Editing Weibull distribution (section)

===Method of moments===
The [[coefficient of variation]] of Weibull distribution depends only on the shape parameter:<ref name="Cohen1965">{{cite journal | url=https://www.stat.cmu.edu/technometrics/59-69/VOL-07-04/v0704579.pdf |title=Maximum Likelihood Estimation in the Weibull Distribution Based on Complete and on Censored Samples |first=A. Clifford |last=Cohen | journal=Technometrics |issue=4 |volume=7 |date=Nov 1965 | pages=579–588|doi=10.1080/00401706.1965.10490300 }}</ref> 

:<math>CV^2 = \frac{\sigma^2}{\mu^2} 
 = \frac{\Gamma\left(1+\frac{2}{k}\right) - \left(\Gamma\left(1+\frac{1}{k}\right)\right)^2}{\left(\Gamma\left(1+\frac{1}{k}\right)\right)^2}.</math>

Equating the sample quantities <math>s^2/\bar{x}^2</math> to <math>\sigma^2/\mu^2</math>, the moment estimate of the shape parameter <math>k</math> can be read off either from a look up table or a graph of <math>CV^2</math> versus <math>k</math>. A more accurate estimate of <math>\hat{k}</math> can be found using a root finding algorithm to solve

:<math>\frac{\Gamma\left(1+\frac{2}{k}\right) - \left(\Gamma\left(1+\frac{1}{k}\right)\right)^2}{\left(\Gamma\left(1+\frac{1}{k}\right)\right)^2} = \frac{s^2}{\bar{x}^2}.</math>

The moment estimate of the scale parameter can then be found using the first moment equation as 

:<math>\hat{\lambda} = \frac{\bar{x}}{\Gamma\left(1 + \frac{1}{\hat{k}}\right)}.</math>