ZScore threshold and low values time-series Example of z-score computation:
1 - E.g.
Time-series: [0, 0, 0, 0, 1]
Current: 1
Mean: 0.2
Std: 0.44721
Z = (1 - 0.2) / 0.44721 ~= 1.7888

2 - E.g.
Time-series: [100, 100, 100, 100, 1000]
Current: 1000
Mean: 280
Std: 402.49224
Z = (1000 - 280) / 402.49224 ~= 1.7888

How can we represent that in case of example 2 - the variation is much more significant than the example 1 because of a higher mean?
Is the only solution is to have z-score thresholds based on the mean value?
 A: It's not clear why you think example 2 has 'more significant variation'. In fact, Example 1 has a larger coefficient of variation (CV); i.e. 
\begin{gather*}
CV(Example1)=0.44721/0.2=2.24,\\
CV(Example2)=402.49/280=1.44.
\end{gather*}
So while example 2 has larger $scale$, it does not really show more $variation$. And as you showed with the deviation scores, both examples have equally anomalous final values when scaled. 
A: Along the same lines as gregory_britten and Glen_b, it depends on what you are measuring.  Here it seems you are comparing two different distributions.  Can you say that the jump from 100 to 1000 in your example 2 is "more significant" than the jump in [1, 1, 1, 1, 10]?
To compare the two, you need to compute the mean and std of the joint distribution of a larger recent section of your time series and compute the z-score from that.  For example: 
Time-series: [0, 0, 0, 0, 1, 100, 100, 100, 100, 1000]
Current: 1000
Mean: 140.1
Std: 290.5
Z = (1000 - 140.1) / 290.5 ~= 2.96

Fixing the mean and std are standards in anomaly detection or series prediction, and you can do it a few ways:


*

*Compute them every time over the whole series, like above. This gives you the true standouts given all data.

*Compute them from a reference baseline and use that for all future points.  This gives you baseline drift detection and stringent criteria for anomalies.

*Compute them from a time window.  This let's you adjust automatically for baseline drift over time if you compute it over the last N data points, for example the last hour or day (depends entirely on what type of signal you are analyzing).  This is what you are doing in your example.  The baseline has drifted from example 1, and so the jump from 100 to 1000 is not really meaningful given the drift.

A: 
How can we represent that in case of example 2 - the variation is much more significant than the example 1 because of a higher mean?

Significance means, in general, the probability of the result when some hypothesis is right (often a null hypothesis). You could see it as a measure for anomalies.
Thus significance, in order to mean something, you need (1) a hypothesis (or more hypotheses plus prior believes about them in the case of Bayesian analysis) and (2) an assumption of the distribution of the data when the hypothesis is right.

In your case the the test would often be presented in terms of a test for a hypothesis that would be a null result (no effect). E.g. you test whether "the variation is the same". If you find a result that would be an anomaly (ie a significant result) then you consider the 'alternative hypothesis', there is not 'no effect'. 
In your case, if you assume that those data are normal distributed then you could use the F-test for equality of variances. 
However, possibly your data is not supposed to be having equal variance in this sense, and you are asking for a different concept of equal variance. Then your data should be scaled somehow in order for a comparison of variance to make sense. 
For instance, if you measure the height of two times five people and estimate the variance of those people in inches for one group and in centimeters for the other than you will likely get a significant result against the hypothesis that those variances are the same. But it would be a nonsense test to make the comparison of variances with different scales.
