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Exam STAM Sample 285 tricky
This is an exercise prepared for you to answer it wrongly. Here is the exercise
"285. You are the producer for the television show Actuarial Idol. Each year, 1000 actuarial clubs audition for the show. The probability of a club being accepted is 0.20. The number of members of an accepted club has a distribution with mean 20 and variance 20. Club acceptances and the numbers of club members are mutually independent. Your annual budget for persons appearing on the show equals 10 times the expected number of persons plus 10 times the standard deviation of the number of persons. Calculate your annual budget for persons appearing on the show. (A) 42,600 (B) 44,200 (C) 45,800 (D) 47,400 (E) 49,000" There are 2 random variables. The first one has a Binomial distribution with parameters m=1000, q=0.2 The second one is just another distribution with known mean 20 and variance 20. So far, so good... Now, which is the frequency, let's call it N, and which the severity, let's call it X to make up the aggregate S? And here is where the trick lies. The one the problem mentions as the number of persons is the severity and not the frequency (like number of claims per policy holder). I smelled a rat from the beginning, but I am not 100% sure of the reasons for calling one frequency and the other the severity. My rationale is that the Binomial shows how many clubs will be accepted, so it acts as claims per exposure, then that would be the primary. I have to add the number of member per club, from 1 to n, and n is the total number of clubs accepted. So it's something like accepted / total. The other random variable, members per club, is like dependent on the first one. In summary it's like S = Sum (Ni) from 1 to n, where Ni is the number of members per club, and n the number of clubs accepted. And then Ni acts as Xi the severity. And then the frequency is the number of clubs accepted. It would not make sense to sum the other way around (like summing how many clubs have 1 member, and then how many have 2 members, from 1 to N.) Specially if the second random variable is continuous and not discrete, and then instead of summing I would integrate. And I guess that's my own answer. Needless to say, one of the possible answers (E) is 49,000 and that is close to the 48,962 if the variables were transposed. (Correct answer is A) And my final question is whether there is a better way to see this. It bothers me that I have reached to this point of the sample and I am having trouble distinguishing Frequency and Severity. I would appreciate your answers.
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