Normal distribution for a given range of numbers.

I have a range in which my parameter can vary for example, 0.38 to 0.5. I need to produce in between points such that the data will be normally distributed. I have thought of the followig way.

    % code
xmin=0.38
xmax=0.5
n=20 
x=xmin+randn(1,n)*(xmax-xmin);

But randn(1,n) gives me random numbers from more than 1 as well. Which does not serve the purpose of putting the range. Then my xmax becomes more than 0.5. How to achieve this for normal distribution and the given range? Thank you :)

Answer 1

[Birdman]

Use rand, instead of randn:

 x=xmin+rand(1,n)*(xmax-xmin);

[Neje]

Thank you :) This will give me uniform distribution though. I want a normal distribution for the numbers in the range 0.38 to 0.5.

[Birdman]

Try normrnd:

 x=xmin+normrnd(0,1,[1 n])*(xmax-xmin)

[Neje]

This gives me values more and less than xmax and xmin respectively. Can I define a range for a normal random number generator somehow? So that it gives me numbers only from 0.38 to 0.5 (normally distributed)

[Stephen23]

"So that it gives me numbers only from 0.38 to 0.5 (normally distributed)"

This is a contradiction: normally distributed data has no bounds.

[Neje]

Yea I think truncated normal distribution works in this case. Thanks.

[Mostafa Nakhaei]

If you draw you will see that x=xmin+rand(1,n)*(xmax-xmin) is far from being normal.

try this:

p = 6;

xmin=0.38;

xmax=0.5;

n=10000;

X = xmin + (xmax - xmin)*sum(rand(n,p),2)/p;

hist(X,50)

you will see this is closer to normal distributation as exlained by John in the next comment.

[Aman Kumar]

what is the role of p in this code.

Answer 2

[John D'Errico]

A normal distribution does not have limits. In theory it is possible to see generated points that lie all the way out to infinity, or at least arbitrarily close to that point.

You might consider a truncated normal distribution. I know there is at least one such utility to be found on the MATLAB Central File Exchange. You can do the search as easily as can I. A truncated normal distribution is not that difficult to sample from either. The stats toolbox would make it fairly easy.

Just as easy is to make use of the central limit theorem. If you want a fairly normal looking distribution of points, that all lie within limits of xmax and xmin, do this:

p = 6;

xmin=0.38;
xmax=0.5;
n=10000;

X = xmin + (xmax - xmin)*sum(rand(n,p),2)/p;

hist(X,50)

Don't go overboard and make p too large however. Large values of p will see you generate very few points near those limits. As well, large values of p will take more memory and time to generate the sample.

If p is smaller, 3 for example, the distribution will look a bit less smooth, but you will more likely get points near the endpoints.

p = 3;

If you have the stats toolbox, then betarnd will work too. So we could generete samples with a distribution that looks fairly like a truncated Normal.

A = 3;
B = 3;
X = betarnd(A,B,1,n);
X = xmin + (xmax - xmin)*betarnd(A,B,1,n);
hist(X,50)

As you make A and B larger, the distribution will start looking vaguely more normally distributed, but the number of events that occur near your limits will start to drop again.

So it all depends on exactly how you want that distribution to behave. But as I have shown, there are lots of ways to do this.

[Neje]

Thanks a lot..both the alternatives work well. I think I will go ahead with betarnd. Thanks again! :)

[John D'Errico]

The nice thing about betarnd is you can control the shape pretty easily, just by playing with A and B.

[Neje]

Yes exactly :) Have one question though, with these plots hist(X,50), what does the y axis represent?

[John D'Errico]

You've never seen a histogram before?

https://en.wikipedia.org/wiki/Histogram

[Neje]

got it thanks :)

[Aman Kumar]

X = xmin + (xmax - xmin)*sum(rand(n,p),2)/p;

Please Define the all variable used in this line

[Aman Kumar]

How you get the value of p variable in this code

Answer 3

[siranjeevi gurumani]

Use rescaling to get a random normal distribution between the desired range if mean and standard deviation is not a concern

xmin=0.38;
xmax=0.5;
n=2000;
temp = normrnd(0,1,1,n);%generating a standard normal distribution
x=xmin+(xmax-xmin)*((temp-min(temp))/(max(temp)-min(temp)));%rescaling the range
hist(x)