First of all you have to tell us if the 100 measurements are independent of not. If they are, the sum of your 100 measurements follows a binomial distribution known with mean and variance
In other words, if $X$ is the rv describing the measurement of a device,
$$Y=X_1+X_2+\dots+X_{100}\sim B(100;0.5)$$
and thus
$$\mathbb{V}[Y]=\frac{100}{4}$$
that is
$$\sigma_Y=5$$
here $Y$ represent the total "1" you have in your 100 measurements.
The random variable describing the total "0" can be written as
$$Z=100-Y$$
with evidently the same st dev.
To calculate $V(Y)$ you have two ways:
Given that your single measurement can be 0 or 1 with the same probability 0.5, the random variable is a bernulli with parameter 0.5.
Thus $V(X_1)=0.5\times0.5=0.25$ and obviously $V(Y)=V(X_1)+\dots+V(X_{100})=\frac{100}{4}=25$
Another way is to observe that the sum of independent bernulli is a binomial with variance $V(Y)=np(1-p)=100\times\frac{1}{2}\times\frac{1}{2}=25$
x1+...x100, how do we implyV[Y] = population size / 4? If you could refer me to the concept, I'd greatly appreciate that ! – Michael Apr 19 '21 at 11:22