exercise:452120ca41

Jan 16'24

Exercise

For a mortality table with a select period of two years, you are given:

[math]x[/math]	[math]q_{[x]}[/math]	[math]q_{[x]+1}[/math]	[math]q_{x+2}[/math]	[math]x+2[/math]
50	0.0050	0.0063	0.0080	52
51	0.0060	0.0073	0.0090	53
52	0.0070	0.0083	0.0100	54
53	0.0080	0.0093	0.0110	55

The force of mortality is constant between integral ages.

Calculate [math]1000_{2.5} q_{[50]+0.4}[/math].

15.2
16.4
17.7
19.0
20.2

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AAdmin

Jan 16'24

Answer: B

[[math]] \begin{aligned} { }_{2.5} q_{[50]+0.4} & =1-{ }_{2.5} p_{[50]+0.4}=1-{ }_{2.9} p_{[50]} /\left(p_{[50]}\right)^{0.4} \\ & =1-\left\{p_{[50]} p_{[50]+1}\left(p_{52}\right)^{0.9}\right\} /\left(1-q_{[50]}\right)^{0.4} \\ & =1-\left\{\left(1-q_{[50]}\right)\left(1-q_{[50]+1}\right)\left(1-q_{52}\right)^{0.9}\right\} /\left(1-q_{[50]}\right)^{0.4} \\ & =1-\left\{(1-0.0050)(1-0.0063)(1-0.0080)^{0.9}\right\} /(1-0.0050)^{0.4} \\ & =0.01642 \end{aligned} [[/math]]

[math]1000_{2.5} q_{[50]+0.4}=16.42[/math]