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

 From "Nick Cox" To Subject RE: st: how to find the integral for a portion of a normal distribution. Date Tue, 4 May 2010 17:06:26 +0100

```This remains a little ambiguous.

If you want the fraction of the simulated data that is in any interval,
just -count- and divide by 100000.

count if <condition involving x>
di r(N) / 100000

If you want the fraction of a normal in any interval, just call
-normal()- to do the work. You can do arithmetic inside the ().

No need for -integ- that I can see.

Nick
n.j.cox@durham.ac.uk

Buzz Burhans

This gets me what I want, but there should be an easier way to express
this
as 2 equations?

clear
drawnorm x , means(2.05) sds(1.74) n(100000) clear
sort x
egen lowresponse = total(x) if _n < (100000*.273)
egen highresponse = total(x) if _n >= (100000*.273)
l in 1/50
l in -60/l

David Muller

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
positive.

On Tue, May 4, 2010 at 2:01 PM, Hollis,Michael E <mhollis@mwdh2o.com>
wrote:

> ...you'll then be able to use the standard normal distribution with
mean
> zero and unit variance.

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```