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Continuous Random Variables

  1. Dec 12, 2011 #1
    Hi all,

    I was having some troubles with a practise question and thought I'd ask here.

    Given an r.v. X has a pdf of f(x) = k(1-x2), where -1<x<1, I found k to be 3/4.

    And I found the c.d.f F(x) = 3/4 * (x - x3/3 + 2/3)

    Now I have to find a value a such that P(-a <= X <= a) = 0.95.

    I thought it would be 2F(a) - 1 =0.95 and just calculate a but it didn't work.

    Any help?
     
    Last edited: Dec 12, 2011
  2. jcsd
  3. Dec 12, 2011 #2

    chiro

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    Hey MiamiThrice and welcome to the forums.

    By your post you have figured out your pdf and cdf which is good.

    So now you are asked to find a given 'a' for your relation.

    Think about the limits of your integration. What is the definition of probability in a continuous distribution? What is the definition for P(X < a) and how can you define P(-a < X < a) in terms of PDF/CDF?

    In saying the above, can you now write an integral expression involving 'a', the PDF, and 0.95?
     
  4. Dec 12, 2011 #3
    I think this is what you mean:

    P(x<a) = F(a).

    I used that to get F(-a) = 1- F(a).

    So for P(-a<x<a) I thought it would be F(a) - (1 - F(a)) = 2F(a) - 1 = 0.95 ?

    Was I doing it correct?
     
  5. Dec 12, 2011 #4

    chiro

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    Almost correct.

    Your first part was correct with P(X < a) = F(a). But P(X < -a) = F(-a). From this we have P(-a < X < a) = P(X < a) - P(X < -a) = F(a) - F(-a) = 0.95. Do you know where to go from here?
     
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