RE: st: how to find the integral for a portion of a normal distribution.

["Buzz Burhans" <buzzb3@earthlink.net>]

Thanks David.

I am with you on this part - I can generate the proportion of responses <1
or >1 easily enough creating Z scores.  What I want is the sum of the
responses in these two proportions - not the count, but the sum of the
response volume produced / accumulated by observations in these two
proportions.

I can do it in several steps - sorting the obs, and summing over each
proportions, i.e. the lowest 27.3% and the remaining 72.7%; but it seems
that there must be an easier way to get these two cumulative amounts in a
single step.  . 

Buzz

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: buzzb3@earthlink.net

-----Original Message-----
From: owner-statalist@hsphsun2.harvard.edu
[mailto:owner-statalist@hsphsun2.harvard.edu] On Behalf Of David Muller
Sent: Monday, May 03, 2010 10:26 PM
To: statalist@hsphsun2.harvard.edu
Subject: Re: st: how to find the integral for a portion of a normal
distribution.

To expand upon Michael's suggestion, -display normal((1 - 2.05)/1.74)-
will give you the proportion of responses <=1, and -display 1 -
normal((1 - 2.05)/1.74)- will give you the proportion of responses
>=1.

Note that assuming this normal model for your response you would
expect a little over 10% of your responses to be less than 0, which
indicates a problem with your assumption if your response is strictly
positive.

David



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