normal distribution from data

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));

[Image Analyst]

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]

thanks Image Analyst, what do you mean by this, 'By the way, I assume you know your x is not a normal distribution', i'm still trying to figure out how it all works.

[Image Analyst]

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]

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]

You're right - I messed up and thought that x was also the Data.

Answer 1

[Youssef Khmou]

Here is another suggestion:

 y=pdf('Normal',x,m,s);
 plot(x,y);

[Image Analyst]

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]

Yes Mr @Image Analyst, the advantage i see is that this function gives a choice for other laws besides the Gaussian,

Answer 2

[Shashank Prasanna]

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.

Answer 3

[Roger Stafford]

You might try the Statistics Toolbox function 'normplot' to see how closely your 'data' comes to a normal distribution.