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RE: st: how to find the integral for a portion of a normal distribution.
From
"Lachenbruch, Peter" <[email protected]>
To
"'[email protected]'" <[email protected]>
Subject
RE: st: how to find the integral for a portion of a normal distribution.
Date
Tue, 4 May 2010 09:43:37 -0700
This seems to be the expectation of x for x>c. I used integration by parts and got (DON'T TRUST ME HERE - I haven't done this for a while)
Exp(-.5*((c-mu)/sigma)^2) + mu*P(x>c)
It's simple enough that it feels right.
Tony
Peter A. Lachenbruch
Department of Public Health
Oregon State University
Corvallis, OR 97330
Phone: 541-737-3832
FAX: 541-737-4001
-----Original Message-----
From: [email protected] [mailto:[email protected]] On Behalf Of Buzz Burhans
Sent: Monday, May 03, 2010 8:59 PM
To: [email protected]
Subject: RE: st: how to find the integral for a portion of a normal distribution.
Mike, thanks for your help. Let me try to clarify what I want. If I have a
trial where average response was 2.05 liters, sd 1.74, What I want to find
is the cumulative volume of all responses >= 1; and the cumulative volume of
all responses <1
Is that any clearer?
Thanks
Buzz Burhans, Ph.D.
Dairy-Tech Group
So. Albany, VT / Twin Falls ID
Phone: 802-755-6842
Cell: 208-320-0829
Fax VT: 802-755-6842
Fax ID: 208-735-1289
Email: [email protected]
-----Original Message-----
From: [email protected]
[mailto:[email protected]] On Behalf Of Hollis,Michael E
Sent: Monday, May 03, 2010 9:36 PM
To: [email protected]
Subject: Re: st: how to find the integral for a portion of a normal
distribution.
I may be missing something here, but can't you simply use the normal
distribution with mean=proportion z >= some threshold, q=1-p and
variance p(1-p)/n? No integration involved.
As I said, I might be missing something!
Sent from my iPhone
On May 3, 2010, at 7:58 PM, "Buzz Burhans" <[email protected]> wrote:
> of the observations <=1 from
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