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Two supposedly simple Standard Normal questions

  1. Jun 20, 2012 #1
    1) If X~N(μ,σ2, find the value of c in terms of σ such that P(μ-c ≤ X ≤ μ + c) = 0.95


    Attempt: Ok so I have a feeling I will eventually need to reference a standard normal table, but I tried to standardize the rv first:

    P (-1 ≤ (X-μ)/c ≤ 1 ) = 0.95

    Now here's where I'm stuck, what trait of normal distributions am I missing to apply here?


    2) If X ~ N(0,σ2), find the density of Y = |X|.

    Attempt: FY(y) = P(Y≤y)
    = P( |X| ≤ y)

    How do I handle the absolute value bars? Would I have to have 2 cases and if so how do I brign those together in order to have a proper change of variables?


    Thanks
     
  2. jcsd
  3. Jun 20, 2012 #2

    Ray Vickson

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    You handle the absolute value by drawing a line (for X) then shading in that portion of the line that satisfies |x| <= y. What region do you get?

    RGV
     
  4. Jun 20, 2012 #3


    If I'm understanding you right, the portion that is shaded in would be everything up to but not including y. In my head I have a picture of the normal distribution graph usually how it's shown in books.
     
  5. Jun 20, 2012 #4

    Ray Vickson

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    Why wouldn't include y: don't we have |y|<=y for y>0?

    If y = 2, would x = -7 be in the set |x|<=y? Would x → -∞ satisfy |x| <= y?

    For now forget about your mental picture of the normal distribution, and just concentrate on getting the region correct. Once you have done that, go back to thinking about the normal distribution.

    RGV
     
  6. Jun 20, 2012 #5



    Ok, I see what your getting at, so I drew the graph for the absolute value of |x| and everything underneath the graph from both sides would be in the region, I just realized too though that |x| ≤ y <==> -y≤ X ≤ y. But back to the graph idea, so with me drawing that graph how does that relate to the normal distribution or any distribution for that matter?


    So the set would be all values of X which are equal to or less in absolute value to y. By FTC I cold switch it into:

    FX(y) - FX(-y)

    then differentiating to get the densities:

    fX(y) + fX(-y)

    Then put these into the normal and I would have the density?
     
  7. Jun 20, 2012 #6

    Ray Vickson

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    Yes.

    RGV
     
  8. Jun 20, 2012 #7
    Awesome. Thanks. I had a quick question in terms of the uniform distribution. So I had to do a similar change of variable process and I arrived at fX(y). Now say I have been given the uniform density, where would I put my "y" value into to obtain the density in terms of y's since the PDF of a uniform is 1/(b-a)?
     
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