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<math display = "block">1000_{2 \mid 3} q_{[60]+0.75}=116.8</math>
<math display = "block">1000_{2 \mid 3} q_{[60]+0.75}=116.8</math>
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Latest revision as of 02:34, 18 January 2024

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.