Using normrnd for generating natural values (without decimal values)

I would like to generate data with average of 27 and standard deviation of 1.41, but I would like the data have no decimal values, ex to be like this 12, 24, 27 ... . Could you please help me how I can do so?

[Walter Roberson]

Do you care about the distribution of the values? You cannot get normally distributed values that way.

[Sahar khalili]

No, the mean and standard deviation are just important for me.

[Walter Roberson]

You cannot use the fact that the sum of identically distributed uniform distribution approximates normal. You can generate 54 binary values and sum them. That will have a mean of 27, but the std will be 3.67.

[Sahar khalili]

As I mentioned I do not care about the distributation of data, I need 6 numbers that just have the mean of 27 and std deviation of 1.41, and what matters for me is just data has no decimal values.

Answer 1

[Sam Chak]

@Sahar khalili

How about this data set {25, 26, 27, 27, 28, 29}?

A = [25, 26, 27, 27, 28, 29]

A = 1×6

25 26 27 27 28 29

M = mean(A)

M = 27

S = std(A)

S = 1.4142

[Walter Roberson]

I calculate that this or a permutation of this is the only set that works.

[Sam Chak]

Wow! I actually don't know the exact steps to find the data set. I just based on the desired mean 27 and used heuristic trial-and-error approach to gradually distribute 6 integers away from the mean, and found this data set. Thanks @Walter Roberson for the confirmation.

[Walter Roberson]

std is sqrt(2). std() involves sqrt(N-1). N=6 so that is division by sqrt(5). In order for that to give sqrt(2) it follows that the numerator must be sqrt(10) and so the sum of exactly 6 squares of values minus mean must be 10.

Can any of the differences from the mean be 4 or more? No that would give a term of 4*4 which exceeds 10.

Can any of the differences be 3? 3*3 = 9 which is less than 10, but that requires that the total sum of squares for the other 5 terms is 1. But you cannot balance a difference of 3 with a single difference of 1 to get the right mean.

Can any of the differences be 2? Yes, make one be -2 and another be +2 and one be -1 and another be +1 and you get a mean balance and sum of squares 2*2 + 2*2 + 1*1 + 1*1 = 10, which works out.

Can exactly one be difference of 2? That contributes 4 to the sum of squares, leaving 6 to be distributed among the remaining 5 values. But under hypothesis none are 2 (squared) and so you have a pigeon-hole problem of distribution of 6 in 5 values that are 0 or 1, and that cannot fit.

Can all the differences be 0 or 1? No, sum of squares for 6 values like that cannot reach 10.

So we conclude that Sam's arrangement (possibly permuted) is the only solution.

[Walter Roberson]

[x1, x2, x3, x4, x5, x6] = ndgrid(27-4:27+4);
meanmask = (x1 + x2 + x3 + x4 + x5 + x6)/6 == 27;
sx1 = x1(meanmask); sx2 = x2(meanmask); sx3 = x3(meanmask);
sx4 = x4(meanmask); sx5 = x5(meanmask); sx6 = x6(meanmask);
sx = [sx1, sx2, sx3, sx4, sx5, sx6];
st = std(sx, [], 2);
inrange = st >= 1.405 & st < 1.415;

values_that_work = sx(inrange,:);

whos values_that_work

  Name                    Size            Bytes  Class     Attributes

  values_that_work      360x6             17280  double              

unique(sort(values_that_work, 2), 'rows')

ans = 1×6

25 26 27 27 28 29

360 matches... but to within permutations they are all the same.

Notice that, as predicted by my analysis, no deviations of 4 or 3 match, only -2, -1, 0, 0, +1, +2

[Sam Chak]

Many thanks to @Walter Roberson for explaning and showing the Permutation search procedure. The permutation-based method is effective.