Is there a function to determine the non-exceedance probability of each value in the table?

14 次查看（过去 30 天）
Macy
Macy 2023-2-12
评论： Macy 2023-2-13
Ran in:
Table = readtable("practice3.xlsx");
x = Table.values
x = 10×1
450 1100 1060 700 350 80 2310 2000 370 50

请先登录，再回答此问题。

采纳的回答

the cyclist
the cyclist 2023-2-12
编辑：the cyclist 2023-2-12
Ran in:
Using info from your near-duplicate of this question:
Table = readtable("https://www.mathworks.com/matlabcentral/answers/uploaded_files/1293540/practice3.xlsx");
x = Table.values
x = 10×1
450 1100 1060 700 350 80 2310 2000 370 50
nonExceedanceProbability = sum(x'<=x,2)/numel(x)
nonExceedanceProbability = 10×1
0.5000 0.8000 0.7000 0.6000 0.3000 0.2000 1.0000 0.9000 0.4000 0.1000
This assumes that you are trying only to do the purely empirical non-exceedance for this particular set of values. If, instead, this is a sample and you are trying to estimate something about a population, you'll need to provide the info that @Torsten mentions.

更多回答（1 个）

Torsten
Torsten 2023-2-12
编辑：Torsten 2023-2-12
Ran in:
You will first have to tell us the type of distribution the values are supposed to follow (e.g. log-normal distribution).
After this, you will have to determine the parameters of this distribution from your measurement data.
After this, you will be able to plot a non-exceedance probability curve.
e.g.
Table = readtable("practice3.xlsx");
x = Table.values
x = 10×1
450 1100 1060 700 350 80 2310 2000 370 50
params = mle(x,'Distribution','LogNormal')
params = 1×2
6.2041 1.2021
hold on
y = linspace(0,max(x));
plot(y,logncdf(y,params(1),params(2)))
legend('Non-Exceedance Probability')
hold off
grid on
  1 个评论
Macy
Macy 2023-2-13
Thank you, I was looking for a command that would help me find emprical non-exeedance values, but this was going to be my second question which you answered.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Descriptive Statistics 的更多信息

标签

产品


版本

R2022b

Community Treasure Hunt

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

Start Hunting!

Translated by