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
Formula for normal distribution with know mean and standard error and n terms
2 messages · Steve_Friedman at nps.gov, GlenB
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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