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Finding unknown coefficients of a quadratic equation describing a PDF

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data

    I have a probability distribution function (PDF) going fromi 0 to a on the y-axis and 0 to 30 on the x-axis. The function describing the PDF is the quadratic equation f(x) = a+bx+cx^2
    I have to find a, b, and c.


    2. Relevant equations



    3. The attempt at a solution

    I know that three unknown require 3 equations.

    1. By definition the area under a PDF = 1 so one equation is intergrating f(x) from 0 to 30 and have that equal to 1. When that's worked out you get 30a+450b+9000C=1

    2. I know that when x = 30 f(x) = 0. Therefore we get a+30b+900 = 0

    but what is the third equation? I know that when x = 0 f(x) = a but that's only the y-intercept

    Any ideas?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Feb 8, 2009 #2

    gabbagabbahey

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    You expect [itex]a[/itex] to be the maximum value of [itex]f(x)[/itex] correct? doesn't that mean that the vertex of your quadratic is at [itex](0,a)[/itex]?...what does that make [itex]b[/itex]?:wink:
     
  4. Feb 8, 2009 #3
    Attached is a picture of the function. We have to determine a, b, c as I described above. I know I have two of the equations write. Based on the schematic can we be certain that the vertex is at (0,a)?

    If it is and if the equation is written in the form y = a+bx+cx^2 then I can come up with a third equation.

    a = -b/2c (what do you think?)
     

    Attached Files:

  5. Feb 8, 2009 #4

    gabbagabbahey

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    I won't be able to see the picture you've attached until it is approved by site admin (which sometime takes days!)

    Try uploading it to imageshack.us instead and the just posting the link to it.
     
  6. Feb 8, 2009 #5
  7. Feb 8, 2009 #6

    gabbagabbahey

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    Yup, that link works fine:smile:

    ....Was that sketch given along with the problem statement, or was that just how you interpreted the question?

    I would have assumed that it would look more like this:
    pic

    In which case, the fact that the maximum value is [itex]a[/itex] and it occurs at [itex]x=0[/itex], tells you that is where the vertex is.
     
    Last edited: Feb 8, 2009
  8. Feb 8, 2009 #7
    I think I figured it out... as I tried the vertex but got lost in calculations. So I just thought some more and I think I answered it:

    so we have f(x) = a+bx+cx^2 (where x goes from 0 to 30) and are asked to determine a, b, and c. The key here is we know that the function is a probability distribution function which has key characteristics:

    1. The area under the curve is 1 (this gives you the first equation by integration)
    2. The value of the pdf is 0 at x=30 (this is just a simple subst of x into the main equation)
    3. This is the one I couldn't get. But I recalled that when you differentiate a continuous probability function you get the cumulative distribution function which always reaches 1 when the at maximum x value (30). Therefore we differentiate

    F(x) (capital F denotes cdf, small f denotes pdf) = b + 2cx
    F(30) = 1 = b+60x

    Then I found a 3-unknown, 3 equation calculator on the web and got a = 15.1 b= -2.01 and c=0.050

    Thoughts?
     
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