How to find distribution of the data?

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Sevil Cansu Yildirim
Sevil Cansu Yildirim el 29 de En. de 2020
Respondida: Jemima Pulipati el 22 de Dic. de 2020
Hello,
I have a code that i calculate pairwise distances of points belonging to same grid on a map. Now, I want to know if those distances are very close to each other or not. In other words, I want to find if these distances in every grid are distributed normally, lognormally or not. I calculated the standard deviation of distances in every grid. But I dont know how to understand if my data is normally distributed in every grid or not.
Here you can find my code and data.
clear all;
addpath('D:\Desktop\BOUN\JOB\Slip_Rates\Slip_Data\MAP');
LAT1=39;LAT2=42;LON1=29.0;LON2=41.0;
sf = 1;
m_proj('albers equal-area','lat',[LAT1 LAT2],'long',[LON1 LON2],'rect','on');m_gshhs_h('color',[.5 .5 .5]);hold on;
m_grid('linewi',1,'linest','none','tickdir','in','fontsize',10);
all = load('all_velocities.txt');
lon1=all(:,1);lat1=all(:,2);ve1=all(:,3);vn1=all(:,4);
% GRIDDING THE AREA
dLO = .5*2; dLA = .3636*2;
lon = [LON1:dLO:LON2];
lat = [LAT1:dLA:LAT2];
% DENSITY OF THE PLOTS & DISTANCES IN EVERY GRID
nlat = length(lat);
nlon = length(lon);
DIST = cell(nlat, nlon);
for i = 1:nlat
for j = 1:nlon
ind = find(abs(lat1-lat(i))<dLA/2 & (abs(lon1-lon(j))<dLO/2)); % FINDING ELEMENTS BELONGING O GRIDS
DENSITY(i,j) = length(ind); % WHICH GRID CONTAINS WHAT AMOUNT OF ELEMENT
points = [reshape(lat1(ind),[],1), reshape(lon1(ind),[],1)];
P{i,j} = points;
DIST{i,j} = pdist(points, 'euclidean'); % FINDING THE DISTANCE BETWEEN ELEMENTS BELONGING TO SAME GRID
end
end
% STANDARD DEVIATION OF THOSE DISTANCES
for i = 1:nlat
for j = 1:nlon
SDIST(i,j) = std(DIST{i,j});
end
end
% PLOT
m_pcolor(lon-dLO/2,lat-dLA/2,DENSITY);colormap(jet);colorbar;
m_quiver(lon1,lat1,sf.*ve1,sf.*vn1,1,'w','filled','AutoScale','off','linewidth',1.5);
hold off;

Respuesta aceptada

Jemima Pulipati
Jemima Pulipati el 22 de Dic. de 2020
Hello,
From my understanding you want to check if the given data is normally distributed or not.
The chi2gof(x) returns a test decision for the null hypothesis that the data in vector x comes from a normal distribution with a mean and variance estimated from x, using the chi-square goodness-of-fit test. The alternative hypothesis is that the data does not come from such a distribution. The result is 1 if the test rejects the null hypothesis at the 5% significance level, and 0 otherwise.
Some of the answers from the community which might be of relevance to you:

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