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Lognormal Distribution Help

  1. Mar 5, 2013 #1
    1. The problem statement, all variables and given/known data


    Q1) It has been observed that contamination level of water [measured in suitable units] in a given waterbody has lognormal distribution with mean = 23.58 units and variance = 3.85 sq units.

    What % of samples would indicate contamination level between 20.65 units and 25.85 units ?


    2. Relevant equations

    Using the Z table and P((ln(x)-mean)/SD)


    3. The attempt at a solution

    P((ln(20.65)-23.58)/sqrt(3.85))<=Z<=((ln(20.65)-23.58)/sqrt(3.85)) = phi(-10.36)-phi(-10.47)

    Values are way too large... whats going on.
     
  2. jcsd
  3. Mar 5, 2013 #2

    Ray Vickson

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    You are using the wrong mean and variance: 25.58 and 3.85 are the mean and variance of X, but you need the mean and variance of Y = ln(X). You will need to solve a set of coupled nonlinear equations to find the mean and variance of Y.
     
  4. Mar 6, 2013 #3
    ok i tried these equations before but they didnt seem to pan out...

    where u=mean & o=variance of X

    U=e^(u+(o^2)/2) = e^(23.58+3.85/2) = 1.1931x10^11

    O=(e^(o^2)-1)e^(2u+o^2)=(e^3.85 -1)e^(2(23.58 + 3.85)) = 6.5472x10^23

    ?????
     
  5. Mar 6, 2013 #4

    haruspex

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    I think you're driving those equations backwards. The given numbers are U and O, not μ and σ.
     
  6. Mar 6, 2013 #5

    Ray Vickson

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    As 'haruspex' has noted, you need to find the solution of
    [tex]e^{m + \frac{1}{2}v} = 23.58, \: (e^v -1) e^{2m + v} = 3.85.[/tex]
     
  7. Mar 6, 2013 #6
    Ahhh... ok so hers what i got now...

    (e^v -1)e^(2m+v) = 3.85

    e^(2(m+v))-e^(2m+v) = 3.85

    2m + 2v - 2m + v = ln 3.85

    v = (ln3.85)/3 = 0.4494 = o



    e^(m+v/2) = 23.58

    m+v/2 = ln23.58

    m = ln23.58 - v/2 = 2.9357 = u

    I made a mistake somewhere cuz i got a z value of 7 something.
     
  8. Mar 6, 2013 #7

    haruspex

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    ln(a-b) is not ln(a)-ln(b).

    You have ##e^{m + \frac{1}{2}v} = 23.58, \: (e^v -1) e^{2m + v} = 3.85.##
    What do you get if you square ##e^{m + \frac{1}{2}v}##?
     
  9. Mar 6, 2013 #8
    uh huhhhhh... didnt see that


    so i squared that and cancel it out and get this...

    e^v - 1 = 3.85/23.58^2, v = .006900

    e^(2m+v) = 23.58^2... 2m+v = ln(23.58^2) = 6.3208 = 2m+.006900, m = 3.1569

    soo my Z values are followed with x values 20.65 & 25.85

    (ln(x)-.0069)/3.1569 = phi(1.0280) - phi(0.9569) = .8485-.8315 = 0.017

    See any issues???
     
  10. Mar 6, 2013 #9

    Ray Vickson

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    Yes: read over what you have done and see where you went wrong.

    Your m and v are OK, but then you went astray. The answer should be somewhere between 75% and 90% probability, but I won't say the exact figure.
     
  11. Mar 11, 2013 #10
    ok it looks like i had the mean and variance flip flopped but still not getting it correct

    so using (ln(x)-mean)/variance = (ln(20.65)-3.1569)/.0069 = -18.72??



    also in the problem it says the variance is 3.85 sq units, so i have to sqrt that to find the actual variance, plug it into the previous equations, which then works out to a variance of .0035???
     
  12. Mar 11, 2013 #11

    haruspex

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    (ln(x)-mean)/SD; SD is sqrt of variance.
     
  13. Mar 11, 2013 #12
    ahhhhh ok i think i got it now.

    got 81.55%... thank you for your help
     
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