exercise:7ea913781d

Jan 16'24

You are given:

i) The following extract from a three-year select and ultimate table:

[math][x][/math]	[math]q_{[x]}[/math]	[math]q_{[x]+1}[/math]	[math]q_{[x]+2}[/math]	[math]q_{x+3}[/math]	[math]x+3[/math]
50	0.020	0.031	0.043	0.056	53
51	0.025	0.037	0.050	0.065	54
52	0.030	0.043	0.057	0.072	55
53	0.035	0.049	0.065	0.091	56
54	0.040	0.055	0.076	0.113	57
55	0.045	0.061	0.090	0.140	58

ii) Mortality follows a uniform distribution of deaths over each year of age

Calculate [math]1000_{\left(0.6 \mid 1.5 q_{[52]+1.7}\right)}[/math].

Jan 16'24

Answer: C

[math][x][/math]	[math]q_{[x]}[/math]	[math]q_{[x]+1}[/math]	[math]q_{[x]+2}[/math]	[math]q_{x+3}[/math]	[math]x+3[/math]
52	0.030	0.043	0.057	0.072	55

[math]l_{[52]}=1000[/math] (arbitrary, for convenience; any other value would work and give the same answer)

[math]l_{[52]+1}=1000(1-0.03)=970[/math]

[math]l_{[52]+2}=970(1-0.043)=928.29[/math]

[math]l_{55}=928.29(1-0.057)=875.3775[/math]

[math]l_{56}=875.3775(1-0.072)=812.3503[/math]

[math]l_{[52]+2.3}=(0.7) l_{[52]+2}+(0.3) l_{55}=912.42[/math]

[math]l_{[52]+3.8}=l_{55.8}=(0.2) l_{55}+(0.8) l_{56}=824.96[/math]

[math]l_{[52]+1.7}=(0.3) l_{[52]+1}+(0.7) l_{[52]+2}=940.80[/math]

[math]1000\left(\left(_{0.6 \mid 1.5} q_{[52]+1.7}\right)=1000 \frac{912.42-824.96}{940.80}=92.96\right.[/math]