> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of David Arnold
> Sent: Tuesday, June 09, 2009 11:49 PM
> To: r-help@r-project.org
> Subject: [R] Creating a specific skewed distribution
>
> All,
> Can someone help me create a skewed distr
tienne B. Racine'
Cc: 'r-help@r-project.org'
Subject: RE: [R] Creating a specific skewed distribution
Here is one way to solve the equation:
require(BB)
f <- function(x) (2*gamma(1+1/x))^x + log (0.10)
Re: [R] Creating a specific skewed distribution
Etienne et al,
This is exactly what I need. So I gave it an algebraic try and set:
Mean=30=b*gamma(1+1/a),
solve for b and substitute into F(60)=0.10. After a little algebra, this
left me with
[ 2*gamma(1+1/a) ]^a = -ln(0.10)
Now, I don't think
Etienne et al,
This is exactly what I need. So I gave it an algebraic try and set:
Mean=30=b*gamma(1+1/a),
solve for b and substitute into F(60)=0.10. After a little algebra,
this left me with
[ 2*gamma(1+1/a) ]^a = -ln(0.10)
Now, I don't think this has a closed form solution, at least not
You could also make some algebra. e.g. :
?rweibull
gives the formula of the mean and of the cumulative distribution function in
the Details section. So using your known parameters (i.e. mean=30 and
p(10)=.10, i.e. cumulative function(10) =.90), I think it is sufficient to
determine the exact val
With skewed unimodal distributions, the mode can't equal the mean, so
assuming you want the mean to be around 30, I find that a weibull
function can get close to what you want:
> mean(rweibull(1e5,1.5,33))
[1] 29.77781
> pweibull(60,1.5,33)
[1] 0.9138475
I'm sure you can play with the parameters
All,
Can someone help me create a skewed distribution, mean = 30, with
probability of selecting a random number from the distribution greater
than or equal 60 equal to 10%?
I need the probability density function to equal zero at zero, and
have a maximum height at or near 30.
Is this possible?
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