excans:5df74bcf35

Answer: B

Let [math]S[/math] denote the number of survivors.

This is a binomial random variable with [math]n=4000[/math] and success probability [math]\frac{21,178.3}{99,871.1}=0.21206[/math]

[math]E(S)=4,000(0.21206)=848.24[/math]

The variance is [math]\operatorname{Var}(S)=(0.21206)(1-0.21206)(4,000)=668.36[/math]

[math]\operatorname{Std} \operatorname{Dev}(S)=\sqrt{668.36}=25.853[/math]

The [math]90 \%[/math] percentile of the standard normal is 1.282

Let [math]S^{*}[/math] denote the normal distribution with mean 848.24 and standard deviation 25.853. Since [math]S[/math] is discrete and integer-valued, for any integer [math]s[/math],

[[math]] \begin{aligned} \operatorname{Pr}(S \geq s) & =\operatorname{Pr}(S \gt s-0.5) \approx \operatorname{Pr}\left(S^{*} \gt s-0.5\right) \\ & =\operatorname{Pr}\left(\frac{S^{*}-848.24}{25.853}\gt\frac{s-0.5-848.24}{25.853}\right) \\ & =\operatorname{Pr}\left(Z\gt\frac{s-0.5-848.24}{25.853}\right) \end{aligned} [[/math]]

For this probability to be at least [math]90 \%[/math], we must have [math]\frac{s-0.5-848.24}{25.853}\lt-1.282[/math]

[[math]] \Rightarrow s\lt815.6 [[/math]]

So [math]s=815[/math] is the largest integer that works.

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.