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Hi there,
Can someone please explain why do have we have to add 0.7x (0.2324-0.2291) when n(d1) is calculated? The same needs to be done when calculating n(d2) – 0.3 x(0.0793-0.0753) needs to be added too.
I have no clue where it comes from.
Thank you in advance,
BrigitaIt is because the tables only let you find N(d1) to 2 decimal places, so the examiner has apportioned between the values on either side.
I would not bother doing this (and I don’t think I would lose any marks) I would just round d1 (and later d2) to 2 decimal places and then use the table figure directly. The answer will end up slightly different, but (again) I don’t think you would lose any marks. (And even if you did, it would not be more than half a mark – I think the time involved in apportioning is not worth wasting for just half a mark 🙂 )
Thank you for a quick reply, John. I thought I missed out on something.;
Have a great evening.:)
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