Forums › Ask ACCA Tutor Forums › Ask the Tutor ACCA AFM Exams › june 2011 Q4 normal distribution
- This topic has 3 replies, 2 voices, and was last updated 10 years ago by John Moffat.
- AuthorPosts
- May 15, 2014 at 3:01 pm #168912
Hi,
In the answer to this question, N(d1) is given as 0.5 + 0.2291 + 0.7 x (0.2324 – 0.2291) and N(d2) is given as 0.5 + 0.0753 + 0.3 x (0.0793 – 0.0753). Can you please explain what the 0.7 and 0.3 are and why it is multiplied as such? Also, why the $2.98m is deducted from $9.53m?Thanks
May 15, 2014 at 4:48 pm #168928The reason for the 0.7 and 0.3 is (using d1 to explain) – d1 is 0.6170. The tables only go to two decimal places and so the examiner has apportioned between the figure from the tables for 0.61 and the figure for 0.62 (its the figure for 0.61 + 70% of the difference between the two).
You don’t really need to do this – just using the nearest one is good enough in the exam.The NPV of the project (ignoring the option to delay) is -2.98
The option to delay is worth 9.53 and so makes the project ‘better’. So the net result with the option today is -2.98 + 9.53
May 16, 2014 at 12:58 pm #169040That was helpful! thanks 🙂 cheers!
May 16, 2014 at 1:28 pm #169046You’re welcome 🙂
- AuthorPosts
- You must be logged in to reply to this topic.