I just took exam P for the first time today, July the 20th, and got a preliminary pass. There was not a single question where my calculated answer did not match one of the options. I suspect I got a 9, maybe even a 10 on the exam. Your videos were a big help, I used a few tricks that I would not have ever learned without your channel, such as the tabular integration by parts. Thank you so much for all the help, and I hope your internship goes well!
That's awesome! I knew you would do well, congrats! I'm glad you found this content helpful, best of luck on your future studies and exams! Thanks for always contributing valuable input to the channel!
I had watched many of ur videos about Exam P (all of them in "MUST KNOW", and many of the othet sample questions) they are really really in details and very very helpful. I got a preliminary pass for my Exam P today. Thank you very much !!!! ur videos did helped a lot !!
I have seen a solution for this that used the double expectation formula for variance. You can make another random variable to represent the 2 sections, one section where the payout is 0, one section where the payout is uniform from 0 to 9b/10. The probability of the payout being 0 is 1/10, the probability of the probability being uniform from 0 to 9b/10 is 9/10. The expected value and variance are both 0 for the section with a payout of 0. The expected value of the payout is 9b/20 for the section with the uniform distribution. The variance of the payout is 81b^2/1200 for that same section. The variance of the conditional expected value is 729b^2/40000 The expected value of the conditional variance is 729b^2/12000 Add those up and you get a unconditional variance of 3159b^2/40000 That gives a standard deviation of 9b(√39)/200 The standard deviation of the original function is (b√3)/6 Divide the 2 to get 27(√13)/100 Which after calculation ends up being approximately 0.9735
So, I fully understand this solution and have worked through it several times. It's pretty straightforward since we're just working from the definition of expected value and of variance. Every time I see this problem though, my first reaction is to try and solve it with the law of total probability conditioning on the loss being less than/greater than the deductible. This never works for me though. Is is possible to solve the problem with law of total prob? If so, any chance you could make a video on it?
