Question: I have a data set with a set of features. Some of them are binary $(1=$ active or fired, $0=$ inactive or dormant), and the rest are real valued, e.g. $4564.342$.

I want to feed this data to a machine learning algorithm, so I $z$-score all the real-valued features. I get them between ranges $3$ and $-2$ approximately. Now the binary values are also $z$-scored, therefore the zeros become $-0.222$ and the ones become $0.5555$.

**Does standardising binary variables like this make sense?**

Answer: 
One nice example where it can be useful to standardize in a slightly different way is given in section 4.2
of Gelman and Hill (<http://www.stat.columbia.edu/~gelman/arm/>). This is mostly when the interpretation of the coefficients is of interest, and perhaps when there are not many predictors.

There, they standardize a binary variable (with equal proportion of 0 and 1) by
$$
\frac{x-\mu\_x}{2\sigma\_x},
$$
instead of the normal $\sigma$. Then these standardized coefficients take on values $\pm 0.5 $ and then the coefficients reflect comparisons between $x=0$ and $x=1$ directly. If scaled by $\sigma$ instead then the coefficient would correspond to half the difference between the possible values of $x$.
What do you want to standardize, a binary random variable, or a proportion?

It makes no sense to standardize a binary random variable. A random variable is a function that assigns a real value to an event $Y:S\rightarrow \mathbb{R} $. In this case 0 for failure and 1 to success, i.e. $Y\in \lbrace 0,1\rbrace$.

In the case of a proportion, this is not a binary random variable, this is a continuous variable $X\in[0,1]$, $x\in \mathbb{R}^+$.