Hi all,(adsbygoogle = window.adsbygoogle || []).push({});

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-x^{2}), where -1<x<1, I found k to be 3/4.

And I found the c.d.f F(x) = 3/4 * (x - x^{3}/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?

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# Continuous Random Variables

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