Precision, why do Matlab and Python numpy give so different outputs?

Maybe the difference comes from the mean and std calls. Compare those first.

There are several definitions for std, some use the sqaure root of

1 / n * sum((xi - mean(x)) ** 2)

others use

1 / (n - 1) * sum((xi - mean(x)) ** 2)

instead.

From a mathematical point: these formulas are estimators of the variance of a normal distributed random variable. The distribution has two parameters sigma and mu. If you know mu exactly the optimal estimator for sigma ** 2 is

1 / n * sum((xi - mu) ** 2)

If you have to estimate mu from the data using mu = mean(xi), the optimal estimator for sigma**2 is

1 / (n - 1) * sum((xi- mean(x))**2)

To answer your question, no, this is not a problem of precision. As @rocksportrocker points out, there are two popular estimators for the standard deviation. MATLAB's std has both available but as a standard uses a different one from what you used in Python.

Try std(Z,1) instead of std(Z):

Za = (Z-repmat(mean(Z),500,1)) ./ repmat(std(Z,2),500,1);Za(1)sprintf('%1.10f', Za(1))

leads to

Za(1) = 21.1905669677

in MATLAB. Read rockspotrocker's answer about which of the two results is more appropriate for what you want to do ;-).

According to the documentation of std at SciPy, it has a parameter called ddof:

ddof : int, optional
Means Delta Degrees of Freedom. The divisor used in calculations is N - ddof, where N represents the number of elements. By default ddof is zero.

In numpy, ddof is zero by default while in MATLAB is one. So, I think this may solve the problem:

std(Z,ddof=1)