Mean and 3-sgima for Lognormal distributions

3 min readMay 23, 2022

Hello,

I am trying to plot the lognormal distribution over 10 iterations and would like to see the mean and 3 sigma outliers. Somehow, doing this for lognormal plots does not look easy. Also, is it possible that I can skip the histogram in my plots and only look at the lognormal curves? This is the code snippet I am using. Thanks!

figure(113);
for i = 1:10
norm=histfit(Current_c(i,:),10,'lognormal'); %%Current_c is a 10*100 matrix %%
[muHat, sigmaHat] = lognfit(Current_c(i,:));
% Plot bounds at +- 3 * sigma.
lowBound = muHat - 3 * sigmaHat;
highBound = muHat + 3 * sigmaHat;
yl = ylim;
%line([lowBound, lowBound], yl, 'Color', [0, .6, 0], 'LineWidth', 3);
%line([highBound, highBound], yl, 'Color', [0, .6, 0], 'LineWidth', 3);
line([muHat, muHat], yl, 'Color', [0, .6, 0], 'LineWidth', 3);
grid on;
set(gcf, 'Toolbar', 'none', 'Menu', 'none');
% Give a name to the title bar.
set(gcf, 'Name', 'Line segmentation', 'NumberTitle', 'Off')
hold on;
end

NOTE:-

Matlabsolutions.com provide latest MatLab Homework Help,MatLab Assignment Help for students, engineers and researchers in Multiple Branches like ECE, EEE, CSE, Mechanical, Civil with 100% output.Matlab Code for B.E, B.Tech,M.E,M.Tech, Ph.D. Scholars with 100% privacy guaranteed. Get MATLAB projects with source code for your learning and research.

to visualize the mean, sigma and 3 sigma on the curve of the probability distribution that approximates your data you have to consider that the bins of the histogram and too wide to directly get accurate marks on the curve, let me explain:

If you try

find(y==my)
 =
   Empty matrix: 1-by-0

no match

But you can find the histogram bin where the mean is contained in:

find(y>my)
 =
  Columns 1 through 14
     3     4     5     6     7     8     9    10    11    12    13    14    15    16
  Columns 15 through 28
    17    18    19    20    21    22    23    24    25    26    27    28    29    30
  Columns 29 through 31
    31    32    33find(y<=my)
 =
  Columns 1 through 14
     1     2    34    35    36    37    38    39    40    41    42    43    44    45
  Columns 15 through 28
    46    47    48    49    50    51    52    53    54    55    56    57    58    59
  Columns 29 through 42
    60    61    62    63    64    65    66    67    68    69    70    71    72    73
  Columns 43 through 56
    74    75    76    77    78    79    80    81    82    83    84    85    86    87
  Columns 57 through 69
    88    89    90    91    92    93    94    95    96    97    98    99   100

Let be ny the reference vector of y:

ny=[1:1:length(y)]

then the mean value is somewhere within the interval

[33 34]

So, to accurately plot my and 3*sy you first have to decide how accurate you need be.

While the y step is, and it varies along the curve

abs(y(33)-y(34))
ans =
   0.258998840336178

the x step is constant

abs(x(33)-x(34))
 =
   0.011562155305573mean(diff(x)) 
=   0.011562155305573max(diff(x))   
=   0.011562155305573min(diff(x))    
=   0.011562155305573

So if you decide that 2 decimals precision (on y) is enough, we have to refine x small enough so that at least one point of y interpolated has the value my truncated after 2nd decimal:

3.02xxxx

for 0.2589 go down to 0.0001 the y step has to be fractioned at least

0.2589/dy=0.0001 hence dy=258.9 , let’s take dy=259

the angle

SEE COMPLETE ANSWER CLICK THE LINK