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P(x=mean) of normal PDF with low sigma - not allowed?

  1. Jan 17, 2013 #1
    P(x=mean) of normal PDF with low sigma -- not allowed?

    About the normal probability distribution: with formula
    P(X=x) = (1/(σ*sqrt(2π))*exp(-(x-μ)2/2σ2), what happens when you look at P(X=μ) if σ<(1/sqrt(2π))? You get P(X=μ)>1, an absurdity. What is going on?

    Second question is one about intuition: suppose μ=0, then why would a scale change of the horizontal axis (say by changing units from meters to kilometers) , which would also change σ, affect the probability of the mean, which it would by the formula?
     
  2. jcsd
  3. Jan 17, 2013 #2
    Re: P(x=mean) of normal PDF with low sigma -- not allowed?

    the probability is 0 since what you looked at is a probability density p(X in x,x+dx) equals f(x)dx, where f is your gaussian function
     
  4. Jan 17, 2013 #3

    HallsofIvy

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    Re: P(x=mean) of normal PDF with low sigma -- not allowed?

    For any continuous PDF, the probability that x is equal to any specific value, rather than in a given interval or set, is 0.
     
  5. Jan 17, 2013 #4
    Re: P(x=mean) of normal PDF with low sigma -- not allowed?

    Thanks, jk22 and HallsofIvy. My reaction upon reading your replies and thinking for a couple of seconds was, "Of course. Stupid of me." So thanks for that destruction of the mental block.
     
  6. Jan 17, 2013 #5

    CompuChip

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    Re: P(x=mean) of normal PDF with low sigma -- not allowed?

    What you can calculate is
    [tex]P(X \le \mu) = \int_{-\infty}^{\mu} p(x) \, dx[/tex]
    where p(x) the PDF. Of course, this will give you 1/2 independent of the mean or standard deviation.
     
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