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Probability Problem

  1. Jan 20, 2010 #1
    Can anyone help me with this?
    If Z ~ N(0,1) determine P(Z^3 > 1)

    I used the formula: x-miu/standard deviation where miu=0 and standard deviation is sq. root of 1.

    I tired:

    P(z^3>1)
    P((x-0/1)^3>1)
    P(x^3>1)
    P(x>1)

    I looked up the area of the standard normal distribution tables and got

    =-3.99, which isn't right

    I also tried

    P(z^3>1)
    P(z>1)

    Looked it up using the tables and got

    =0.15866

    I don't think the question would be that easy though.
     
  2. jcsd
  3. Jan 20, 2010 #2

    statdad

    User Avatar
    Homework Helper

    You must have used the table incorrectly, since -3.99 is not a probability (or you have a typo here).
    If two inequalities are equivalent, they represent events that have the same probability. thus, for example,

    [tex]
    P(3Z+5 > 7) = P(Z > 2/3)
    [/tex]

    since [tex] 3Z+5>7 [/tex] and [tex] Z > 2/3 [/tex] are equivalent inequalities.

    So, what inequality is equivalent to [tex] Z^3 > 1 [/tex]? Perhaps you were on the correct track and it simply seemed too easy.
     
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