# [ncl-talk] conversion of Z-score to Probability

Dennis Shea shea at ucar.edu
Mon Nov 21 08:44:20 MST 2016

If you have calculated the Z-scores, then you must have the *population*
means and standard deviations at each grid point,

http://www.ncl.ucar.edu/Document/Functions/Built-in/cdfnor_p.shtml

x         = -1.64485
mean  = 0.0
sd       = 1.0

P = cdfnor_p(x,mean,sd)
print("P="+P)

====
An alternative function is the student-t function. For large 'n' (say,
n>25) the student-t is the Z-score

http://www.ncl.ucar.edu/Document/Functions/Built-in/student_t.shtml

t  = -2.12
P  = student_t(t, df)   ; P = 0.05
print(P)

In the unlikely event that you have a Z-score but no idea about size use:

P = student_t(z, 100)

=============

http://www.ncl.ucar.edu/Document/Functions/cumul_dist.shtml
http://www.ncl.ucar.edu/Document/Functions/statistics.shtml

On Mon, Nov 21, 2016 at 7:28 AM, Seshagirirao Kolusu <seshukolusu at gmail.com>
wrote:

> Dear all,
> I am new to NCL. I have negative Z-score values at each grid point but I
> wanted to convert Z to  the probability values (see:
> Could you please guide me how to convert them?
>
> Thanks,
> Seshu
>
>
>
> --
> ****************************************************
> Dr.rer.nat. K.Seshagirirao
> PDRA
> Institute for Climate and Atmospheric Science (ICAS)
> <http://www.see.leeds.ac.uk/research/icas/>,
> School of Earth and Environment (SEE) <http://www.see.leeds.ac.uk/>,
> University of Leeds <http://www.leeds.ac.uk/>,
> Leeds,
> LS2 9JT;UK
> http://www.see.leeds.ac.uk/people/s.kolusu
> ****************************************************
>
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