Exercise
ABy Admin
Jan 19'24
Answer
Answer: C
Let [math]C[/math] be the annual contribution, then [math]C=\frac{{ }_{20} E_{45} \ddot{a}_{65}}{\ddot{a}_{45: 20}}[/math]
Let [math]K_{65}[/math] be the curtate future lifetime of (65). The required probability is
[[math]]
\operatorname{Pr}\left(\frac{C \ddot{a}_{45: 20}}{{ }_{20} E_{45}}\gt\ddot{a}_{\overline{K_{65}+1}}\right)=\operatorname{Pr}\left(\frac{{ }_{20} E_{45} \ddot{a}_{65}}{\ddot{a}_{45: 20}} \frac{\ddot{a}_{45: 20}}{{ }_{20} E_{45}}\gt\ddot{a}_{\overline{K_{65}+1}}\right)=\operatorname{Pr}\left(\ddot{a}_{65}\gt\ddot{a}_{\overline{K_{65}+1}}\right)=\operatorname{Pr}\left(13.5498\gt\ddot{a}_{\overline{K_{65}+1}}\right)
[[/math]]
Thus, since [math]\ddot{a}_{\overline{21}}=13.4622[/math] and [math]\ddot{a}_{\overline{22}}=13.8212[/math] we have
[[math]]
\operatorname{Pr}\left(\ddot{a}_{K_{65}+1}\lt13.5498\right)=\operatorname{Pr}\left(K_{65}+1 \leq 21\right)=1-_{21} p_{65}=1-\frac{l_{86}}{l_{65}}=1-\frac{57,656.7}{94,579.7}=0.390
[[/math]]