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More Distributions

  1. Jan 4, 2005 #1
    How would I solve for a on a question like this: P(0 < z < a)=0.2 ?

    I know that for a question like P(z < a)= 0.85 I would find the inverse-norm of 0.85 to solve for a. I've tried the same thing for the first question, but of course it doesn't work and I'm out of ideas as to how else I should try and solve it. :yuck: Could someone please help? o:)
  2. jcsd
  3. Jan 5, 2005 #2


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    I assume you are talking about the normal distribution. How is your table of normal distribution set up? Some of them, for example, the one at
    http://people.hofstra.edu/faculty/Stefan_Waner/RealWorld/normaltable.html [Broken]
    give the value under the curve from 0 to a which is exactly what you want.

    If your table gives from negative infinity to a, then, since the distribution is symmetric about 0, the value from negative infinity to 0 is 1/2 and you just have to subtract 1/2 from the table value.

    In general, with either kind of table, to find P(a< z< b), look up the values for a and b separately and subtractP: P(a< z< b)= P(z< b)- P(z< a).
    Last edited by a moderator: May 1, 2017
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