Continuous Random Variables

[MiamiThrice]

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?

 

[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?

 

[MiamiThrice]

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?

 

[blue_raver22]

Hi there,

It looks like you are working with continuous random variables and trying to find a specific value for a given probability. Your approach of using the c.d.f. to solve for a is correct, but it seems like there may be a mistake in your calculation. It's difficult to provide specific help without seeing your work, but here are some general steps you can follow to solve for a:

1. Start with the c.d.f. you found: F(x) = 3/4 * (x - x^3/3 + 2/3)

2. Plug in the limits of integration (-a and a) and set the resulting equation equal to 0.95. This will give you an equation in terms of a.

3. Use algebra to solve for a. You may need to use a calculator or numerical methods to solve for a.

Make sure to check your work and make sure it makes sense. It's always a good idea to double check your calculations to make sure you didn't make any mistakes. If you are still having trouble, I suggest reaching out to your professor or a tutor for additional guidance.

Hope this helps!