5000·0.86·0.14}=\\ l^{-}_{55} = 4259.642 \end{aligned}\\ \]
extype: cloze exclozetype: verbatim|verbatim exsolution: :NUMERICAL:=0.932405023502521:0.1~%50%4559.642:0.1#Normal-based instead of t-based interval; for small samples, intervals based on the normal approximation are too narrow.|:NUMERICAL:=0.934782608695652:0.1~%50%4340.358:0.1#Normal-based instead of t-based interval; for small samples, intervals based on the normal approximation are too narrow. exname: Confidence interval extol: 0.01