Incorrect mean and SD of Pascal distribution?

GegznaV shared this problem 4 years ago
Solved

In https://wiki.geogebra.org/e... it is written:

"The Pascal distribution models the number of failures before the nth success in repeated mutually independent Bernoulli trials, each with probability of success p."

And this definition is in alignment with the definition used in program R (see "Details" in documentation http://stat.ethz.ch/R-manua...)

If X ~ Pascal(n, p) and X is a number of failures, n is a number of successes and p is a probability of one success, then the mean of X should be n(1-p)/p and the variance of X should be n(1-p)/(p^2) (as defined in R's manual). Standard deviation (SD) is the square root of variance. It seams that GeoGebra calculates the mean as np/(1-p) and the standard deviation as sqrt(np/((1-p)^2) which is not the mean and the SD of failures.

Are these calculations intentional or is it a bug?

Comments (3)

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Thanks, fixed for v455

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mi ingles hace que yo no entienda bien qué diferencia hay entre Pascal(n,p) del post anterior y esta pagina de la wiki

https://en.wikipedia.org/wi...


From Wikipedia, the free encyclopedia

(Redirected from Pascal distribution)

¿en qué son diferentes? ¿está errónea la wiki?

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Please try the new version (v455)

@mathmagic: replace (p) with (1-p) in the Wikipedia page to get consistency with GeoGebra


Different texts adopt slightly different definitions for the negative binomial distribution. They can be distinguished by whether the support starts at k = 0 or at k = r, whether p denotes the probability of a success or of a failure, and whether r represents success or failure,[1] so it is crucial to identify the specific parametrization used in any given text.

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