Noise quantification from sampled data
Yes, those equations are correct and the same.

In the first equation, you just replace \$x\$ with its expression following from \$y=\bar{y}+x\$.

For the second equation, you can do the following:
The square of the rms value of y is equal to
$$
\begin{aligned}
rms(y)^2\
&amp;= \frac{1}{N}\sum_{n=1}^N (y[n])^2\\
&amp;= \frac{1}{N}\sum_{n=1}^N (\bar{y} + x[n])^2 \\
&amp;= \frac{1}{N}\sum_{n=1}^N (\bar{y}^2 + 2 \bar{y} x[n] + (x[n])^2)
\end{aligned}
$$

The first term (\$\frac{1}{N}\sum\limits_{n=1}^N \bar{y}^2\$) is just equal to \$\bar{y}^2\$.

The second term (\$\frac{1}{N}\sum\limits_{n=1}^N 2 \bar{y} x[n]\$) is zero since you can take the constant factor \$2 \bar{y}\$ out of the sum and \$\bar{x}=\frac{1}{N}\sum\limits_{n=1}^N x[n]\$ is zero.

The third term (\$\frac{1}{N}\sum\limits_{n=1}^N (x[n])^2\$) is the square of the rms value of x.

Thus
$$
rms(y)^2 = \bar{y}^2 + rms(x)^2
$$
and thus
$$
rms(x) = \sqrt{rms(y)^2 - \bar{y}^2}
$$



The example of Marcus Müller in the comments is still valid:
\$\bar{y}=2\$, \$rms(x)=0\$, and thus \$rms(y)=2\$. Note that an rms value is not the same as a standard deviation if the signal (\$y\$ here) has a non-zero mean value.