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**genericname****Member**- Registered: 2012-05-16
- Posts: 52

*For the exponential density f(x) = 3e^-3x, compute the actual probability for k=3 and k=4.*

Expected value = 1/3. The variance is 1/9 and the standard deviation is 1/3 so for k=3 I got:

(1/3) - 3*(1/3) < X < (1/3) + 3*(1/3) (Wasn't sure about this, all I did was sub 2 with 3 from the problem where k=2.)

= -2/3 < X < 4/3

So to get the probability, you integrate f(x) from -2/3 to 4/3? Would this be correct?

*Last edited by genericname (2012-11-04 06:27:59)*

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,453

hi genericname

I'm not fully following this.

For a probability density function the integral over all popssible values must be 1.

If I integrate from - ∞ to + ∞ that doesn't happen, so I'm assuming it should be from 0 to + ∞. then it works.

That fits with

http://en.wikipedia.org/wiki/Exponential_distribution

and my calculations for mean and variance are 1/3 and 1/9 as you have. So far so good.

But k ?

According to

http://en.wikipedia.org/wiki/Chebyshev%27s_inequality

k is an value indicating how many 'standard deviations from the mean. Your calculations for what x values to use follow ok.

But if x > 0 for the pdf what are we to do with x = -2/3 ?

The integral up to zero will be zero, so I'm thinking you should integrate from 0 to 4/3.

Hope that's correct.

Bob

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi genericname;

Isn't Chebychev's an inequality that bounds an unknown distribution? What does it have to do with this question since you know the distribution? And why are you integrating?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**genericname****Member**- Registered: 2012-05-16
- Posts: 52

The example for when k=2 said that you have to integrate f(x) from something to something to compute p( _ < X < _ ) so I assumed that I had to. It was listed as an Exponential distribution problem and it said to do the same with k= 3 and 4.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

One more question. If you want the probability and this is a Chebychev problem why not just use the inequality?

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**genericname****Member**- Registered: 2012-05-16
- Posts: 52

The next part of the problem wanted us to compare the answers to the predictions of Chebyshev's theorem.

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**bobbym****bumpkin**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 109,606

Hi;

I understand now. Thanks.

**In mathematics, you don't understand things. You just get used to them.****If it ain't broke, fix it until it is.**** Always satisfy the Prime Directive of getting the right answer above all else.**

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**bob bundy****Administrator**- Registered: 2010-06-20
- Posts: 8,453

The question does ask for the 'actual' probability.

Maybe the next lesson is to evaluate how much quicker it would have been to have used the inequality.

(that's me trying to get inside the head of the teacher. Dangerous activity that! )

Bob

EDIT: Hey. It seems reality has got ahead of my thoughts.

Children are not defined by school ...........The Fonz

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Sometimes I deliberately make mistakes, just to test you! …………….Bob Bundy

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