Why do I ask this?
In a video they discuss "Uncertainty in measurements"; just have to watch 2:30 minutes to get context.
For that they compare an analog with a digitial scale.
On the analog scale they say, that in 130 pounds 1,3 are certain and 0 is a uncertain digit (3 sigfigs).
Compared to that on the digital scale 133.6 pounds 1,3,3 are certain and 6 is uncertain (4 sigfigs).
They determine the position of the uncertain digit in the number of sigfigs as the decimal place, in which the result of $\frac{\Delta x}2$ lies.
They define $resolution=\frac{\Delta x}2$.
Now I get the following contradiction:
Intuitive resolution=0.1 pounds (display of digital scale).
Mathematical resolution=1 pounds
Why mathematical resolution 1 pounds?:
$\frac{1}{2}pounds$ is the decimal tenths place for the uncertain digit in the sequence of sigfigs.
I would than note a reading this way:
$133.6pound \pm 0.5pounds$ and this has 4 sigfigs.
To collect the communities opinion about this topic I had asked the following question instead:
If we have a digital scale, with 100g increments.
When we read 133.1 kg, does it mean that the true value is
$133.1 kg \pm 0.05kg$ or
$133.1 kg \pm 0.1kg$ or
$133.1 kg \pm 0.5kg$
I tend to the last guess, because the display resolution of 0.1 kg doesn't reflect the measure resolution of $\Delta x = 1 kg$.
If that guess is true, then the number of sigfigs is 4, isn't it?
So the rule is:
y is normalized notation!
Example with $y=0.5=5 \times 10^{-1}$.
$x \pm \frac{resolution}{2} = x \pm (y \times 10^t)$
and the #sigfigs is:
$\lceil{\log_{10}(\frac{x}{10^t})}\rceil$
Example: $\lceil{\log_{10}(\frac{133.1}{10^{-1}})}\rceil=4$
Thx in advance.