Question

Consider the following life table and calculate TABLE 1. Extract from a life table. lz de 30 10000.00 34.78 31 9965.22 38.10 32 9927.12 41.76 33 9885.35 45.81 34 9839.55 50.26 35 9789.29 55.17 36 9734.12 60.56 37 9673.56 66.49 38 9607.07 72.99 39 9534.08 80.11 (a) 144 (b) 9P31 (c) 433 (d) 5935 (e) the probability that a life currently aged exactly 30 dies between ages 35 and 36.

Answer 1

a) $l_{44}$ cannot be obtained from the table given above as it does not contains the mortality value for the age more than 40.

b) $_{9}p_{31}$ =  $\frac{l_{40}}{l_{31}}$ = 9453.97 / 9965.22 = 0.9457
where  $l_{40}$ is obtained as follows :
We know that, dx = l(x) - l(x+1)
Thus,  
d39 39 140.

i.e. $l_{40}$ = 9534.08 - 80.11 = 9453.97

c) $q_{39}=\frac{l_{33}-l_{34}}{l_{33}}$ $=\frac{9885.35-9839.55}{9885.35}$ = 0.0046

d) $_{5}q_{35}=\frac{l_{35}-l_{40}}{l_{35}}$ $=\frac{9789.29-9453.97}{9789.29}$ =   0.0343

e) $_{5}p_{30}*q_{35}$ =  $\frac{l_{35}}{30}*\frac{l_{35}-l_{36}}{l_{35}}$ =  $\frac{9789.29-9734.12}{10000}$ = 0.0055

Hope this answers your query!