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Finding μ and σ from a normal cumulative distribution function

  1. Aug 2, 2016 #1
    1. The problem statement, all variables and given/known data
    The relationship between the expected value and the variance for a particular normal CDF is known to follow the rule ##E(X)=arcsin(ln(Var(X)))##. Given that ##Pr(0.32<Z<2.32)=Pr(12.9<X<74.275)##, determine the possible values of the mean and the standard deviation correct to 4 decimal places.

    2. Relevant equations
    ##Z=\frac {X-μ} {σ}##
    ##Var(X)=σ^2##

    3. The attempt at a solution
    When expressed in the form ##normCDf(lower limit, upper limit, σ, μ)##, ##Pr(0.32<Z<2.32)=Pr(12.9<X<74.275)## becomes ##normCDf(0.32, 2.32, 1, 0)=normCDf(12.9, 74.275, σ, μ)##. After also taking into account ##E(X)=arcsin(ln(Var(X)))## and ##Var(X)=σ^2##, we get ##normCDf(0.32, 2.32, 1, 0)=normCDf(12.9, 74.275, σ, arcsin(ln(σ^2)))##. I attempted to solve this last equation on my calculator, but my calculator simply will not output a numerical answer. I'm not sure what to do from here, can anyone help?

    Thank you for your time.
     
  2. jcsd
  3. Aug 2, 2016 #2

    haruspex

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    I put all this into a spreadsheet, using the unknown mean as the variable, and played around with it until the probabilities matched to several significant figures. Only took a few minutes.
     
  4. Aug 2, 2016 #3
    I'm not familiar with how to use spreadsheets with probabilities. I doubt I am expected to do that though because we've never used spreadsheets in class.
     
  5. Aug 2, 2016 #4

    haruspex

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    Ok.
    If you plug in trial values for mu and compute Pr(12.9<X<74.275) for each on your calculator, it's the same, just a bit slower.
    It will be a lot faster if you can pick a good starting point. A promising guess here would be to consider the X points 12.9 and 74.275 as being a direct mapping (by shifting the mean and scaling by variance) of the Z points 0.32 and 2.32. Indeed, it gives mu to within a couple of percent.
     
  6. Aug 2, 2016 #5
    Ah thank you that helps a lot. I also just worked out if you rearrange ##μ=arcsin(ln(σ^2))## for σ, then you get ##σ=e^\frac {sin(μ)} {2}##. Then when I solve ##normCDf(0.32,2.32,1,0)=normCDf(12.9,74.275,e^\frac {sin(μ)} {2},μ## my calculator actually gives me the values for μ; 12.555 and 74.510.

    Anyway, thanks for your help :)
     
  7. Aug 3, 2016 #6

    haruspex

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    The reason your calculator couldn't handle the arcsin form is that the arcsin function is defined only to return values in the range (−π/2, π/2]. mu here is outside that range. Switching to the sin form solved that.
     
  8. Aug 3, 2016 #7
    Okay that's good to know. I'll remember that for the future. Thanks again for your help :)
     
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