Prashant Kumar answered . 2024-05-18 02:06:09

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    33

find(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))
 =
   0.011562155305573

mean(diff(x)) 
=   0.011562155305573

max(diff(x))   
=   0.011562155305573

min(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

alpha=atand(abs(y(33)-y(34))/abs(x(33)-x(34)))
=
 87.443914593184061

    dx=.0001/atand(alpha)
    =
         1.119259323130124e-06

x_step=abs(x(33)-x(34))/dx

   =
       1.033018449490165e+04

make it

x_step=10331

then

x2=linspace(x(1),x(end),x_step);
y2=interp1(x,y,x2);

overlap, check both y and y2 are the same:

figure(1);plot(x,y,x2,y2);grid on;grid minor

where in y2 is my located?

find(y2>my)
=
..
..
      3396        3397        3398        3399        3400        3401        3402
  Columns 3228 through 3234
        3403        3404        3405        3406        3407        3408        3409
  Columns 3235 through 3238
        3410        3411        3412        3413

x_mean=max(find(y2>my))

the actual mean value is going to be approximated with y2(3413) = 3.021925056586602

and this error is acceptable

 abs(my-y2(3413))
 =
    1.358549030059386e-04

put what looks like the mean on the curve:

hold on;plot(x2(x_mean),y2(x_mean),'bo')

sum(y)
 =
     3.021789201683596e+02
length(y)
 =
   100
mean(y)
=
   3.021789201683596

and already found mean located between [33 34]

the upper tail cannot even accommodate half single sigma (34.1%)

sum(y([33:end]))/sum(y)*100
 =
  13.483260209259832

Let's find mu summing samples:

pc_target=50
n=2
pc=sum(y([1:n]))/sum(y)*100
while pc

So it turns out mu is within [15 16]

the x boundaries for y~mu

x([15 16])
=
0.201607447590281   0.213169602895854

Using y2 for more detail

pc_target=50
n=2
pc=sum(y2([1:n]))/sum(y2)*100
while pc

for y2, mu is within [1600 1601]

n_mu=1600
and x2 boundaries

x2([1600 1601])
ans =
   0.216920307874458   0.217031116526467

n_mu=1600

the upper y2 boundary for 1 sigma (+34.1%)

pc_target=34.1
n=1
pc=sum(y2([1600:1600+n]))/sum(y2)*100
while pc

the numeral distance for +1sigma is

   n_up_1sigma=1475;

And the location on x2 of +1sigma is:

x2_up_1sigma=x2(1600+n_up_1sigma)
x2_up_1sigma =
   0.380363069587555

the lower y2 boundary for 1 sigma (-34.1%)

pc_target=34.1
n=1
pc=sum(y2([1600-n:1600]))/sum(y2)*100
while pc

the numeral distance for -1sigma is

   n_down_1sigma=842;

And the location on x2 of -1sigma is:

x2_down_1sigma=x2(1600-n_down_1sigma)
 =
   0.123619422882982

Repeating for 3 sigma, understanding that all +- sigma interval cover 89%, then up it is:

pc_target=44.5
n=1
pc=sum(y2([1600:1600+n]))/sum(y2)*100
while pc

and down we go:

pc_target=44.5
n=1
pc=sum(y2([1600-n:1600]))/sum(y2)*100
while pc

and the plot you asked for:

plot(x2,y2,x2(1600),y2(1600),'ro',...
x2(n_mu+n_up_1sigma),y2(n_mu+n_up_1sigma),'rd',x2(n_mu-n_down_1sigma),y2(n_mu-n_down_1sigma),'rd',...
x2(n_mu+n_up_3sigma),y2(n_mu+n_up_3sigma),'gd',x2(n_mu-n_down_3sigma),y2(n_mu-n_down_3sigma),'gd')
grid on;grid minor;

additional comments:

1. do not mix lognormal mu sigma and normal mu sigma parameters: mind the notation gap. Your x is a reference vector and your y is the actual distribution.

Y=lognpdf(X,mu,sigma)

MATLAB function lognpdf calculates the lognormal distribution Y out of the normal distribution X, where X has mean mu and standard variance sigma. In lognpdf help page, both Y and X are random variables, data, each with their respective reference vectors.

As explained in lognpdf help page, the 'mu' and 'sigma' of Y are related to those belonging to X by:

m=exp(mu+sigma^2/2)
v=exp(2*mu+sigma^2)*(exp(sigma^2)-1)

and

mu=log(m^2/(v+m^2)^.5)
sigma=(log(v/m^2)+1)^.5

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