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

[Jeph Herrin <junk@spandrel.net>]

So you want to integrate xf(x), so to speak; let's try to
integrate using centiliters (cls):

 drawnorm x , means(2.02) sds(1.75) n(100000) clear
 gen cl=round(x,0.01)              // integrate over centiliters
 bys cl: gen cls=_N*cl/100000      // expected cls in this cl

 egen liters=sum(cls*(x<=1)/100)   // total liters over x < = 1 liter
                                   // the 100 is cl->l factor
 replace liters=liters*10          // 10 cows!
 sum liters


hth,
Jeph


Buzz Burhans wrote:
> Let me try again, I am sorry I am so dense here.
>
> Suppose I have an intervention applied to cows with a demonstrated mean milk
> yield response of +2.05 liters, sd 1.74.
>
> Suppose I am interested a 1 liter cut point.  I know how to find the
> proportion of responses at or below the 1 liter response; and the proportion
> of responses at or above the one liter response.  This is what you are doing
> here, or it can be done with a z score.
>
> My question is, if the response is normally distributed, given n cows, how
> many total liters were included or accumulated in the only responses below 1
> liter, and how many total liters were accumulated in only the responses at
> or above 1 liter.  (Negative responses are possible). My interest is in the
> total liters, not the fraction of the total population either above or below
> the cutpoint.
>
> This is not the same as the proportion of responders in either category; it
> is the total accumulated response above and below my cut point.
>
> I am off to a meeting, so thanks for any further responses - I won't be able
> to see them until this evening.
>
> 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 Jeph Herrin
> Sent: Tuesday, May 04, 2010 10:15 AM
> To: statalist@hsphsun2.harvard.edu
> Subject: Re: st: how to find the integral for a portion of a normal
> distribution.
>
> The total area under any normal curve is always 1, so you just need
> to count obs:
>
>   drawnorm x , means(2.02) sds(1.75) n(100000) clear
>   gen byte lessthan1=x<=1
>   sum lessthan1
>
> The mean value of -lessthan1- is the area under the curve
> to the left of 1.
>
> hth,
> Jeph
>
>
> Buzz Burhans wrote:
> > My weak math skills are showing up here - 
> >
> > For demonstration purposes, I have generated a set of 100,000 observations
> > with a normal distribution, mean 2.02, SD 1.74 and visualized this data
> and
> > the portion of observations above 1  with a histogram
> >
> > drawnorm x , means(2.02) sds(1.75) n(100,000) clear
> > hist, xline(1)
> >
> > I also computed the proportion of the observations <=1 from the Z score.
> >
> > What I don't know how to do in Stata is find the integral for the
> > proportions of the distribution above and below 1.  
> >
> > How can I determine those two integrals in Stata?
> >
> > 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: buzzb3@earthlink.net
> >
> >
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