fitdist is givin me mu of 6.43686 with this code. Am i doing something wrong. I'm trying to get mu and sigma values to compare this data with log normal distribution
Add=16
Add = 16
%%info given in question 4
River_flood_level=[516 678 802 683 491 582 625 531 451 726 652 556 470 570 774 670 432 604 881 707 735 614 698 403 548 504 758 1000 419];
River_flood_level=River_flood_level+Add;
%%solution of question 4
fitdist(River_flood_level.','Lognormal')
ans =
LognormalDistribution Lognormal distribution mu = 6.43686 [6.35256, 6.52116] sigma = 0.22162 [0.175873, 0.29973]

1 Kommentar

Why do you think the result is incorrect? What are you expecting mu to be?
The fitted distribution looks pretty reasonable given the limited amount of data provided.
Add=16;
%%info given in question 4
River_flood_level=[516 678 802 683 491 582 625 531 451 726 652 556 470 570 774 670 432 604 881 707 735 614 698 403 548 504 758 1000 419];
River_flood_level=River_flood_level+Add;
%%solution of question 4
p = fitdist(River_flood_level.','Lognormal');
figure
histogram(River_flood_level,'Normalization','pdf')
hold on
plot(400:1200,pdf(p,400:1200))

Melden Sie sich an, um zu kommentieren.

 Akzeptierte Antwort

Star Strider
Star Strider am 5 Jan. 2024

0 Stimmen

The μ and σ values are in log units. Calculatee their exponentials —
Add=16
Add = 16
%%info given in question 4
River_flood_level=[516 678 802 683 491 582 625 531 451 726 652 556 470 570 774 670 432 604 881 707 735 614 698 403 548 504 758 1000 419];
River_flood_level=River_flood_level+Add;
%%solution of question 4
df = fitdist(River_flood_level.','Lognormal')
df =
LognormalDistribution Lognormal distribution mu = 6.43686 [6.35256, 6.52116] sigma = 0.22162 [0.175873, 0.29973]
Mean = exp(df.mu)
Mean = 624.4451
Sigma = exp(df.sigma)
Sigma = 1.2481
figure
hf = histfit(River_flood_level.', 10, 'Lognormal');
hold on
Line = findobj(hf,'Type','line');
ymu = interp1(Line.XData, Line.YData, exp(df.mu));
plot(exp(df.mu)*[1 1], [0 ymu], ':r', 'LineWidth',3)
hold off
Ax = gca;
Ax.XTickLabelRotation = 60;
.

6 Kommentare

Emmeirrt
Emmeirrt am 6 Jan. 2024
Make sense, thank you!
Star Strider
Star Strider am 6 Jan. 2024
As always, my pleasure!
I have not worked with lognormal distrbutions in a while (although all physiological parameters are lognormally distributed) so I had to rediscover that.
Paul
Paul am 6 Jan. 2024
Mean is not the mean of the distribution and Sigma is not the sqrt of the variance of the distribution, so I'm still not sure what the OP is, or was, looking for. Perhaps @Emmeirrt can clarify.
Corrected —
Add=16
Add = 16
%%info given in question 4
River_flood_level=[516 678 802 683 491 582 625 531 451 726 652 556 470 570 774 670 432 604 881 707 735 614 698 403 548 504 758 1000 419];
River_flood_level=River_flood_level+Add;
%%solution of question 4
df = fitdist(River_flood_level.','Lognormal')
df =
LognormalDistribution Lognormal distribution mu = 6.43686 [6.35256, 6.52116] sigma = 0.22162 [0.175873, 0.29973]
m = exp(df.mu + df.sigma^2/2)
m = 639.9699
nu = exp(2*df.mu+df.sigma^2)*(exp(df.sigma^2)-1);
figure
hf = histfit(River_flood_level.', 10, 'Lognormal');
hold on
Line = findobj(hf,'Type','line');
ymu = interp1(Line.XData, Line.YData, m);
plot([1;1]*m, [0;1]*ymu, '--r', 'LineWidth',2)
hold off
Ax = gca;
Ax.XTickLabelRotation = 60;
.
Emmeirrt
Emmeirrt am 29 Jan. 2024
Thank you! ı just needed the info that they are in lognormal. I fixed the rest myself. Sorry for not answering early
Star Strider
Star Strider am 29 Jan. 2024
As always, my pleasure!
No worries!
Thank you for the follow-up (and for Accepting my Answer).

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Produkte

Version

R2023b

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by