excans:5df74bcf35

Exercise

Jan 16'24

Answer

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.