normal distribution from data
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is there a more efficient way to derive a normal distribution.
% Deriving Normal Distribution From the Data
x=0:1:12;
m=mean(Data);
s=std(Data);
p=(1/(s*sqrt(2*pi)))*exp((-(x-m).^2)/(2*s^2));
5 个评论
Image Analyst
2013-9-10
Looks pretty efficient to me. Just how much speed do you need? By the way, I assume you know your x is not a normal distribution.
harley
2013-9-10
Image Analyst
2013-9-10
编辑：Image Analyst
2013-9-10
Just plot it and look at it: plot(x). Does it look like a Gaussian shape to you? No, it's a triangle, so it's a uniform distribution - a box, a flat distribution. You have equal probabilities of having any number. No numbers are more likely than any others - that is unlike what you'd see in a Gaussian Distribution.
Roger Stafford
2013-9-10
Image Analyst, it isn't 'x' that Harley is stating has the normal distribution. It is 'data' which isn't being specified here. The 'x' is the independent variable in the hypothesized normal distribution. A plot of
plot(x,p)
would give the theoretical normal distribution pdf values as functions of x for the mean and std which have been computed from 'data'.
Image Analyst
2013-9-10
You're right - I messed up and thought that x was also the Data.
采纳的回答
Youssef Khmou
2013-9-10
Here is another suggestion:
y=pdf('Normal',x,m,s);
plot(x,y);
2 个评论
Image Analyst
2013-9-10
It's fewer characters, so it's simpler to look at, but I doubt it's faster or more efficient (since there is more than one line of code inside that function), but I doubt he really wanted/needed more efficiency or speed anyway.
Youssef Khmou
2013-9-11
Yes Mr @Image Analyst, the advantage i see is that this function gives a choice for other laws besides the Gaussian,
更多回答（2 个）
Shashank Prasanna
2013-9-10
编辑：Shashank Prasanna
2013-9-10
Since this is normal distribution, the mean and std of the data are the maximum likelihood estimates for the normal distribution from the data.
Once you have the PDF, like you have in the last line of code as 'p', you could plot the PDF using x to span -4*sigma to +4*sigma:
x = -4*s:0.01:4*s
p=(1/(s*sqrt(2*pi)))*exp((-(x-m).^2)/(2*s^2));
plot(x,p)
You could use a wider range if you wanted to.
Roger Stafford
2013-9-10
0 个投票
You might try the Statistics Toolbox function 'normplot' to see how closely your 'data' comes to a normal distribution.
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