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Lognormal distribution

  1. Apr 28, 2008 #1
    A labour force's annual incomes are lognormally distributed. If the labour force is arranged in order of decreasing annual incomes and the top 10% earns 37% of the total annual incomes, what proportion of the total annual income does the bottom 10% earn?

    Kindly help on this one. It looks simple until you start solving and you realise that there is no mean or variance!
    Also take note that all the information is provided. There is nothing missing in this question,
  2. jcsd
  3. Apr 28, 2008 #2
    Welcome to PF

    What is the density of the log normal distribution and how would you rephrase the given information in terms of this density?
  4. Mar 22, 2011 #3

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    proportion bottom 10% = [TEX]1 - \Phi ( \Phi^{-1}(0.9) + \Phi^{-1}(0.37) - \Phi^{-1}(0.1) )[/TEX] = 1.28%
    Last edited: Mar 22, 2011
  5. Mar 28, 2011 #4
    I gave the exact solution and was given a big X. Could my professor be wrong?
  6. Mar 28, 2011 #5

    I like Serena

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    What I gave is my 2 cents, which I derived using the formulas given on wikipedia.
    If you came out to the same answer that should be enough reason to go talk to your professor I guess.
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