B How to calculate a random measurement error?

AI Thread Summary
The discussion centers on two formulas for measuring random error: the standard deviation and the mean absolute deviation. The standard deviation formula, which is the first one presented, is the conventional method for calculating variability in data sets. The mean absolute deviation, while simpler and sometimes used in machine learning for efficiency, is less common in fields like experimental physics. Participants agree that for accurate scientific measurements, the standard deviation is the preferred choice. Overall, the standard deviation is emphasized as the more reliable metric for error calculation.
Lotto
Messages
251
Reaction score
16
TL;DR Summary
How to calculate a random measurement error? If we did a measurement and want to calculate a random error, what formula are we to use?
I have seen this formula

$$\sigma=\sqrt{\frac {\sum_{i=1}^{N}{(X_i- \bar{X})^2}}{N(N-1)}}$$

but also this formula $$\sigma =\frac{\sum_{i=1}^{N}{|X_i- \bar {X}|}}{N}.$$ Which of them is correct?
 
Last edited:
Physics news on Phys.org
The first one :smile:

##\ ##
 
  • Like
Likes Dale and vanhees71
The first one is the standard deviation. That is the usual one. The second one is the mean absolute deviation and it is rarely used at all.
 
Dale said:
The second one is the mean absolute deviation and it is rarely used at all.
Some machine learning applications seem to use it - I think because it's cheaper to calculate, so it gives you a time saving if you can live with the less mathematically nice behaviour. But in experimental physics I agree it's a no, you want the standard deviation.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Back
Top