Are "uncertainties" in Heisenberg Uncertainity just standard deviations? Can someone confirm that the uncertainties in Heisenberg's uncertainty relation are really just standard deviations based on the expectation values?
For example, the $\Delta x$ can be computed by $\sqrt{(\bar x^2 - (\bar x)^2)}$, and the $\Delta p$ can be computed by $\sqrt{(\bar p^2 - (\bar p)^2)}$, then $\Delta x \Delta p \geq \hbar /2$.
 A: The quantity on the right side of the expression for the product of uncertainties basically depends on the mathematical definition of "uncertainty" one used. Without a rigid mathematical definition of this quantity one often just say that the product of uncertainties in position and momentum is of the order of Planck constant (or the reduced Planck constant; since they are proportional to each other it does not matter)
$$ \Delta x \Delta p \sim \hbar \, .$$
To make the statement more precise, one have to define what is actually meant by "uncertainty" and usually uncertainties are defined as the standard deviations.
$$
\Delta x = \sigma_x = \sqrt{\langle (x - \langle x \rangle)^2 \rangle} = \sqrt{\langle x^2 \rangle - \langle x \rangle^2} \, , \\
\Delta p = \sigma_p = \sqrt{\langle (p - \langle p \rangle)^2 \rangle} = \sqrt{\langle p^2 \rangle - \langle p \rangle^2} \, , \\
$$
where angle brackets $\langle \phantom{x} \rangle$ stand for the expectation value. And with these definitions one can indeed prove that
$$
\Delta x \Delta p \geq \hbar / 2 \, .
$$
Actually, we usually even favour the usual notation for the standard deviation in this case to avoid an ambiguity
$$
\sigma_x \sigma_p \geq \hbar / 2 \, .
$$

In fact, you can prove a more general Robertson uncertainty relation that for any two observables $A$ and $B$ represented by self-adjoint operators $\hat{A}$ and $\hat{B}$ 
$$
    \sigma_{A}\sigma_{B} \geq \left| \frac{1}{2i}\langle[\hat{A},\hat{B}]\rangle \right| = \frac{1}{2}\left|\langle[\hat{A},\hat{B}]\rangle \right|.
$$
Now, for $x$ and $p$ it is postulated by so-called canonical commutation relation that $[\hat{x},\hat{p}]=i\hbar$, thus, leading to 
$$
    \sigma_x \sigma_p \geq \hbar / 2 \, .
$$

Note that the term "expectation value" has a number of other names and few possible notations. In physics we almost exclusively use the term expectation value and denote it with angle brackets $\langle \phantom{x} \rangle$, so I will continue to do so.
So, both position and momentum are continuous random variables, the expectation value of which is defined as follows,
$$
    \langle X \rangle = \int_{-\infty}^\infty x f(x) \mathrm{d}x \, , \tag{1} 
$$ 
where $f(x)$ is the probability density function. In quantum mechanics, the probability density function is given by the square modulus of the corresponding wave function. I say "corresponding" because if you would like to calculate expectation values according to the above mentioned mathematical definition, yo have to use the square modulus of the position-space wave function $\Psi(x, t)$ for the case of position 
$$
    \langle x \rangle = \int_{-\infty}^\infty x \left|\Psi(x, t)\right|^2 \mathrm{d}x \, ,
$$ 
and the square modulus of the momentum-space wave function $\Phi(p, t)$ for the case of momentum
$$
    \langle p \rangle = \int_{-\infty}^\infty p \left|\Phi(p, t)\right|^2 \mathrm{d}x \, .
$$
Usually though, since both position- and momentum-space wave functions contain the same information about a quantum system, we prefer to work with just one of them (most often with the first one) and calculate the expectation value of any observable $A$ represented by a self-adjoint operator $\hat{A}$ as postulated in quantum mechanics
$$
    \langle A \rangle = \int_{-\infty}^\infty \Psi^{*}(x, t) \hat{A} \Psi(x, t) \mathrm{d}x \, , \tag{2}
$$
rather than by using its mathematical definition mentioned above. Since position operator $\hat{x}$ in position space is just a multiplicative one, $\hat{x} \Psi(x, t) = x \Psi(x, t)$, (2) leads to the exact same expression as (1). For momentum operator $\hat{p}$ the situation is a little bit more involved, but definitions (1) and (2) can be shown to give exact same result for the expectation value of momentum, as well as any other observable.

Anyway, irregardless of the way we calculate expectation values of $x$ and $p$, the crucial thin is that, as we have already said, angle brackets $\langle \phantom{x} \rangle$ in the definition of standard deviations stand for the expectation value. And both expectation value, and consequently, standard deviations are theoretically calculated quantities which can be calculated once we known the state of a system, i.e. the wave function $\Psi(x, t)$.
If, however, we would like to experimentally test the uncertainty principle, we have to do a number of simultaneous measurements of position and momentum on an ensemble of identically prepared quantum systems all in the same state $\Psi(x, t)$.
As a results of such experiment on the ensemble of $n$ quantum systems, we will have sets of $n$ simultaneously measured positions $\{ x_i \}_{i=1}^{n}$ and momenta $\{ p_i \}_{i=1}^{n}$. 
With these results at hands we could calculate the average of the results, usually referred to as the sample average, as follows,
$$
    \overline{x}_{n} = \frac{1}{n}\sum_{i=1}^{n} x_i \, , \quad
    \overline{p}_{n} = \frac{1}{n}\sum_{i=1}^{n} p_i \, ,
$$
and by the law of large numbers, which (somewhat loosely) states that the sample average converges to the expectation value,
$$
    \overline{x}_n \to \langle x \rangle \quad \textrm{and} \quad
    \overline{p}_n \to \langle p \rangle \quad \textrm{when} \quad
    n \to \infty. 
$$
use them to estimate expectation values, and consequently, standard deviations. And as more and more trials are performed we will get closer and closer to the theoretically predicted numbers.
The only problem is that the instrument for such experiment should be incredibly accurate in independently measuring position and momentum. But no matter how accurate our measuring device is in independently measuring position and momentum, the product of uncertainties in position and momentum will always be non-zero. And this has nothing to do with the inaccuracy of independent measurements of position and momentum, rather it is the way nature works. Even with a hypothetical perfect instruments which can independently measure both position and momentum with an infinite accuracy, we can not measure them both simultaneously with that accuracy. There is a theoretical limit.
A: It depends what you mean as "standard deviations" if you mean the classical statisctical mechanics definition arising from the assumption of randomness, giving rise to the normal distribution,  the answer is NO.
The distributions leading to the wavefunctions  are deterministic, from the quantum mechanical solutions using the potential and boundary conditions of the problem, since the  Heisenberg uncertainty mathematically evolves from the commutators of quantum mechanical operators.

