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 one I
have the skill to find. So, how would I solve this using R?
David.
On Jun 10, 2009, at 7:09 AM, Etienne B. Racine wrote:
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 values you want shape and scale to be.
And you can do that with any probability function (some easier the
calculate) as most (if not all) have these equations. Of course, it
might be
more pleasing to play with the parameters.
Etienne
Mike Lawrence wrote:
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 to try to get even closer
to
what you want.
On Wed, Jun 10, 2009 at 2:48 AM, David Arnold<dwarnol...@suddenlink.net
>
wrote:
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?
And if it is possible, how can I adjust the distribution so that the
probability of selecting a random number greater than or equal to 60
is p.
Thanks. No idea how to start.
David
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