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VaR

Forums › Ask ACCA Tutor Forums › Ask the Tutor ACCA AFM Exams › VaR

  • This topic has 11 replies, 5 voices, and was last updated 12 years ago by John Moffat.
Viewing 12 posts - 1 through 12 (of 12 total)
  • Author
    Posts
  • October 31, 2012 at 12:07 pm #54954
    harripool
    Member
    • Topics: 13
    • Replies: 33
    • β˜†

    Hi John,

    Can you explain something to me please?

    A 95% confidence level produces a 1.65 normal distribution value.
    I can’t understand how this value is produced. I know this means a 5% probability level but not how I extract the -1.65 value from the distribution table!!!

    Help please?????

    October 31, 2012 at 3:28 pm #106240
    muhaimin
    Member
    • Topics: 1
    • Replies: 9
    • β˜†

    Hi harri!

    95% means 0.95, we must have arrived at 0.95 by adding “some value” to 0.5 (since d1 > 0). If we calculate that “value” by subtracting 0.5 from 0.95, we will have 0.45. Now just go to the Normal distribution table and see the corresponding value of 0.45, you’ll get 1.65.

    Hope john likes it πŸ˜‰

    October 31, 2012 at 6:28 pm #106241
    harripool
    Member
    • Topics: 13
    • Replies: 33
    • β˜†

    Hi muhaimin,

    Thanks for your response but if I look up 0.45 I get 0.1736?

    I’m confused. Am I not using the table correctly?

    October 31, 2012 at 6:38 pm #106242
    harripool
    Member
    • Topics: 13
    • Replies: 33
    • β˜†

    Do I find 0.4505 and that is equal to 1.65?
    I think so!

    November 1, 2012 at 7:11 am #106243
    muhaimin
    Member
    • Topics: 1
    • Replies: 9
    • β˜†

    yes harri, closer to 0.45 is 0.4505 and 0.4495, so it can either be 1.65 or 1.64.

    Since the examiner rounds it to the lower side so do we.

    November 1, 2012 at 2:56 pm #106244
    harripool
    Member
    • Topics: 13
    • Replies: 33
    • β˜†

    Thanks Muhaimin, you helped solve my headache!
    Much appreciated

    November 1, 2012 at 3:55 pm #106245
    muhaimin
    Member
    • Topics: 1
    • Replies: 9
    • β˜†

    most welcome!

    November 1, 2012 at 5:12 pm #106246
    John Moffat
    Keymaster
    • Topics: 57
    • Replies: 54699
    • β˜†β˜†β˜†β˜†β˜†

    Good answer Muhalmin.
    (Some of the exam/revision kits try to be clever and start approximating the tables, but there is no need to in the exam!)

    November 4, 2012 at 2:32 pm #106247
    muhaimin
    Member
    • Topics: 1
    • Replies: 9
    • β˜†

    Thanks john, I would be pleased to help the other students as well if i get the opportunity to do so.

    I would appreciate if you see my post (a question from pilot paper) which is still unanswered. I know your answer would be “Time constraints! time constraints! ;)” ….

    Dear Harripool! if you can help me out? πŸ™‚
    Thanks

    November 10, 2012 at 9:27 am #106248
    sukaina1786
    Member
    • Topics: 5
    • Replies: 12
    • β˜†

    at 90% confidence level, is the value from the table 1,29???

    November 10, 2012 at 9:36 am #106249
    dazhong0703
    Member
    • Topics: 44
    • Replies: 130
    • β˜†β˜†

    You should find 0.9-0.5=0.40 (cell), then you roughly get 1.285.

    November 10, 2012 at 9:29 pm #106250
    John Moffat
    Keymaster
    • Topics: 57
    • Replies: 54699
    • β˜†β˜†β˜†β˜†β˜†

    correct πŸ™‚

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