exercise:2b56a6c321

Jan 16'24

Exercise

The SULT Club has 4000 members all age 25 with independent future lifetimes. The mortality for each member follows the Standard Ultimate Life Table.

Calculate the largest integer [math]N[/math], using the normal approximation, such that the probability that there are at least [math]N[/math] survivors at age 95 is at least [math]90 \%[/math].

800
815
830
845
860

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AAdmin

Jan 16'24

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.