# VAR with a difference

Hi,

have been taking a good few practice exams and been inputting results into a spreadsheet.

then i started messing around with the figures with averages and standard deviations. then shortfall risk came to mind return mean - 2*standard deviation assuming normal distribution my worst case scenario will be 55%…must study harder.

then VAR came to mind: mean result - Z(standard deviation) = 65%

what are chances of getting 65% or less in exam? maybe you must assume a mean return of 65% on practice exams.

so say your mean result is 68% with a standard deviation of 6%. Look up z score for 0.5 [(68-65)/6] gives you .1915 and subtract from 0.5 to get 0.31. you got a 31% chance of getting less then 65% and risking failure.

so you can take your chances or keep studying to improve those odds.

just a bit of playing around this morning on a break, thought i would show that VAR has a practical use beyond finance.

let me know what u think.

Wrote a program in VB.net to simulate it, wasted a day.

So many assumptions you can make, I ran teh assumptions with 120 of the total los tested and with the assumption of if you know the los you get the point

turned out you need to know 75% + of the material to pass with a 95 confidence assuming a pass avg of 70. So while some people think knowing only 70% will pass them, not for sure…

Anyway, changing any assumption would change the results.