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

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

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].

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Copyright 2024. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.