(i) An excerpt from a select and ultimate life table with a select period of 3 years:

[math]x[/math]	[math]l_{[x]}[/math]	[math]l_{[x]+1}[/math]	[math]l_{[x]+2}[/math]	[math]l_{x+3}[/math]	[math]x+3[/math]
60	80,000	79,000	77,000	74,000	63
61	78,000	76,000	73,000	70,000	64
62	75,000	72,000	69,000	67,000	65
63	71,000	68,000	66,000	65,000	66

(ii) Deaths follow a constant force of mortality over each year of age

Calculate [math]1000_{2 \mid 3} q_{[60]+0.75}[/math].

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Answer: B

Under constant force over each year of age, [math]l_{x+k}=\left(l_{x}\right)^{1-k}\left(l_{x+1}\right)^{k}[/math] for [math]x[/math] an integer and [math]0 \leq k \leq 1[/math].

[[math]] \begin{aligned} & { }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}} \\ & l_{[60]+0.75}=(80,000)^{0.25}(79,000)^{0.75}=79,249 \\ & l_{[60]+2.75}=(77,000)^{0.25}(74,000)^{0.75}=74,739 \\ & l_{[60]+5.75}=(67,000)^{0.25}(65,000)^{0.75}=65,494 \end{aligned} [[/math]]

[[math]]{ }_{2 \mid 3} q_{[60]+0.75}=\frac{l_{[60]+2.75}-l_{[60]+5.75}}{l_{[60]+0.75}}=\frac{74,739-65,494}{79,249}=0.11679[[/math]]

[[math]]1000_{2 \mid 3} q_{[60]+0.75}=116.8[[/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.