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Re: A guess of the Probability density function from percentile values
Posted:
Dec 27, 2010 7:41 AM
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On 27 déc, 13:19, Allamarein <matteo.diplom...@gmail.com> wrote:
> On 27 Dic, 12:23, Valeri Astanoff <astan...@gmail.com> wrote:
>
>
>
>
>
> > On 27 déc, 11:17, Allamarein <matteo.diplom...@gmail.com> wrote:
>
> > > I have just posted a similar thread in another group.
> > > I hope to be more lucky here.
>
> > > I know three percentile values.
> > > Let's say they are:
> > > 95% 82.1
> > > 50% 80.3
> > > 5% 77.8
>
> > > I would to get a Probability density function.
> > > I presume I would guess the shape of this curve.
> > > Since these data refer to a scientific measuring, I would find a t-
> > > Student or a Gaussian distribution that is consistent with the
> > > previous percentiles.
>
> > > Any suggestions?
>
> > Good day,
>
> > A suggestion (for a normal distribution) :
>
> > FindMinimum[
> > (Quantile[NormalDistribution[m, s], 5/100] - 77.8)^2 +
> > (Quantile[NormalDistribution[m, s], 50/100] - 80.3)^2 +
> > (Quantile[NormalDistribution[m, s], 95/100] - 82.1)^2,
> > {m, 1}, {s, 1}]
>
> > {0.0816667, {m -> 80.0667, s -> 1.30711}}
>
> > --
> > Valeri
>
> It seems you write a code in Mathematica.
> I am not so familiar with this language.
> May you give me more details?- Masquer le texte des messages précédents -
>
> - Afficher le texte des messages précédents -
I'm not a specialist. I just use Mathematica to solve
such problems. All I can say is that
you get the same result using derivatives wrt m and s,
instead of FindMinimum :
q[x_] = Quantile[NormalDistribution[m, s], x]
m + Sqrt[2]*s*InverseErf[2*x-1]
Solve[
D[(q[0.05] - 77.8)^2 + (q[0.50] - 80.3)^2 + (q[0.95] - 82.1)^2, m] ==
0 &&
D[(q[0.05] - 77.8)^2 + (q[0.50] - 80.3)^2 + (q[0.95] - 82.1)^2, s] ==
0,
{m, s}]
{{m -> 80.0667, s -> 1.30711}}
--
Valeri
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