The 'meaning' of interval estimates is a controversial topic on
applied statistics. So there is no universally accepted answer to your
important question.
Let's just use a proposed sample of size $n = 31$ from a normal population with
unknown population mean $\mu$ and unknown variance $\sigma^2.$
Then $T = \frac{\bar X - \mu}{S/\sqrt{n}} \sim \mathsf{T}(n-1),$ so that
$P(-2.042 \le T = \frac{\bar X - \mu}{S/\sqrt{n}} \le 2.042 )=0.95.$
Here $\bar X$ and $S$ are the sample mean and variance, respectively.
Manipulating inequalities in the event, we get
$P(\bar X - 2.042\frac{S}{\sqrt{n}} \le \mu \le \bar X + 2.042\frac{S}{\sqrt{n}}) = 0.95.$ This is purely a probability statement. Specifically, it is a
probability statement about the behavior of the random variable $\bar X$ and $S$: the random interval $(\bar X - 2.042\frac{S}{\sqrt{n}}, \bar X + 2.042\frac{S}{\sqrt{n}})$ has a 95% probability of covering (including) the unknown constant $\mu.$
Now suppose we take the sample and obtain $\bar X = 21.3$ and $S^2 = 1.44.$ Then the random interval becomes $\left(21.3 - 2.042(0.1249), (21.3 + 2.042(0.1249)\right)$ or $(20.860, 21.740).$
But now we are dealing with observed quantities. According to the usual
frequentist interpretation of probability, this is no longer a probability
statement: Either the interval $(20.860, 21.740)$ includes $\mu$ or it
does not. Accordingly, the interval $(20.860, 21.740)$ is called a 95% confidence interval.
The confidence interval is a statement about the data. Over the long run,
we will obtain data so that
the manipulation in the emphasized paragraph will produce an interval
that includes the true population $\mu$ in 95% of such experiments.
The reason for calling the interval estimate a 'confidence' interval
instead of a 'probability' interval has to do with a strict interpretation
by frequentist statisticians of the word 'probability'.
Bayesian statisticians treat $\mu$ as a random variable, begin with a
'prior' distribution on $\mu$, combine the data with the prior distribution
to get a 'posterior' distribution, and use the posterior distribution to
get a probability interval for $\mu$ (some say a credible interval). If the prior distribution is
"flat" (containing little information), then the Bayesian and frequentist
interval estimates will be numerically very similar. But philosophies as to
the "meaning" of the interval estimate differ.
Both frequentists and Bayesians have their critics. Strictly speaking,
frequentists are are not saying anything about the experiment at hand--only
about what 'works' over the long run. A Bayesian is addressing the experiment
at hand, but needs to explain how
the prior distribution was obtained and what effect it has on the interval estimate.
Best Answer
You are told that several hundred measurements on a known mass have been made and the resulting SD is 18 ug. What this is telling you is that they have estimated the SD of the errors very well, and so you can assume that the true SD is, in fact, 18 ug ($\pm$ some small value). This means that your standard error will be $\frac{18}{\sqrt{50}}$ -- your standard error was 11% too high.
As a theoretical aside, we know the standard deviation is essentially exact because we are told to assume the errors are gaussian with no bias. Therefore, the standard deviation distribution for this quantity will be essentially normally distributed (for N in the hundreds), with mean equal to $b(N)\sigma$, where $b(N)$ is the bias as given in (6) of the link; with large N, this will be extremely close to 1 (e.g. it is 0.996 for N=200). The variance of this estimate will be as given in (10) in the link; for large N, say, N=200, the variance of this estimate will be $0.0025\sigma^2$ and the standard deviation will be $0.05\sigma$. What this means is that the estiamte of the standard deviation (i.e., 18 us) will be within 10% of the true standard deviation 95% of the time! So you know with a high degree of confidence that your standard devation is somewhere between 16.2 and 19.8 ug...and that is just for 200 measurements. If you have HUNDREDS e.g., 600 then the accracy will increase so that you know the standard devation within 6%.