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probability and statistics
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Akilah Seecharan
science forum beginner


Joined: 14 Jul 2006
Posts: 1

PostPosted: Fri Jul 14, 2006 6:09 pm    Post subject: probability and statistics Reply with quote

I'm having a problem finding the mean (U) of this distribution function, f(x) = x to the 4, 0<x<1 How would you solve this problem using the formula:
u=[xf(x)dx] from 0<x<1?
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Lynn Kurtz
science forum Guru


Joined: 02 May 2005
Posts: 603

PostPosted: Fri Jul 14, 2006 6:42 pm    Post subject: Re: probability and statistics Reply with quote

On Fri, 14 Jul 2006 14:09:33 EDT, Akilah Seecharan
<akilah16@hotmail.com> wrote:

Quote:
I'm having a problem finding the mean (U) of this distribution function, f(x) = x to the 4, 0<x<1 How would you solve this problem using the formula:
u=[xf(x)dx] from 0<x<1?

Do you mean f is x cumulative distribution function given by this? :

f(x) = 0 for x <= 0
f(x) = x^4 for 0 < x < 1
f(x) = 1 for x > 1

Your suggestion of int[0..1] x f(x) dx suggests you are thinking of
f(x) as a probability density function, which it isn't because the
integral from 0 to 1 of f(x) isn't 1.

Now, *if* f(x) is the cumulative distribution function I propose
above, then its density function would be f'(x) which is 4x^3 on (0,1)
and 0 elsewhere. In that case to calculate the expected value you
would integrate x * 4x^3 = 4x^5.

--Lynn
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Lynn Kurtz
science forum Guru


Joined: 02 May 2005
Posts: 603

PostPosted: Fri Jul 14, 2006 6:47 pm    Post subject: Re: probability and statistics Reply with quote

On Fri, 14 Jul 2006 18:42:24 GMT, "[Mr.] Lynn Kurtz"
<kurtzDELETE-THIS@asu.edu> wrote:


Two miserable typos below

Quote:
Do you mean f is x cumulative distribution function given by this? :

....f is a cumulative...

Quote:
f(x) = 0 for x <= 0
f(x) = x^4 for 0 < x < 1
f(x) = 1 for x > 1

Your suggestion of int[0..1] x f(x) dx suggests you are thinking of
f(x) as a probability density function, which it isn't because the
integral from 0 to 1 of f(x) isn't 1.

Now, *if* f(x) is the cumulative distribution function I propose
above, then its density function would be f'(x) which is 4x^3 on (0,1)
and 0 elsewhere. In that case to calculate the expected value you
would integrate x * 4x^3 = 4x^5.

.... integrate.. 4x^4.... over (0,1)

--Lynn
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