R help - Jan 2010 - Formula for normal distribution with know mean and standard error and n terms

Hello,

I am searching for a method to calculate a normal distribution.

For example this equation is used to calculate the normal curve when the
mean and standard deviation are know.
p(x) = (1/?*sqrt(2?)) x exp (- (x-?)2/2?2)


or
(Embedded image moved to file: pic27350.jpg)Normal Probability Distribution
Formula


However, some of the literature I'm reading (I'm building an ecological
niche model for vegetation along several ecological gradients) report the
standard error instead and n sample size.  Is there an equivalent formula ?
If so, how can I also normalize the p(x) term to be within the 0-1 range?


Thank you all
Steve


Steve Friedman Ph. D.
Spatial Ecological Analyst
Everglades and Dry Tortugas National Park
950 N Krome Ave (3rd Floor)
Homestead, Florida 33034

Steve_Friedman at nps.gov
Office (305) 224 - 4282
Fax     (305) 224 - 4147

Steve_Friedman at nps.gov wrote:
> 
> I am searching for a method to calculate a normal distribution.
> 
> For example this equation is used to calculate the normal curve when the
> mean and standard deviation are know.
> p(x) = (1/?*sqrt(2?)) x exp (- (x-?)2/2?2)
> 
> However, some of the literature I'm reading (I'm building an
ecological
> niche model for vegetation along several ecological gradients) report the
> standard error instead and n sample size.  Is there an equivalent formula
> ?
> If so, how can I also normalize the p(x) term to be within the 0-1 range?
> 
What you have there (p) is a density rather than the distribution.

note that p(x) is NOT a probability, so it doesn't lie between 0 and 1 

(integrals of p(x).dx are probabilities and do lie between 0 and 1)

The function to compute p is dnorm. Try ?dnorm in R.

if you're given the standard error of a mean (which I'll call
"se") and n,
then sigma = sqrt(n)*se

(because se = sigma/sqrt(n) ).

If it's the standard error of something other than the mean you'll need
to
give
more details.


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