T Tan New member Joined Jul 20, 2006 Messages 3 Aug 3, 2006 #1 Use integration to determine first and second moments of the Weibull distribution with parameters a, b, and c: f(y) = (c/b)*((y - a)/b)^(c - 1) *exp{-((y - a)/b)^c) , a<y
Use integration to determine first and second moments of the Weibull distribution with parameters a, b, and c: f(y) = (c/b)*((y - a)/b)^(c - 1) *exp{-((y - a)/b)^c) , a<y
stapel Super Moderator Staff member Joined Feb 4, 2004 Messages 16,550 Aug 3, 2006 #2 Tan said: f(y) = (c/b)*((y - a)/b)^(c - 1) *exp{-((y - a)/b)^c) Click to expand... Just to clarify, does the snippet quoted above mean the following? . . . . .\(\displaystyle \L f(y)\,= \,\left(\frac{c}{b}\right)\, {\left(\frac{y\,-\,a}{b}\right)}^{c\,-\,1}\, e^{\left(\frac{-(y\,-\,a)}{b}\right)}^c\) Thank you. Eliz.
Tan said: f(y) = (c/b)*((y - a)/b)^(c - 1) *exp{-((y - a)/b)^c) Click to expand... Just to clarify, does the snippet quoted above mean the following? . . . . .\(\displaystyle \L f(y)\,= \,\left(\frac{c}{b}\right)\, {\left(\frac{y\,-\,a}{b}\right)}^{c\,-\,1}\, e^{\left(\frac{-(y\,-\,a)}{b}\right)}^c\) Thank you. Eliz.