Exam Feedback May 2019 Part 2 Exam Feedback
- Thread starter Nicole Seaman
- Start date
-
- Tags
- exam-feedback
- Status
If I remember I had the purple but yes in the first 10 or so questions lots of Bs. Then somewhere around question 30, there were lots of Ds. Very few Ds in first 20 questions. That’s what I recall anyway and it did stand out to me.
amit.m.sharma
Member
Here's my thoughts on the exam score and possible passing score:
Regardless of the cut off point, I feel the distribution of prices would be bimodal or even multimodal. This is because you expect to do better if you are sitting for the exam a second time (multimodal if there are more forces acting such as bionic turtle help, able to prepare much better, etc). The number of registrations for the exams has been increasing and while the percentage of people passing the second exam is high, we don't really know what percentage of the exam takers and passes were first time test takers and second or multiple time test takers. I can only imagine that your score will improve if you take the exam more than once but the improvement is upto a limit.
If the percentage of the first time test takers is p (I think 0.5 < p < 0.66), X is the distribution of scores of first time test takers, and Y is the distribution of scores of the multiple times test takers then, the combined distribution of all test takers Z would be,
Z = pX + (1-p)Y.
Given no correlation between multiple times test takers and first time test takers, rho(x,y) = 0.
So, expected value of Z, E(Z) = pE(X) + (1-p)E(Y) and Variance(Z) = p^2*Variance(X) + (1-p)^2*Variance(Y)
My intuition says E(Y) > E(X) and that X probably is positively skewed and Y is negatively skewed. I don't know how close E(Y) and E(X) can be but say E(Y) - E(X) = 15. So, E(Z) = pE(X) + (1-p)(E(X)+15) = E(X) + (1-p)*15.
I see some members are suggesting 40-50 as the passing mark. I assume they are first time test takers and appreciate that this is what their intuition says. I'll assume the percentage of the first time test takers is 60%, so as per my model the expected value of Z = 45 + 0.4*15 = 51. Since the percentage of people passing changes each year I can only assume the cutoff is the mean of the total distribution.
Regardless of the cut off point, I feel the distribution of prices would be bimodal or even multimodal. This is because you expect to do better if you are sitting for the exam a second time (multimodal if there are more forces acting such as bionic turtle help, able to prepare much better, etc). The number of registrations for the exams has been increasing and while the percentage of people passing the second exam is high, we don't really know what percentage of the exam takers and passes were first time test takers and second or multiple time test takers. I can only imagine that your score will improve if you take the exam more than once but the improvement is upto a limit.
If the percentage of the first time test takers is p (I think 0.5 < p < 0.66), X is the distribution of scores of first time test takers, and Y is the distribution of scores of the multiple times test takers then, the combined distribution of all test takers Z would be,
Z = pX + (1-p)Y.
Given no correlation between multiple times test takers and first time test takers, rho(x,y) = 0.
So, expected value of Z, E(Z) = pE(X) + (1-p)E(Y) and Variance(Z) = p^2*Variance(X) + (1-p)^2*Variance(Y)
My intuition says E(Y) > E(X) and that X probably is positively skewed and Y is negatively skewed. I don't know how close E(Y) and E(X) can be but say E(Y) - E(X) = 15. So, E(Z) = pE(X) + (1-p)(E(X)+15) = E(X) + (1-p)*15.
I see some members are suggesting 40-50 as the passing mark. I assume they are first time test takers and appreciate that this is what their intuition says. I'll assume the percentage of the first time test takers is 60%, so as per my model the expected value of Z = 45 + 0.4*15 = 51. Since the percentage of people passing changes each year I can only assume the cutoff is the mean of the total distribution.
Amit, you didn’t pass back in Nov 2018 on your first try. Obviously you’re better prepared now and should be able to pass.
But going back to six months ago, how much do you think you scored of 80? I know someone with a 32212 who passed but was certain he scored no more than 45 correct, and likely less than that. You must’ve been close to passing in Nov 2018 going by your quantiles scores...if you had to guess one figure and a narrow range, how did you score back then out of 80?
theapplecrispguy
New Member
amit.m.sharma
Member
Unfortunately I didn’t prepare as well as I wanted to this time due to really long hours at the workplace and other commitments at home. But, overall I feel I did better because of bionic turtle. Fingers crossed for the results!
In the Nov 2018 exam I definitely guessed on more questions than I wanted to. But following the thread for Nov 2018 part exam I was able to confirm that I solved several numerical problems correctly. So worst case I felt I scored in the low 40s. I have a very hard time believing I scored less than 40 in November.
Ok but going on one number, what you have expected your result to be out of 80 (or rather, the best estimate of what you actually scored)? I’d imagine you missed the cutoff by one or two points only.
And on the whole, which paper was trickier? Granted this paper would be easier since you put in more work, but which one had more tricky question you say (i.e. which paper had the higher overall level of difficulty)?
I had yellow one boss
kumaranimesh09
New Member
If you are talking about the question where there was a merger due to take place, I marked it as frown because of price jump.
Ohhh i marked smirk
amit.m.sharma
Member
If I had only one number to pick it would be 42
The difficulty level was consistent. I can’t say one paper was more tricky than the other.
gprisby
Active Member
The one thing I have learned from looking at these threads is that this is a select group of high performers. I feel like the people here have the motivation and resources to do better than the greater population of FRM test takers.
gprisby
Active Member
Hopefully! Thinking I'm around 46-50.
amit.m.sharma
Member
42 is what I think I scored last time which was not enough for a pass.....I feel I did better this time with less guesswork...I can only hope I pass this time
I marked " C" , which is volatility frown but most of my friends mark " smirk " for this one...
I also marked frown. believe it's correct.
A risk analyst estimates that the hazard rate for a company is 0.12 per year. Assuming a constant hazard rate
model, what is the probability that the company will survive in the first year and then default before the end
of the second year?
A. 8.9%
B. 10.0%
C. 11.3%
D. 21.3%
Correct Answer: B
Explanation: B is correct.
The joint probability of survival up to time t and default over (t, t+ τ) is:
P[t* > t ∩ t* < t+τ] = 1-e-λ(t+ τ)-(1-e-λt) = e-λt
(1-e-λτ
)
The joint probability of survival the first year and default over the first year and the
second year is:
P[t* > 1 ∩ t* < 1+1] = e-0.12*1(1-e-0.12*1) = 10.03%
- Status
Similar threads
