Uniform Variate

**zorro** · December 2nd 2009, 04:17 PM

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

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

**mr fantastic** · December 2nd 2009, 08:58 PM

Originally Posted by zorro

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

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

What have you tried? Where are you stuck?

**matheagle** · December 2nd 2009, 09:52 PM

use the fact that those are complementary events.

**zorro** · December 4th 2009, 02:40 PM

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

**mr fantastic** · December 5th 2009, 01:21 AM

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:

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

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

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

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

Hence solve for k.

**zorro** · December 8th 2009, 08:08 PM

Originally Posted by mr fantastic

Draw a picture and use basic geometry.

Alternatively:

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

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

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

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

Hence solve for k.

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

**CaptainBlack** · December 8th 2009, 11:05 PM

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

**zorro** · December 9th 2009, 12:11 AM

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 .....

**CaptainBlack** · December 9th 2009, 03:02 AM

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

**mr fantastic** · December 9th 2009, 03:15 AM

Originally Posted by zorro

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

Originally Posted by Mr Fantastic

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

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

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

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

etc.

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