En una tabla de mortalidad \(lo = 10000\) , observando que siendo \(l_{30}=9900\) y siendo \(l_{42}=9300\) . Hemos recargado la tabla con \(lo =5000\) para calular la prima de riesgo de un seguro de vida, resultando dichos valores recargados \(l^{-}_{30}=4940.114\) y siendo \(l^{-}_{42}=4624.65\) . Calcular el riesgo asociado a dicha prima
\[ l^{-}_{30}\phantom{1cm} se\phantom{1} calculó\\ p_{0}=\frac{9900}{10000}=0.99 \phantom{1cm} q_{0}=1-0.99=0.01\phantom{rrr} lambda=1.4050716\\ \begin{aligned} l^{-}_{30} = l_{0}·p_{0} - \lambda·\sqrt{l_{0}·p_{0}·q_{0}}\\ l^{-}_{30}=5000· 0.99-1.4050716\sqrt{5000·0.99·0.01}=\\ l^{-}_{30} = 4940.114 \end{aligned}\\ luego\\ p_{0}=\frac{9900}{10000}=0.99 \phantom{1cm} q_{0}=1-0.99=0.01\\ \begin{aligned} l^{-}_{30}=4940.114 \longrightarrow \lambda=\frac{l^{-}_{30}-l_{0}·p_{0}}{\sqrt{l_{0}·p_{0}·q_{0}}}=\\ =\lambda=\frac{4940.114-(5000·0.99)}{\sqrt{5000·0.99·0.01}}=-1.4050846\\ con\phantom{1}tabla \phantom{1} de\phantom{1} Normal\\ riesgo=7.9998065% \end{aligned}\\ \]
extype: cloze exclozetype: verbatim|verbatim exsolution: :NUMERICAL:=7.99980646486252:0.1~%50%4924.65:0.1#Normal-based instead of t-based interval; for small samples, intervals based on the normal approximation are too narrow. exname: Confidence interval extol: 0.002