exercise:Ec571433ec

ABy Admin

Jan 15'24

Exercise

You are given:

i)[math] q_{80}=0.04[/math]

ii)[math] q_{81}=0.06[/math]

iii)[math]q_{82}=0.08[/math]

iv) Deaths between ages 80 and 81 are uniformly distributed

v) Deaths between ages 81 and 82 are subject to a constant force of mortality

Calculate the probability that a person aged 80.6 will die between ages 81.1 and 81.6.

0.0294
0.0296
0.0298
0.0300
0.0302

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ABy Admin

Jan 15'24

Answer: C

Let [math]l_{80}=1000 \Rightarrow l_{81}=960 \Rightarrow l_{82}=902.4[/math]

[[math]]\textrm{Answer} \, =\frac{l_{81.1}-l_{81.6}}{l_{80.6}}=\frac{(960)^{0.9}(902.4)^{0.1}-(960)^{0.4}(902.4)^{0.6}}{(0.4)(1000)+(0.6)(960)}=0.02978[[/math]]

The value for [math]l_{80}[/math] is arbitrary. Any other starting value gives the same result.

Another form for the survivors between 81 and 82 would be

[math]\mu=-\ln (902 / 960)=0.0095833[/math];

[math]l_{81.1}=960 \times e^{-0.1 \times 0.0095833}=959.08 ;[/math]

[math]l_{81.6}=960 \times e^{-0.6 \times 0.0095833}=954.496[/math].