Example 9, Module 3, Page 23 ASM

[Actuario]

I feel like the horse:

I'm having some trouble with a seemingly easy problem, Example 9, Module 3, Page 23, ASM. I've tried this problem every which way possible but cannot get the answer.

A Town sells lottery tickets; winning ticket pays $1 with probability .7, $5 with probability .2, $10 with probabiliy .1, and the number of winning tickets has a Poisson distribution with a rate 50 per month. Using normal approximation, find the 95th percentile of the total payout in a year.

The answer is given as $1,779.66

HELP!

[carrytheCrøss]

"Monthly" Frequency ~ Poisson(50), so "Annual" Frequency (let's call it N) ~ Poisson(600).

Severity (let's call it X) has E(X) = 2.7 and Var(X) = 8.41 (Work through this.)

Total Payout (let's call it S) has E(S) = E(N)*E(X) = 1620 and Var(S) = E(N)*Var(X)+E(X)^2*Var(N) = 600*8.41+ 2.7^2 = 9420.

95th Percentile, p, is such that P("Total Payout" < p) = P(S < p) = P((S - E(S))/sqrt(Var(S)) < (p - E(S))/sqrt(Var(S))) [<-- from normal approximation] = P(Z < (p - 1620) / sqrt(9420)) = .95

This implies that (p - 1620) / sqrt(9420) = 1.645 (From Exam M table) and p = 1779.66.

Hope this helps.

[Actuario]

I had miscalculated E[X] and hence there was no way I could get the answer.

I no longer feel like the horse.

Thanks!


TGIF!

[DudeMan]

Aggregate Loss Problems.......Anytime you are given a claim frequency model and a claim severity model and asks for normal approx. of aggregate claims use these formulas.

N = Frequency variable
X = Severity Variable
S = Aggregate (total) loss variable

E(S) = E(N)*E(X)
Var(S) = E(N)*Var(X) + E(X)^2*Var(N)

Z_confidence = [Y-E(S)]/StDev(s)

Look up Z in the normal chart, and solve equation for Y

[Actuario]

Thank you both very much!