# Continuous Random Variables

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:

chiro
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(c) - 1 =0.95 and just calculate c but it didn't work.

Any help?

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?

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?

chiro
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?

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?