Revision as of 12:29, 18 January 2024 by Admin (Created page with "'''Answer: A''' The median of <math>K_{48}</math> is the integer <math>m</math> for which <math display="block"> P\left(K_{48} < m\right) \leq 0.5 \text { and } P\left(K_{48} > m\right) \leq 0.5 \text {. } </math> This is equivalent to finding <math>m</math> for which <math display="block"> \frac{l_{48+m}}{l_{48}} \geq 0.5 \text { and } \frac{l_{48+m+1}}{l_{48}} \leq 0.5 </math> Based on the SULT and <math>l_{48}(0.5)=(98,783.9)(0.5)=49,391.95</math>, we have...")
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Exercise


Jan 18'24

Answer

Answer: A

The median of [math]K_{48}[/math] is the integer [math]m[/math] for which

[[math]] P\left(K_{48} \lt m\right) \leq 0.5 \text { and } P\left(K_{48} \gt m\right) \leq 0.5 \text {. } [[/math]]


This is equivalent to finding [math]m[/math] for which

[[math]] \frac{l_{48+m}}{l_{48}} \geq 0.5 \text { and } \frac{l_{48+m+1}}{l_{48}} \leq 0.5 [[/math]]


Based on the SULT and [math]l_{48}(0.5)=(98,783.9)(0.5)=49,391.95[/math], we have [math]m=40[/math] since [math]l_{88} \geq 49,391.95[/math] and [math]l_{89} \leq 49,391.95[/math].

So: [math]A P V=5000 A_{48}+5000_{40} E_{48} A_{88}=5000 A_{48}+5000 \cdot{ }_{20} E_{48}{ }_{20} E_{68} \cdot A_{88}[/math] [math]=5000(0.17330)+5000(0.35370)(0.20343)(0.72349)=1126.79[/math]

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

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