random number generation within the defined range
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Hii,
First of all thanks to everybody for giving time to respond to my question, which is as follows:
I am trying to generate a random number between -.4776 and .5004 with the mean as .02528 and standard deviation as .116. I tried using the following command but its violating the specified range.
r = -.4776 +(.5004+.4776).*normrnd(.02528, .116,179290, 1)
Any help in this regard??
Thanks a lot!!
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Réponse acceptée
Daniel Shub
le 10 Avr 2012
You cannot use a Gaussian/normal distribution and expect all the values to fall into a particular range. When you specify a range the most common distribution is a uniform distribution. For your range a uniform distribution does not have the desired mean or standard deviation.
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Plus de réponses (2)
Junaid
le 10 Avr 2012
As Daniel suggested that we might not get exactly what you want. But there can be several tricks. Once possible trick is that you generate random numbers by defining the range;
m + s to m - s;
where m, and s are required mean and standard deviation.
ex.
m = .02528;
s = .116;
a = m + s;
b = abs(m - s);
% now generating the numbers between a and b.
g = a + (b-a).*rand(100,1);
You get mean and standard deviation very close to you required values.
1 commentaire
Daniel Shub
le 10 Avr 2012
I think you reversed the m and the s, but either way this does not really give the desired range (although it doesn't violate the range either), and has the incorrect standard deviation. The standard deviation of a uniform distribution is |a-b|/sqrt(12).
Dyno
le 10 Avr 2012
2 commentaires
Daniel Shub
le 10 Avr 2012
You should edit your question to include this information instead of posting it as an answer.
Image Analyst
le 10 Avr 2012
Huh? You want to try to model your fixed pattern noise of your sensor (basically differences in the sensitivity or responsivity of the detectors) by creating a simulated noise pattern rather than using the actual noise pattern that you can measure? Why would you want to do that when you have the actual pattern? It's fairly simple: Just divide by your actual pattern to correct your signal, assuming you've collected enough data to give you a good estimate of the "true" pattern and it's not just some noisy mess.
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