How to Generate Randomized Data Set with Given Mean/Median/Standard Deviation

Need to generate a data set thats randomIZED but not completely random with the following criteria:

Mean: 55.4

Median: 54.8

Standard Deviation: 10

I appreciate any help I can get!

[Ive J]

Suppose you know your sample distribution, you can make use of different functions (such as random).

% Assuming a normal distribution
rng('default') % For reproducibility
r = normrnd(55.4, 10, 1e4, 1);
fprintf('Sample mean: %.2f, and sample SD: %.2f\n', mean(r), std(r))

[Jeff Miller]

It is not completely clear from your question whether you want those three criteria to be met for the true underlying distribution as a whole (so each random sample deviates slightly from those criteria by chance), or if you want the criteria to be met exactly for each individual sample. If you want the former, Bruno's answer gives a solution. If you want the latter, the problem is much harder.

[Julianna Sims]

The criteria need not be met for each exact individual sample. I'm looking for something to give me an array of generated values that collectively meet the three criteria as a data set. Sorry this is such a wacky question! I appreciate any assistance you all can provide.

[Bruno Luong]

It's not "much harder".

Sorry but such request just does look any pratical sense statistically, and usually asked by person who does not really know about random process and distribution.

Still this is a code, try to make use it if you can, good luck.

meantarget = 55.4;
mediantarget = 54.8;
starget = 10;

n = 100;
midp = 500/n;
xmid = midp*randn(1,n-2);
xmid = xmid-median(xmid);
m = meantarget-mediantarget;
s = sum(xmid);
s2 = sum(xmid.^2);
b = (m*n-s);
c = (b^2+s2-n*(m^2+starget^2))/2;
delta = b^2-4*c;
if delta < 0
    error('reduce midp')
end
xhi = (b+sqrt(delta))/2;
xlo = b-xhi;
x = mediantarget+[xlo xmid xhi];

medianx = median(x)
meanx = mean(x)
stdx = std(x,1)

[Jeff Miller]

Very clever, Bruno. I didn't think of generating n-2 scores and computing the last 2. I guess I should have said, "much harder for me".

Answer 1

[Bruno Luong]

There is infinity of random variables with the same mean/median/standard deviation

Here is one that meet what you ask (three discrete values).

meantarget = 55.4;
mediantarget = 54.8;
starget = 10;

pm = 0.1;
pout = (1-pm)/2;
p = [pout pm pout];
dmx = meantarget-mediantarget;
v = starget^2;
a = dmx;
b = v+dmx^2;

delta = 2*b/pout-a^2/pout^2;
xhi = (a/pout+sqrt(delta))/2;
xlo = a/pout-xhi;

x3 = mediantarget + [xlo 0 xhi];
c = cumsum(p);
c = c/c(end);
n = 1e6;
x = rand(1,n);
[~,i] = histc(x, [0 c]);
x = x3(i);

meanx = mean(x)
medianx = median(x)
stdx = std(x)