For examples, if I measure length with a meter stick with the smallest unit of 1mm, can I use the uncertainty of 0.5mm in this formula: $$\Delta C = \sqrt{\left(\frac{\partial C}{\partial x_1}\Delta x_1\right)^2 + \left(\frac{\partial C}{\partial x_2}\Delta x_2\right)^2 + \hspace{0.3cm}...}$$ The reason I ask is that according to my understanding the uncertainties used in the formula carry with them a probabilistic nature. Assume the measurements are distributed normally, then $\Delta x_1,\Delta x_2,...$ can be thought of as 1 standard deviation from the mean, which gives the range that the true value can fall into 68% of the time. In contrast, the 0.5mm estimated is the maximum error that we can have(true value fall into this range 100% of the time). So how can I fit this uncertainty of 0.5mm into this formula? Assume I use 0.5mm directly into the formula, what will the result ($\Delta C$) mean? I doubt it can be considered to be either the maximum error possible or 1 standard deviation from the mean
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