How Do You Find 'a' for P(-a ≤ X ≤ a) = 0.95?

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Discussion Overview

The discussion revolves around finding a value 'a' such that P(-a ≤ X ≤ a) = 0.95 for a random variable X with a specified probability density function (pdf). Participants explore the relationship between the cumulative distribution function (cdf) and the probability expression, delving into the integration limits and definitions of probability in continuous distributions.

Discussion Character

  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant states they found the pdf and cdf for the random variable X, with the pdf given as f(x) = k(1-x²) and k calculated to be 3/4.
  • Another participant proposes using the relationship 2F(a) - 1 = 0.95 to find 'a', but indicates that this approach did not yield the correct result.
  • A participant clarifies the definition of probability in a continuous distribution and suggests writing an integral expression involving 'a', the pdf, and 0.95.
  • There is a discussion about the correct formulation of P(-a < X < a) in terms of the cdf, with one participant asserting that P(-a < X < a) can be expressed as F(a) - F(-a) = 0.95.
  • Another participant questions whether their understanding of the relationship between the cdf and the probabilities is correct, specifically regarding the calculations involving F(a) and F(-a).

Areas of Agreement / Disagreement

Participants express uncertainty about the correct approach to find 'a', with some agreeing on the use of the cdf but differing on the interpretation and application of the probabilities involved. The discussion remains unresolved regarding the correct method to calculate 'a'.

Contextual Notes

Participants have not fully resolved the mathematical steps necessary to derive 'a', and there are indications of missing assumptions regarding the integration limits and definitions of the probabilities involved.

MiamiThrice
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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?
 
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MiamiThrice said:
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?
 
MiamiThrice said:
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?
 

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