A life is subject to the following 3 -year select and ultimate table:

[math][x][/math]	[math]\ell_{[x]}[/math]	[math]\ell_{[x]+1}[/math]	[math]\ell_{[x]+2}[/math]	[math]\ell_{x+3}[/math]	[math]x+3[/math]
55	10,000	9,493	8,533	7,664	58
56	8,547	8,028	6,889	5,630	59
57	7,011	6,443	5,395	3,904	60
58	5,853	4,846	3,548	2,210	61

You are also given:

(i) [math]e_{60}=1[/math]

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

Calculate [math]\stackrel{\circ}{e}_{[58]+2}[/math].

• 1.5
• 1.6
• 1.7
• 1.8
• 1.9

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

Answer: B

[[math]] \stackrel{\circ}{e}_{[58]+2}=e_{[58]+2}+0.5 [[/math]]

[[math]]e_{[58]+2}=p_{[58]+2}\left(1+e_{61}\right)=p_{[58]+2}\left[1+\frac{e_{60}}{p_{60}}-1\right][[/math]]

[[math]]=\frac{l_{61}}{l_{[58]+2}} \times \frac{e_{60}}{p_{60}}=\frac{2210}{3548} \times \frac{1}{(2210 / 3904)}=\frac{3904}{3549}=1.100338[[/math]]

[[math]]\stackrel{\circ}{e}_{[58]+2}=1.100338+0.5=1.6[[/math]]