En una tabla de mortalidad \(lo = 10000\) , siendo \(l_{35}=9700\) Recargar \(\textbf{-}\) el valor \(l_{35}\) para un riesgo del \(0.03\)%
Estableciendo un nuevo \(lo =5000\)
\[ p_{0}=\frac{9700}{10000}=0.97 \phantom{1cm} q_{0}=1-0.97=0.03\phantom{rrr} lambda=1.8807936\\ \begin{aligned} l^{-}_{35} = l_{0}·p_{0} - \lambda·\sqrt{l_{0}·p_{0}·q_{0}}\\ l^{-}_{35}=5000· 0.97-1.8807936\sqrt{5000·0.97·0.03}=\\ l^{-}_{35} = 4827.313 \end{aligned} \]
extype: cloze exclozetype: verbatim|verbatim exsolution: :NUMERICAL:=4827.313:0.1~%50%4827.313:0.1#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