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


From   David Muller <davidmull@gmail.com>
To   statalist@hsphsun2.harvard.edu
Subject   Re: st: how to find the integral for a portion of a normal distribution.
Date   Tue, 4 May 2010 14:25:31 +1000

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


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.
>
> Sent from my iPhone
>
> On May 3, 2010, at 8:49 PM, "Buzz Burhans" <buzzb3@earthlink.net> wrote:
>
>> Doesn't your suggestion fit only if the curve is standard normal?, i.e.
>> mean
>> 0, SD 1?
>>
>> Perhaps my use of integral is incorrect; what I want is the area under the
>> curve of a normal distribution, mean 2.05, sd 1.75 for the portion >=
>> threshold =1, and then the portion < 1
>>
>> Buzz
>>
>>
>> statalist@hsphsun2.harvard.edu] On Behalf Of Hollis,Michael E
>> Sent: Monday, May 03, 2010 9:36 PM
>> To: statalist@hsphsun2.harvard.edu
>> 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
>>
>> .ucla.edu/stat/stata/
>>
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