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Difference between Z notation and sigma

  1. May 30, 2015 #1
    1. The problem statement, all variables and given/known data
    What is the difference between this statement being expressed in z-notation and the notation shown? What are the benefits of the two and how do you do the conversion?
    Nevermind about answering the question itself.

    if X ~N(mu,sigma) show that P(|X-mu|<= 0.675*sigma) = 0.5

    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. May 30, 2015 #2

    LCKurtz

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    The statement that$$
    P(|X-\mu|\le .675\sigma)=.5$$is completely equivalent to$$
    P\left(\frac{|X-\mu|}{\sigma}\le .675\right)=.5$$which is equivalent to$$
    P\left(-.675\le \frac{X-\mu}{\sigma}\le .675\right)=.5$$Is that what you are asking?
     
  4. May 30, 2015 #3
    is that due to symmetry?
     
  5. May 30, 2015 #4

    LCKurtz

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    No, it is just algebra. To get the second statement you just divide both sides of the inequality by the positive ##\sigma## and the last form just uses properties of absolute values, specifically that ##|x|\le a## is the same thing as ##-a\le x \le a##.
     
  6. May 31, 2015 #5

    pasmith

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    The only normal distribution whose cdf and inverse cdf you will find tabulated is N(0,1). If [itex]X \sim N(\mu, \sigma^2)[/itex] then to use tables you have instead to work with [itex]Z = (X - \mu)/\sigma \sim N(0,1)[/itex].
     
  7. Jun 1, 2015 #6
    what is function N()?
     
  8. Jun 1, 2015 #7

    Ray Vickson

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    It is exactly the same N( ) that you yourself wrote in post #1. It is actually not a function, but, rather, a name (short for "Normal" or "Normal distribution").
     
  9. Jun 1, 2015 #8

    WWGD

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    True, but doesnt the symmetry in the algebra follow from the symmetry of the normal distribution?
     
    Last edited: Jun 1, 2015
  10. Jun 1, 2015 #9

    Ray Vickson

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    No:. For a general density we have
    [tex] P(|X-\mu| \leq a) = P(-a \leq X- \mu \leq a) = \int_{\mu-a}^{\mu+a} f(x) \, dx,[/tex]
    whether or not ##f## possesses any symmetry properties.
     
  11. Jun 1, 2015 #10

    vela

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    I don't think so. There's absolutely no information about the normal distribution that goes into the algebraic manipulations LCKurtz showed above. The same algebra would have been used if X obeyed a completely different distribution.
     
  12. Jun 1, 2015 #11

    WWGD

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    What I meant is that the original formula, in the OP is satisfied by the normal, i.e. ##, P((|X- \mu|)/\sigma <0.675) ## reflects the symmetry of the normal. Obviously, ##|x|< a \rightarrow -a<x<a ## follows straightforward from definition of absolute value. But I guess the OP was referring to the derivation ##P(|x- \mu|/\sigma< 0.675 \rightarrow -0.675 \sigma < P(|x-\mu| )< 0.675 \sigma ##, so my bad, I jumped in without reading carefully.
     
  13. Jun 1, 2015 #12

    HallsofIvy

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    It follows from the "symmetry" in the absolute value.
     
  14. Jun 1, 2015 #13

    WWGD

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    Please check my last post, since I corrected myself. I was referring to something different.
     
  15. Jun 3, 2015 #14

    micromass

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    Why would that hold? What does ##\mathbb{P}(|X-\mu)|)## even mean???
     
  16. Jun 3, 2015 #15

    WWGD

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    ##\mathb {P}(|x- \mu|)## doesn't mean anything, what I wrote was ##P(|x-\mu|<\sigma (.675))## , which means ##P( -0.675 \sigma < x- \mu < 0.675\sigma ) ##. Anyway, I jumped in without reading carefully and I wrote something that is pointless here.

    All I was trying to say (replying to no one but my own head) was that , by symmetry of the normal, ##P( x-\mu< \sigma) =P(x-\mu > -\sigma )## , which is not true for all distributions.
     
    Last edited: Jun 3, 2015
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