Why are uncertainties calculated as an average? Imagine there is a cube with length $l = (10 \pm 0.2) \; \text{cm}$. The volume $V$ would be $l^3 = 1000 \; \text{cm}^3$ and the uncertainty in $V$, $\Delta V$ would be $\frac{10.2^3 - 9.8^3}{2} = 60.008 \; \text{cm}^3$. I would then write $V = (1000 \pm 60.008) \; \text {cm}^3$.
Now, calculating the forward uncertainty $\Delta V_f$ and backward uncertainty $\Delta V_b$ yields the following:
$$\Delta V_f = 10.2^3 - 10^3 = 61.208 \; \text{cm}^3$$
$$\Delta V_b = 10^3 - 9.8^3 = 58.808 \; \text{cm}^3$$
Hence, isn't $V = (1000 \pm 60.008) \; \text {cm}^3$ not a true representation of reality since the maximum value of $V$ is $(1000 + 61.208) \; \text{cm}^3$ and not $(1000 + 60.008) \; \text{cm}^3$? Shouldn't the notation for uncertainty be something like $(1000 \pm (61.208, 58.808)) \; \text{cm}^3$ instead? Wouldn't the discrepancy be extremely large for $\Delta V_f >>> \Delta V$?
 A: This question has been answered (in the comments). I am writing this so that this question no longer remains in the unanswered section.
Here is a brief summary of the answer (from the comments):

*

*Yes, $V = (1000 \pm 60.008) \; \text{cm}^3$ is not a true representation of reality. $V = (1000 \pm (61.208, 58.808)) \; \text{cm}^3$ is more precise.

*Using $\Delta V$ as $60.008 \; \text{cm}^3$ usually stems from the "physicist" point of view, in contrast to the "interval arithmetic" approach which keeps track of any asymmetries.

*Generally $\Delta V$ is so tiny such that $\Delta V_f - \Delta V_b \approx 0$. In this case, $\Delta V$ can be well approximated by $V^{\prime}(l)\cdot \Delta l$, assuming that $V(l)$ is differentiable.

*Avoiding dividing by $2$ in $\frac{10.2^3 - 9.8^3}{2}$ can be more useful since it can give a better understanding of the range in a practical setting.

Credits: Ian, FShrike
A: You say you have a cube whose edge length is $(10 \pm 0.2)\ \mathrm{cm}.$
Did you actually mean to say that the edge length can be anywhere
between $9.8$ cm exactly and $10.2$ cm exactly?
That is, it is possible that the length is $9.80001$ cm or $10.19999$ cm,
but it is absolutely impossible that the length is $9.79999$ cm
or $10.20001$ cm?
If that is what you meant, then it may indeed make sense to say that the volume is strictly bounded by $941.142 < V < 1061.208.$
That is how we mathematically interpret the information that
$9.8 < \ell < 10.2$ and $V = \ell^3.$
However, the notation $\ell = 10 \pm 0.2$ is not generally used in mathematics
to signify that $9.8 < \ell < 10.2.$
Instead, if you want a compact notation for use in an exact mathematical context
that signifies that $9.8 < \ell < 10.2,$
try $\lvert \ell - 10 \rvert < 0.2.$
In most mathematical contexts, for example when solving a quadratic equation,
$\ell = 10 \pm 0.2$ signifies that either $\ell = 9.8$ or $\ell = 10.2,$
and that $\ell$ is not any other value in between those two possibilities.
The people who write $\ell = (10 \pm 0.2)\ \mathrm{cm}$ to signify that
$\ell$ is "approximately $10$ cm but possibly up to $0.2$ cm larger or smaller"
are physicists, chemists, engineers, and other practical people.
Those same people also usually keep track of the significant digits
in their calculations, and in $10 \pm 0.2$ you have only three significant digits
(if even that many; really it should be written $10.0 \pm 0.2$).
Hence, from the point of view of the kind of person who normally would write
$\ell = (10.0 \pm 0.2)\ \mathrm{cm}$ to express uncertainty among more than two possible exact values,
the correct answer is that $V = (1000 \pm 60)\ \mathrm{cm}^3,$
or more explicitly $V = (1.00 \pm 0.06)\times 10^3\ \mathrm{cm}^3.$
Such people do sometimes recognize different error bounds above and below the nominal value, however.
For example, if $\ell = (10.0 \pm 0.5)\ \mathrm{cm},$ then
$V = \left( 1.00 \begin{matrix}\scriptsize{}+0.16 \\[-1ex]
      \scriptsize{}-0.14\end{matrix}\right) \times 10^3\ \mathrm{cm}^3.$
If your application is such that you are writing things like $10.0 \pm 0.2$
to signify a range of uncertainty and you need to propagate an asymmetric range,
you could adopt that notation.
Note that in the notation $1.00 \begin{matrix}\scriptsize{}+0.16 \\[-1ex]
      \scriptsize{}-0.14\end{matrix}$
it is very clear which number is added for the upper bound and which is subtracted for the lower bound, unlike in the notation $1000 \pm (61.208, 58.808).$