In physics, this is an important overarching principle in quantum mechanics. The commutator of two operators acting on a Hilbert space is a central concept in quantum mechanics, since it quantifies how well the two observables described by these operators can be measured simultaneously. The uncertainty principle is ultimately a theorem about such commutators, by virtue of the Robertson–Schrödinger relation. In phase space, equivalent commutators of function star-products are called Moyal brackets, and are completely isomorphic to the Hilbert-space commutator structures mentioned.

EDIT after the "hold".
The question says:

are really just standard deviations based on the expectation values?

Two different frameworks are mixed in this question, the classical statistics with"standard deviations" and quantum mechanics with "expectation values".
There is a distinction on delta(x), delta(p) entering in the Heisenberg Uncertainty Principle, HUP, and in the standard deviations accompanying measurements.  The wikipedia quote :

The formal inequality relating the standard deviation of position σ_x and the standard deviation of momentum σ_p 

is talking about measurements of position and momentum and refers to the normal statistical definition of standard deviation as coming from the underlying uncertainties of the experimental measurement. Experimental measurements depend on the accuracy of our instruments, and classically can be made as small as possible by technology.
In contrast, the  HUP states  that no matter how good the technology, the value of momentum and position are tied up with the relationship. They are tied up with the probabilistic formulation of Quantum mechanics, through expectations values which due to the nature of the operators involved are limited through the HUP. 
So if one looks at the measurements and their errors, one can treat them as classical standard deviations. If one looks at the HUP relationship the measurements are constrained by it no matter how small the standard deviations of measurement. Looking at it experimentally, it is as if a systematic error enters for certain pairs of variables ( systematic errors are treated  a differently in careful measurements, than the standard errors and do not obey a normal distribution)
I think it is an important distinction because the confusion is probably the beginning for looking for classical underpinnings of quantum mechanics. The probabilities for finding a particle at x are controlled by quantum mechanical probabilistic functions, and not by underlying statistical conditions. 
A: They are in the sense that they measure the mean(!) derivation from a expectation value, which is the definition of standard derivation.
The Definition of the sandard derivation is  reproduced one to one in QM terms. An operatot $\Omega$ with eigenvalues $\omega_i$ and eigenvectors $|\omega_i \rangle$ and a probability of $P(\omega_i)$ to measure $\omega_i$.  Has the expectation value:
$\langle \Omega \rangle := \sum \limits_i P(\omega_i)\omega_i=\sum \limits_i |\langle \omega_i|\psi\rangle|^2 \omega_i$
which can easyly be shown to be equal to the more usual $\langle \Omega \rangle:=\langle \Psi | \Omega |\Psi \rangle$
With this you get to the Deifnition of the uncertanty (standard derivation) (as in Shankars Principles of Quantum mechanics) as the difference of each value $\omega_i$ to the expectation value weight by the probability of measuring it (squared to get rid of the sign).
As in any other function of a statistically distributed variable its expectation value is given by: 
$\Delta \Omega  = [\sum \limits_i P(\omega_i)(\omega_i-\langle \Omega \rangle)^2]^{1/2}\\
=[\underbrace{\sum \limits_i P(\omega_i) \omega_i^2}_{\langle \Omega^2 \rangle}- 2 \langle \Omega \rangle \underbrace{\sum \limits_i P(\omega_i)\omega_i}_{=\langle \Omega \rangle}+\langle \Omega \rangle]^{1/2}
\\=[\langle \Omega^2 \rangle-\langle \Omega \rangle^2]^{1/2}$ 
which is the classical form of the standard derivation.
A: They are standard deviations in the following sense. Let $\omega$ be a state, i.e. a positive, linear and normalised map from observables to $\mathbb R$. Then for any observable $A$ define
$$\Delta_\omega(A)^2 := \omega((A-\omega(A))^2).$$
This is well defined because the new observable $(A-\omega(A))^2$ is legitimate functional calculus on a self-adjoint element from the C*-algebra of observables, and indeed it is an element in the unital C*-algebra generated by $A$ itself. Hence $\Delta_\omega(A)^2$ can be interpreted as a variance, and since this is just a real number you can take its square root and declare it to be the standard deviation. If you then follow von Neumann's approach to quantum mechanics, where the value of $\omega$ on $A$ is defined as the average of the outcomes of measurements of $A$ over a statistically significant number of exact copies (ensemble) of the same system in the same state, you can then realise that this really is the usual standard deviation.
