Suppose we have $n$ numbers $x_1, x_2, \ldots, x_n$ ranging from $a$ to $b$, meaning
that $x_i = a$ for at least one $i$, $1 \leq i \leq n$, and $x_j = b$ for at least
one $j$, $1 \leq j \leq n$. Duplicates are allowed, meaning that two or more of the
$n$ numbers could possibly have the same value in $[a, b]$. Define the mean $\bar{x}$ and the variance $\sigma_x^2$ of the set of $n$ numbers as
$$\bar{x} = \frac{1}{n}\sum_{i=1}^n x_i, ~~
\sigma_x^2 = \frac{1}{n}\sum_{i=1}^n (x_i - \bar{x})^2
= \left(\frac{1}{n}\sum_{i=1}^n x_i^2\right) - \bar{x}^2.$$
For convenience, let us change the set to $y_1, y_2, \ldots, y_n$ where
$y_i = x_i - a$, so that the new set has values ranging from $0$ to $b-a$.
The mean $\bar{y}$ is just $\bar{x}-a$, while the variance is unchanged:
$\sigma_y^2 = \sigma_x^2$. Now, it is shown in the answers to
this question on stats.SE
that the ratio
$\sigma_y/\bar{y} = \sigma_x/(\bar{x}-a)$ can be no larger than $\sqrt{n-1}$.
Note that the value does not depend on $b$ at all
You don't say what the value of $n$ is, but given the numbers in your
answer, the upper bound on the standard deviation is
$$\sigma_x = 554 \leq (464-100)\sqrt{n-1}$$
which is certainly satisfied except in the unusual circumstance
that there are only $3$ peers in the experiment and thus
only $3$ numbers $x_1$, $x_2$, and $x_3$ are being described
in terms of mean/standard-deviation/min-max: certainly overkill!
In summary, there are no obvious problems with the standard deviation
being larger than the mean.
General Solution
To compute mean, variance, and standard deviation you only need to keep track of three sums $s_0, s_1, s_2$ defined as follows for a set of values $X$:
$$(s_0, s_1, s_2) = \sum_{x \in X} (1, x, x^2)$$
In English, $s_0$ is the number of values, $s_1$ is the sum of the values, and $s_2$ is the sum of the square of each value. Given these sums, we can now derive mean (average) $\mu$, variance (population) $\sigma^2$, and standard deviation (population) $\sigma$:
$$\mu = \frac{s_1}{s_0} \qquad \sigma^2 = \frac{s_2}{s_0} - \left(\frac{s_1}{s_0}\right)^2 \qquad \sigma = \sqrt{\frac{s_2}{s_0} - \left(\frac{s_1}{s_0}\right)^2}$$
In English, the variance is the average of the square of each value minus the square of the average value.
Your particular case
You have $s_0, \mu, \sigma$, so you need to compute $s_1$ and $s_2$ by solving the above for those variables:
$$s_1 = s_0\mu \qquad s_2 = s_0\left(\mu^2 + \sigma^2\right)$$
Once you have $s_0, s_1, s_2$ for each data set, aggregation is just a matter of adding the corresponding sums together and deriving the desired aggregate values from those sums.
Variance Equation Derivation
We start with the standard equation for variance (population) and go from there:
$$\sigma^2 = \frac{1}{n}\sum_{x \in X} \left(x - \mu\right)^2
= \frac{1}{s_0}\sum_{x \in X} \left(x - \frac{s_1}{s_0}\right)^2$$
$$= \frac{1}{s_0}\sum_{x \in X} \left(x^2 - 2x\frac{s_1}{s_0} + \left(\frac{s_1}{s_0}\right)^2\right)
= \frac{1}{s_0}\sum_{x \in X} x^2 - 2\frac{s_1}{s_0^2}\sum_{x \in X} x + \frac{s_1^2}{s_0^3}\sum_{x \in X} 1
$$
$$= \frac{1}{s_0}(s_2) - 2\frac{s_1}{s_0^2}(s_1) + \frac{s_1^2}{s_0^3}(s_0)
= \frac{s_2}{s_0} - 2\frac{s_1^2}{s_0^2} + \frac{s_1^2}{s_0^2}
= \frac{s_2}{s_0} - \left(\frac{s_1}{s_0}\right)^2
$$
Best Answer
It is not possible to give an optimal answer to your question without definitions of terminology 'normalized finger flex' and 'absolute finger flex'. (These are not terms immediately clear to a statistician, and I am curious about the 'independence' you mention.) What do these two terms mean in terms of finger flexing and why are they important to you?
Also, it would be helpful to see what the data look like: are neighboring values in the sequence of 1000 near to each other (for instance, suggesting a sinusoidal curve), and about how many finger flexes are there in a 10-second period.
Lacking this information I have to guess, and the results of guessing may not be useful to you. It is a little like asking a physician to diagnose a disorder if you say on the phone, "Sometimes the twitch in my left eye is very disturbing."
If the data move fairly regularly up and down about 10 times in 10 seconds, then you might look for several highest local maxima (about 10) and average, and to look for several lowest local minima and average them. Do individual flexes result in widely different max and min values? Do you want to average over all individual flexes, or only get averages for the several most extreme. (By any definition, there would be ways to automate this search pretty well, so after deciding an a strategy by by looking at a couple of data streams of 1000, you could have a computer search additional such streams.)
Another approach (simpler, more general, possibly less satisfactory) would be just to average the top 50 values (about) to get the 'average global maximum' and similarly for the bottom 50 to get the 'average global minimum'. (Of course, "50" is a mildly educated guess without seeing the data.)
The reason I'd like to know what you mean by 'absolute flex value' and how you derive that from data, is that your word 'independent' indicates you want the 'normalized finger flex' to carry fundamentally different information than the 'absolute flex value' and I cannot be sure that is true without knowing what 'absolute flex' means.
If this answer sets you on the right track, then please so indicate. If not, please leave a Comment with more information, and maybe I (or someone else) can do better.