Generate random numbers with conditions (min, max, mean, and specific values)

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Hi! I'm very new at Matlab and I would like some tips to solve the following issue ...
I would like to generate a sequence of 40 random numbers (floats) (min=8, max=20, mean=10). However, 4 of these numbers must be equal to 16. How do you solve this question?
  6 Comments
the cyclist
the cyclist on 21 Nov 2022
I will also try to be clearer. You cannot have both of these things:
  • numbers between 8 and 20
  • numbers from a normal distribution
These are mathematically contradictory facts. It is like saying you want a number that is both bigger than 3, and smaller than 2. It just cannot be.
You need to figure out how to reconcile these facts. Often, people don't really understand what a normal distribution is, and they might be OK with some other random distribution. Or sometimes they are OK with a truncated normal. But, we cannot figure that out for you.

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Answers (2)

Dev
Dev on 28 Nov 2022
You might be aware that at least 4 numbers are going to be 16, so you have to generate 36 random numbers with mean 28/3. If type of distribution doesn’t matter, You can use the following code to generate such distribution.
n = 36;
xmean = 28/3;
xmin = 8;
xmax = 20;
xeps = 0.01;
x = randi([xmin xmax],n,1);
while abs(xmean - mean(x)) >= xeps
if xmean > mean(x)
x(find(x < xmean,1)) = randi([ceil(xmean) xmax]);
elseif xmean < mean(x)
x(find(x > xmean,1)) = randi([xmin floor(xmean)]);
end
end

Bruno Luong
Bruno Luong on 28 Nov 2022
Use this fex
https://fr.mathworks.com/matlabcentral/fileexchange/9700-random-vectors-with-fixed-sum
r = randperm([16+zeros(4,1); randfixedsum(36,1,40*10-16*4,8,20)]);
p=randperm(40);
r=r(p)
Most of numbers are around 10 since the MEAN pull the numbers arouns that value, so the chance to get big number (20) is tiny.

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