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Thread: Is there are more advanced way? (SORTED!)

  1. #1

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    Fanatic Member Matt_T_hat's Avatar
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    Is there are more advanced way? (SORTED!)

    When answering this post please remember that I dropped Advanced Level Maths and although good with the subject am baddly out of practice, I will follow vb(ish), Spreadsheetish, SQL(ish) examples much better if there is an option.

    Now for the question: I have a database that tracks the sales of items. If MR25T sold for £35.50 in EX+ condition what would it have sold for in mint? Also what does a Mint MR25T go for nowdays? This is what the database answers.

    To do this we have constructed a value chart that looks like this:

    Mint: 100%
    NrMint: 95%
    Ex+: 90%

    etc. The condition names are standard trader discriptions and have fairly rigid boundries to them. The percentages are also fairly standard as a "rule of thumb" and is from a well respected value chart divised by experts in this particular subject.

    So now we know that MR25T @ £35.50 is only 90% of the possible price for a Mint condition copy.

    We use this as a way of working out a base figure to quote values on.

    Dim BV as BaseValue
    Set X = Any Given Item
    Let aP = Average(Percentages(X))
    Let aS = Average(SalePrice(X))

    Access stores percentages as decimals of 1.

    BV = (aS / (aP * 100) )*100

    now given the limits of an access database query can any maths gurus come up with a more revieling method of creating a value that we can work on.

    I figure we really should take into account a three month moving average (however we might be able to do that is another story) and at least some standard deviation.

    There is a clear case here for useing the data to mine information about market trends and so forth in the collectables market but now we have only to figure out how.

    any ideas?
    Last edited by Matt_T_hat; Oct 29th, 2003 at 05:13 AM.
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  2. #2
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    Statistics

    I do believe your question is more in the realm of Financial and Statistical Analysis, then it is for this Maths Forum. However, if you give the data for an example, and post a straight, understandable qustion in this Maths Forum, then, some mathematically-inclined person will be able to offer a mathematial solution.

  3. #3
    So Unbanned DiGiTaIErRoR's Avatar
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    Well...

    BV = (aS / (aP * 100) )*100

    simplifies to:

    BV = aS / aP

  4. #4
    transcendental analytic kedaman's Avatar
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    the average is the sum of the elements divided by the number of elements
    the standard deviation is the root of
    [(the sum of the squared differences between each element and their average) divided by the number of elements ]
    sometimes the number of elements-1, for more statistical accuracy.
    I just know the basics in statistic, so anything beyond the formulas and some statistical tests are above me, but i can help with what i know.
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