Question

See below for a link to a screenshot of the question.

Answer 1

http://i.imgur.com/5BgKpHa.png

Answer 2

Not smart enough to help with this sorry ;-;

Answer 3

No problemo. Thanks anyway!

Answer 4

I'm trying to apply the method outlined in the following article:
http://mathmodelsblog.wordpress.com/2010/01/05/an-insurance-example-of-a-mixed-distribution-%E2%80%93-ii/

However, when I try to apply the method, I get the wrong answer. The correct answer is E.

Answer 5

haha i've actually taken this exam before
ok the claim "y" = loss(x) - deductible
--> y = x-4 or x = y+4
any loss less than 4 has a 0 claim payment so its not considered

\[E(y)= \sum_{}^{} y*p(y)\]
\[p(y) = .02(y+4) , 0<y<6\]
\[E(y) = .02 \int\limits_0^6 y(y+4)dy = 2.88\]

Answer 6

Thanks dumbcow for your reply! I did what you did and got the same answer. However, for the some reason the solution manual reads:
http://i.imgur.com/fh1pk49.png

I don't know if we're doing it wrong or if the solution manual is wrong :(

Answer 7

ohh well it has been a few years and i never passed exam 2

seems we have to adjust p(y) based on how likely it is for loss to be more than 4
\[P(x>4) = .02 \int\limits_4^{10} x dx = .84\]
\[p(y) = \frac{.02(y+4)}{.84}, 0<y<6\]

Answer 8

Oh, I suppose they didn't use conditional expressions in the article because an exponential distribution was used as an example and exponential distributions are memoryless:

http://i.imgur.com/llyOk98.png

Makes sense now! Thanks for your help dumbcow.