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Math Help - Normally Distributed Random Variable-Plzzzz Help

  1. #1
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    Red face Normally Distributed Random Variable-Plzzzz Help

    A Normally distributed random variable with mean [mu] has a probability density function given by

    [gamma]/2 Pi *sigma exp(-gamma^2/sigma*x-[mu]^2/2).

    Its standard deviation is given by:



    *Sorry I'm still learning how to use Latex
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  2. #2
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    Quote Originally Posted by KayPee View Post
    A Normally distributed random variable with mean [mu] has a probability density function given by

    [gamma]/2 Pi *sigma exp(-gamma^2/sigma*x-[mu]^2/2).

    Its standard deviation is given by:



    *Sorry I'm still learning how to use Latex
    Sorry but I can't make sense of the expression you have posted for the pdf.

    If it's a normal distribution then it has the usual form found here: Normal distribution - Wikipedia, the free encyclopedia

    So unless you can re-post a clearer expression I suggest you read the link, fit your pdf to the normal pdf given in the link and then read off what the variance and hence standard deviation is.
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  3. #3
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    Edited Question

    Quote Originally Posted by mr fantastic View Post
    Sorry but I can't make sense of the expression you have posted for the pdf.

    If it's a normal distribution then it has the usual form found here: Normal distribution - Wikipedia, the free encyclopedia

    So unless you can re-post a clearer expression I suggest you read the link, fit your pdf to the normal pdf given in the link and then read off what the variance and hence standard deviation is.
    γ/Sqrt(2πσ)exp(-γ/σ*(x-μ)^2/2))
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  4. #4
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    Quote Originally Posted by KayPee View Post
    γ/Sqrt(2πσ)exp(-γ/σ*(x-μ)^2/2))
    Is it \frac{y}{\sqrt{2 \pi \sigma}} e^{-\frac{y (x - \mu)^2}{2 \sigma}} ?

    That can't be right. It has to be either \frac{y}{\sqrt{2 \pi \sigma}} e^{-\frac{y^2 (x - \mu)^2}{2 \sigma}} or \frac{\sqrt{y}}{\sqrt{2 \pi \sigma}} e^{-\frac{y (x - \mu)^2}{2 \sigma}}

    Assuming the former expression, then I would have thought that the comparison with the pdf in the link I gave you is straightforward:


    \frac{y}{\sqrt{2 \pi \sigma}} = \frac{1}{\sigma_X \sqrt{2 \pi}} therefore \sigma_X = \, ....

    If it's the latter expression, then you do a similar thing.
    Last edited by mr fantastic; January 10th 2009 at 03:05 PM.
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