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Normal distribution

  1. Aug 27, 2008 #1
    A normal distribution can be completely defined by two parameters - the mean and the standard deviation. Given a normal distribution however, say X, how can I use just the mean and the standard deviation to give me conditional expected values for X<=0 and for X>0? Im guessing the distribution can be standardised to obtain a z-statistic.
     
  2. jcsd
  3. Aug 27, 2008 #2

    statdad

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    "Given a normal distribution however, say X" - I assume you mean that the variable [tex] X [/tex] has a normal distribution. Are both [tex] \mu [/tex] and [tex] \sigma [/tex] known?

    "how can I use just the mean and the standard deviation to give me conditional expected values for [tex] X \le 0 [/tex] and for [tex] X>0 [/tex] ?"
    This doesn't make sense to me as it stands. In statistics we take expected values of some function of a random variable - can you elaborate on what it is you seek?
     
  4. Aug 27, 2008 #3

    HallsofIvy

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    If X has normal distribution with mean [itex]\mu[/itex] and standard deviation [itex]\sigma[/itex] then [itex]z= (x- \mu)/\sigma[/itex] has the standard normal distribution. As statdad said, "conditional expected values for X< 0 and X> 0" makes no sense." I might interpret as "suppose X a standard normal distribution, restricted to be larger than 0. What is the the expected value of X?"

    That would be
    [tex]\frac{1}{\sqrt{\pi}}\int_0^\infty x e^{-x^2}dx= \frac{1}{2\sqrt{\pi}}[/tex]
    The general problem, with non-zero mean or standard deviation not 1 would be a much harder integral.
     
    Last edited: Aug 27, 2008
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