A towing company provides all towing services to members of the City Automobile Club. You are given:

1. The automobile owner must pay 10% of the cost and the remainder is paid by the City Automobile Club.
2. The number of towings has a Poisson distribution with mean of 1000 per year.
3. The number of towings and the costs of individual towings are all mutually independent.

Calculate the probability that the City Automobile Club pays more than 90,000 in any given year using the normal approximation for the distribution of aggregate towing costs.

• 3%
• 10%
• 50%
• 90%
• 97%

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

Key: B

First restate the table to be CAC’s cost, after the [math]10 \%[/math] payment by the auto owner:

Towing Cost, [math]x[/math]	[math]p(x)[/math]
72	[math]50 \%[/math]
90	[math]40 \%[/math]
144	[math]10 \%[/math]

Then

[[math]]\operatorname{E}(X)=0.5(72)+0.4(90)+0.1(144)=86.4[[/math]]

[[math]]E\left(X^{2}\right)=0.5\left(72^{2}\right)+0.4\left(90^{2}\right)+0.1\left(144^{2}\right)=7905.6[[/math]]

[[math]]\operatorname{Var}(X)=7905.6-86.4^{2}=440.64[[/math]]

Because Poisson,

[[math]]\operatorname{E}(N)=\operatorname{Var}(N)=1000[[/math]]

[[math]]\operatorname{E}(S)=\operatorname{E}(X) \operatorname{E}(N)=86.4(1000)=86,400[[/math]]

[[math]]\operatorname{Var}(S)=\operatorname{E}(N) \operatorname{Var}(X)+\operatorname{E}(X)^{2} \operatorname{Var}(N)=1000(440.64)+86.4^{2}(1000)=7,905,600[[/math]]

[[math]]\operatorname{Pr}(S\gt90,000)+\operatorname{Pr}\left(\frac{S-\operatorname{E}(S)}{\sqrt{\operatorname{Var}(S)}}\gt\frac{90,000-86,400}{\sqrt{7,905,600}}\right)=\operatorname{Pr}(Z\gt1.28)=1-\Phi(1.28)=0.10[[/math]]

Since the frequency is Poisson, you could also have used

[[math]]\operatorname{Var}(S)=\lambda E\left(X^{2}\right)=1000(7905.6)=7,905,600[[/math]]

. That way, you would not need to have calculated [math]\operatorname{Var}(X)[/math].