Computer maintenance costs for a department are modeled as follows:

1. The distribution of the number of maintenance calls each machine will need in a year is Poisson with mean 3.
2. The cost for a maintenance call has mean 80 and standard deviation 200.
3. The number of maintenance calls and the costs of the maintenance calls are all mutually independent.

The department must buy a maintenance contract to cover repairs if there is at least a 10% probability that aggregate maintenance costs in a given year will exceed 120% of the expected costs.

Calculate the minimum number of computers needed to avoid purchasing a maintenance contract using the normal approximation for the distribution of the aggregate maintenance costs.

• 80
• 90
• 100
• 110
• 120

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Key: C

Let [math]N=[/math] number of computers in department

Let [math]X=[/math] cost of a maintenance call

Let [math]S=[/math] aggregate cost

[math]\operatorname{Var}(X)=[\operatorname{Standard~Deviation~}(X)]^{2}=200^{2}=40,000[/math]

[math]\operatorname{E}(X^{2})=\operatorname{Var}(X)+[\operatorname{E}(X)]^{2} =40,000+80^{2}=46,400[/math]

[math]\operatorname{E}(S)=N \lambda \operatorname{E}(X)=N(3)(80)=240 N[/math]

[math]\operatorname{Var}(S)=N \lambda \times \operatorname{E}(X^{2})=N(3)(46,400)=139,200 N[/math]

We want

[[math]] \begin{aligned} 0.1 \geq \operatorname{Pr}(S\gt1.2 \operatorname{E}(S)) \geq \operatorname{Pr}\left(\frac{S-\operatorname{E}(S)}{\sqrt{139,200 N}}\gt\frac{0.2 \operatorname{E}(S)}{\sqrt{139,200 N}}\right) \Rightarrow \frac{0.2(240) N}{373.1 \sqrt{N}} & \geq 1.282 \\ &=\Phi(0.9)N \\ &\geq\left(\frac{1.282(373.1)}{48}\right)^{2} \\ &=99.3 \end{aligned} [[/math]]

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.