Hi Ajay,apparently the reverse is true for '11 and '12 and the two %ages have been almost identical in '10 and '14,if the chart at the top of the page is assumed as accurate.So other factors are also at play for sure..
Kavita honestly it seems like this depends on the shape of the distribution of marks.If the tail is thin(a distance of few marks from mean covers more %age of candidates than normal) then in the interest of fairness,GARP would want to pass this larger set of people because they're close in terms...
Wow so if we look at it,55 odd pass %age is more of a downside..generally this has touched 60% throughout..now the question is whether pass %age varies directly or inversely with difficulty level....
Hehehehehe....yeah,anyways people don't write XYZ,FRM(4th quartiler) in their resume do they? :D....not to say I want that for any of us but hey, a ",FRM" is a ",FRM"... :-)
Mate infact now if they put up a simple paper that'll be an anamoly.I prepared for part 1 via a different institute(non BT,non shwescher) and could crack 75%+ in 2.5 hrs..but part 1 last year was a horror story...could attempt 89/100 only after consuming the whole 4 hours but ended up passing...
Guys,since we're talking of FRM P2 here,let's appreciate the fact that the marks obtained will follow a probability distribution with thin tails(very few will nail it out and out or flunk horribly).There's no linear logic that says 75%==gold medal and hence 60%==pass.What if few dozen people get...
One aspect is the number of questions outside of all study packages(5-6 atleast),where even the most prepared folks would need to half-guess.Plus not may would have genuinely attempted 77-80.And even if we take the absentees out(not sure if that's valid),still more than half would be passing.So...
Even the top most scorer could have only attempted 80,and with so many close calls,would easily get 15 wrong so 65-60 would be a top 5% score and not any lower.Now we have to consider the fact that many people would have have genuinely attempted 65-70 odd to begin with,and also that there would...
Hi David,
A silly doubt on concept actually. From the formula sheet page 55,
1)Since return= beta*m +sqrt(1-beta^2)*epsilon,I am assuming that whether (k-return)>or<0 will decide default or no default.So not sure why epsilon<k-beta*m denotes default.Should'nt this be the opposite?
2)Again if...
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