1)-Question 6 in paper March 2006 Explain the solution ,I am not understand properly the equation of x & y. 2)Question 8 ,Part(i) in same paper, Why, we take X=(>=)3
I found this one difficult when I did the paper. Try and think about how you would standardize the X rv x -- n(u, s^2) (x-u)/sqrt(s^2) -- N(0,1) This is so you can use the Z values in the tables as they are N(0,1) ie P(X<x) = P(Z< (x-u)/sqrt(s^2) ) u = mu s^2= sigma sqaured sqrt is square root normally you look for (x-u)/sqrt(s^2) in the tables, well now we have simulated this by looking up the values in the table from the U(0,1) random numbers. Examiners have given us a "slight" HINT: Notice how the values correspond nicely to values in the tables. We are equating z = (x-u)/sqrt(s^2) Rearrange x = z*sqrt(s^2) + u so from the answer line 1 is x = 200 +10(0.180) = 201.80 Take this value and use in Y|X=x -- N(x,1) P(Y<y) = P(Z < (y-x)/1) Equate z = y - x Rearrange y = z + x y = 0.930 + 201.80 and so on. Imagine a the discrete disbn of x - poi(u) if we sum over all possible x x=0, 1, 2, .... we get a total of 1 i.e. P(0) + P(1) + P(2) + P(3) + P(4) + P(5)....... = 1 we require P(0) + P(1) + P(2) + P(everything else) = 1 P(everything else) = P(X>=3) = 1 - P(X<= 2) Hope I have helped.
Thanks for the help, Actually ,I want to know why we consider '>3' value ,According to the question ,TO determine the probability of only 0,1,2,or 3 ,but not mention >3. So why we consider >3
The question states "However, the policy is limited such that only the first three claims in any one year are paid." If we let C be the number of potential claims, and N be the number of claims that are paid, then: C ~ Poi(0.8) P(N = 0) = P(C = 0) P(N = 1) = P(C = 1) P(N = 2) = P(C = 2) P(N = 3) = P(C >= 3) as a maximum of 3 claims will get paid, regardless of how many more potential claims there are.
