Classical vs quantum definition of observables Let's suppose the probability distribution for position of a particle is very sharply centered about 0. Then surely I know it has little deviation about the mean? If it has little deviation about the mean then its momentum should be near 0 as its moving slowly. Inversely, the particle must be whizzing about to be measured over a large spread. How is this in line with the Heisenberg Uncertainty Principle where small position uncertainties yield large momentum uncertainties?
I think this question fails to appreciate momentum as an instantaneously defined quantity independent of position. In order to solve the problem, we must first define momentum rigorously. If q.m is to make predictions, surely it should be possible to define the rules of the game? We take classical observables such as momentum which are classically defined by measurements of time delay, position between two times etc. Time isn't even considered an observable in q.m, so what do all these* quantities actually mean.
*let's start with: 1) Time of measurement/arrival 2) position 3) momentum
 A: 
Let's suppose the probability distribution for position of a particle is very sharply centered about 0. Then surely I know it has little deviation about the mean?

Just because the probability distribution is sharply centered at 0 right now, doesn't mean it will continue to be sharply centered at that (or any) location in the near future.
The uncertainty principle is telling you that the more certain you are about the particle's location at some specific time, the less certain you can be about where it is going. That means beyond some limit, even if you have a particle located at a specific location, you can't actually keep the particle confined there for any period of time.
There is an analogous effect in optics where if you pass light through a narrow slit (ensuring you know the "location" of the wave as it passes through the slit) then the narrower the slit, the broader the diffraction pattern it makes after passing through (the less certain you are about the direction the light energy will travel after exiting the slit).
A: This answers version 2 before another question was added into the post.
It looks like you have mixed some classical physics into your quantum physics.
The uncertainties in position and momentum are not due to us just being ignorant about where the particle is and how fast it is moving. A small uncertainty in position doesn't mean the particle is slowly moving around, and a large uncertainty in position doesn't mean the particle is whizzing around. Before measurement, the particle does not have a well-defined position or momentum. There is no trajectory you can follow for the particle. Therefore, the uncertainties of these quantities doesn't determine what their values will actually be upon measurement, and before measurement they don't have set values at all.
The uncertainties are just standard deviations of repeated measurements. So if you say that "the probability distribution for position of a particle is very sharply centered about $0$", this means that if you prepared a bunch of particles in the same way and measured their position, you would find the distribution of measurements to be a tight distribution about $0$. The "position uncertainty" then would be the standard deviation of this distribution. The same thing holds true for momentum measurements.
The HUP then gives a lower bound on the product of these uncertainties: $\Delta x\cdot\Delta p\leq\hbar/2$. This is not due to some artifact of measurement, it isn't due to "limited information", and it doesn't mean a localized particle is "whizzing around". This is just something that must be true for all quantum states: If the uncertainty in position is sufficiently "small", then the uncertainty in momentum must be sufficiently "large" so that this inequality holds. But depending on what you mean by "small" and "large", you could have any of the 4 possible combinations of "small" and "large" position and momentum uncertainties as long as the inequality is valid.
So therefore, if all you know is that the position probability distribution is sharply centered about $0$ with some uncertaintly $\Delta x$, then all you can guarantee is that $\Delta p$ must be at least as large as $\hbar/(2\Delta x)$. You don't know anything about the particle's actual momentum though (some interpretations of QM say the particle doesn't have an actual momentum until it is "measured").
A: The fact that your probability distribution is sharply centered around the zero actually restricts the range of possible outcomes for a measurement of the particle’s position. As a consequence, the standard deviation of position $\sigma_{x}$ associated to your probability distribution is “small”, which implies, according to the uncertainty principle formulated as
$\sigma_{x}\sigma_{p}\geq\frac{\hbar}{2}$
that the standard deviation of momentum $\sigma_{p}$ has to be large enough to at least make for a product of $\frac{\hbar}{2}$. In turn, this means that the momentum probability distribution is smeared enough.
