# Uniform Variate

• Dec 2nd 2009, 04:17 PM
zorro
Uniform Variate
Question :
Let X be a Uniform Variate defined on (-k,k). Determine k so that
$\displaystyle P(|X| < 2) = P(|X| > 2)$

Please could please tell me what formula should i use to solve this question?
• Dec 2nd 2009, 08:58 PM
mr fantastic
Quote:

Originally Posted by zorro
Question :
Let X be a Uniform Variate defined on (-k,k). Determine k so that
$\displaystyle P(|X| < 2) = P(|X| > 2)$

Please could please tell me what formula should i use to solve this question?

What have you tried? Where are you stuck?
• Dec 2nd 2009, 09:52 PM
matheagle
use the fact that those are complementary events.
• Dec 4th 2009, 02:40 PM
zorro
I dont know where to start
Quote:

Originally Posted by mr fantastic
What have you tried? Where are you stuck?

Mr fantastic i dont know where to start ....No clue at all,,,,If possible coldu please guide me
• Dec 5th 2009, 01:21 AM
mr fantastic
Quote:

Originally Posted by zorro
Mr fantastic i dont know where to start ....No clue at all,,,,If possible coldu please guide me

Draw a picture and use basic geometry.

Alternatively:

$\displaystyle \Pr(|X| < 2) = \Pr(|X| > 2)$

$\displaystyle \Rightarrow \Pr(|X| < 2) = 1 - \Pr(|X| < 2)$ (using matheagle's suggestion)

$\displaystyle \Rightarrow \Pr(|X| < 2) = \frac{1}{2}$.

And you really should be able to use either the pdf or simple geometry to calculate $\displaystyle \Pr(|X| < 2)$.

Hence solve for k.
• Dec 8th 2009, 08:08 PM
zorro
Quote:

Originally Posted by mr fantastic
Draw a picture and use basic geometry.

Alternatively:

$\displaystyle \Pr(|X| < 2) = \Pr(|X| > 2)$

$\displaystyle \Rightarrow \Pr(|X| < 2) = 1 - \Pr(|X| < 2)$ (using matheagle's suggestion)

$\displaystyle \Rightarrow \Pr(|X| < 2) = \frac{1}{2}$.

And you really should be able to use either the pdf or simple geometry to calculate $\displaystyle \Pr(|X| < 2)$.

Hence solve for k.

Mr fantastic could u please tell me how have u put in the value of................. $\displaystyle 1- \Pr(|X| < 2) = \frac{1}{2}$
• Dec 8th 2009, 11:05 PM
CaptainBlack
Quote:

Originally Posted by zorro
Mr fantastic i dont know where to start ....No clue at all,,,,If possible coldu please guide me

I think it is time to ask: Why is it you seem to have no clue where to start with so many of these question? That seems to be the real problem here.

You have at least some of: notes, a text and Google at your disposal.

CB
• Dec 9th 2009, 12:11 AM
zorro
I am sorry for these annoying quetions
Quote:

Originally Posted by CaptainBlack
I think it is time to ask: Why is it you seem to have no clue where to start with so many of these question? That seems to be the real problem here.

You have at least some of: notes, a text and Google at your disposal.

CB

I am sorry for annoying u .....
• Dec 9th 2009, 03:02 AM
CaptainBlack
Quote:

Originally Posted by zorro
I am sorry for annoying u .....

You are not annoying me, but it is best to address the real problem which we are not at present.

Cb
• Dec 9th 2009, 03:15 AM
mr fantastic
Quote:

Originally Posted by zorro
Mr fantastic could u please tell me how have u put in the value of................. $\displaystyle 1- \Pr(|X| < 2) = \frac{1}{2}$

Quote:

Originally Posted by Mr Fantastic

$\displaystyle \Rightarrow \Pr(|X| < 2) = 1 - \Pr(|X| < 2)$

Add $\displaystyle \Pr(|X| < 2)$ to both sides:

$\displaystyle \Rightarrow 2 \Pr(|X| < 2) = 1$

$\displaystyle \Rightarrow \Pr(|X| < 2) = \frac{1}{2}$

etc.