exercise:45e946d2d0

Jan 16'24

Exercise

You are given the following information from a life table:

[math]x[/math]	[math]l_{x}[/math]	[math]d_{x}[/math]	[math]p_{x}[/math]	[math]q_{x}[/math]
95	-	-	-	0.40
96	-	-	0.20	-
97	-	72	-	1.00

You are also given:

(i) [math]\quad l_{90}=1000[/math] and [math]l_{93}=825[/math]

(ii) Deaths are uniformly distributed over each year of age.

Calculate the probability that (90) dies between ages 93 and 95.5.

0.195
0.220
0.345
0.465
0.668

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AAdmin

Jan 16'24

Answer: C

We need to determine [math]{ }_{3 \mid 2.5} q_{90}[/math].

[[math]] { }_{3 \mid 2.5} q_{90}=\frac{l_{90+3}-l_{90+3+2.5}}{l_{90}}=\frac{l_{93}-l_{95.5}}{l_{90}}=\frac{l_{93}-\left(l_{95}-0.5 d_{95}\right)}{l_{90}}=\frac{825-[600-0.5(240)]}{1,000}=0.3450 [[/math]]

where [math]l_{90}=1,000, l_{93}=825, l_{97}=\frac{d_{97}}{q_{97}}=\frac{72}{1}=72, l_{96}=\frac{l_{97}}{p_{96}}=\frac{72}{0.2}=360[/math],

[math]l_{95}=\frac{l_{96}}{p_{95}}=\frac{360}{1-0.4}=600[/math], and [math]d_{95}=l_{95}-l_{96}=600-360=240[/math].