One concept essential to understand this process is that whatever we are measuring in the lab is basically a sample out of the large number of experiments needed to determine the actual mean $\mu$. So, the mean obtained from the experiment, i.e. the mean of a value measured $N$ times is basically a sample mean.
We can get the actual mean by repeating the measurement infinite number of times, which gives us a normal curve with mean $\mu$ and variance $\sigma^2$. Since this is not feasible in practice, we take a finite number of measurements $X_1, X_2, ... 915
, X_N$ in the lab, and take the sample mean $\bar{X}$, defined as: \begin{equation} \bar{X}= \frac{X_1+X_2+...+X_N}{N} \end{equation}
This is stated as the 'mean obtained in the experiment'. It is worth-noting that $\bar{X}$ is not a certain value but a random variable, and has a standard deviation ($\sigma_\bar{X}$) associated with it, which is of our interest.
Each of the individual measurements $X_1, X_2, ... , X_N$ are also normal random variables, independent and identically distributed with mean $\mu$ and variance $\sigma^2$. It should be made clear that this standard deviation ($\sigma$) is different from standard deviation of the (sample) mean ($\sigma_\bar{X}$).
Since all $X_i$'s are identical and independent, variance of $N\bar{X}\left(=X_1+X_2+...+X_N\right)$ is \begin{equation} \sigma^2_{N\bar{X}}=N\sigma^2 \implies N^2\sigma^2_{\bar{X}}=N\sigma^2 \end{equation}
\begin{equation} \therefore \sigma^2_{\bar{X}}=\frac{\sigma^2}{N} \text{, or } \sigma_{\bar{X}}=\frac{\sigma}{\sqrt{N}} \end{equation}
The key motivation behind this error-analysis is considering the whole set of measurements (needed for actual mean) to be the population, and the set of finite recorded observation (taken in the lab) as a sample. Then, this is nothing but finding the standard error in the sample mean.
One of the implication of this formula is that: "To decrease the error in the mean by $k$, the number of observation taken should be increased to $k^2$." 915