Gaussian distribution with randn

Is it possible and how can i obtain a Gaussian distribution with randn for

   mean= 0.126 
and it varies by+- 0.02, max=0.146 and min=0.106, 
to generate 1300 random values.

[Rik]

Have you read the documentation (just type doc randn)? You'll need to shift the mean, change the width of the distribution and do something about values outside your range. Think carefully about the order in which you do these.

[Joseph Lee]

I tried but generated some negative values instead, which is why i am asking how and whether is it possible or is there a better function to use.

Answer 1

[Akira Agata]

If you want to generate Gaussian distribution with the given mean and variance (not std), and then extract the values in [min max] range, the following code can do it.

va = 0.02;
mu = 0.126;
ul = 0.146;
ll = 0.106;

x = mu + randn(20000,1)*sqrt(va); % Generate sufficient random numbers
idx = (ll <= x) & (x <= ul); % Extract the value in the given range [min max]
x = x(idx);

x = x(1:1300); % Extract 1300 numbers

[Joseph Lee]

Thanks!

[Rik]

If your limit values are getting closer to the mean, you will need to progressively generate more values. There will be a function that estimates how many values you will need, but it is probably just easier to hard-code it into the solution given by Akira:

va = 0.02;
mu = 0.126;
ul = 0.146;
ll = 0.106;
nvals=1300;

multiplier=10;%Akira started with 15, for this example, 9 is not always enough, 10 is
x = mu + randn(multiplier*nvals,1)*sqrt(va); % Generate sufficient random numbers
idx = (ll <= x) & (x <= ul); % Extract the value in the given range [min max]
while sum(idx)<nvals
    multiplier=multiplier+1;
    x = mu + randn(multiplier*nvals,1)*sqrt(va); % Generate sufficient random numbers
    idx = (ll <= x) & (x <= ul); % Extract the value in the given range [min max]
end

x = x(idx);
x = x(1:nvals); % Extract 1300 numbers

[Joseph Lee]

Thanks for the explanation

Answer 2

[Guillaume]

By definition, a gaussian distribution covers the whole range [-∞, +∞]. If your distribution has a min and max it's not gaussian anymore.

Furthermore, a gaussian distribution is defined by a mean and a standard deviation, not a mean and a range. If a gaussian distribution has a standard deviation of 0.02, you'll still find about 32% of the samples outside of that ±0.02 range.

[Joseph Lee]

+- 0.02 that can be considered the variance