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In BSOP – For calculation of N(d1) , If value of d1 is 0.6548 , can we directly consider d1 as 0.65 and take its normal distribution (0.2422) and add 0.5 to it instead of finding the the proportionate normal distribution of 0.6548 in between 0.65 and 0.66.
So if you consider d1 as 0.65 – N(d1) = 0.2422 +0.5 = 0.7422
and if you consider d1 as 0.6548, you have to invest more time in finding N(d1) by following method :-
value of normal distribution at 0.6548 = 0.2422 + ( [0.0048*0.0032]/0.01) = 0.2437.
N(d1) = 0.5 + 0.2437 = 0.7437.
Kaplan uses the 2nd method (N(d1) as 0.7437.
Strictly you should apportion between the values (the examiner does in his answers and Kaplan simply copies out the examiners answers). However it takes time and you are unlikely to lose more than 1 mark (and quite likely not even that) if you do not bother apportioning.
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