3
$\begingroup$

I need to experimentally determine some value $Z = 3X+Y$ where $X$ and $Y$ are measurements. I was given the value for $X$, let's say $X = 14.2\pm 0.1$.

Up to now I thought that if I read something like $X=14.2 \pm 0.1$, it was assumed that the measurements were normally distributed with a mean of $14.2$ and a standard deviation of $0.1$.

Is this correct?

My measurements of $Y$ resulted in a mean of let's say $9.5$ and a standard deviation of $0.2$.

I proceeded as follows: I assumed $X$ and $Y$ are both normally distributed and independent with the given means and standarddeviations. So $Z$ is the sum of two normally distributed values, so it must be normally distributed as well with the mean $3\cdot 14.2 + 9.5 = 52.1$ and a standard deviation of $\sqrt{ 3^2 \cdot 0.1^2+0.2^2} = 0.36$ So I could write $Z=52.1 \pm 0.36$.

Again is this the correct way to do this, or is my understanding of this notation wrong?

$\endgroup$
0

2 Answers 2

7
$\begingroup$

Unless stated otherwise, the $\pm$ refers to the standard error which is indeed the standard deviation of the measurement.

However, whether you may assume it to be normally distributed depends on the context (e.g. how the measurements have been made). For example, if the measurements involved a Bernoulli trial then you'd be wrong to assume the error to be normally distributed.

Generally though, if it's not explicitly stated and not clear what the distribution should be, then it's reasonable to take it as normally distributed, in which case your analysis is correct.

$\endgroup$
2
  • $\begingroup$ Thank you very much! So if we assume that some values follow some distribution, we can caluclate with them the same way as we would with random variables with those distributions? $\endgroup$
    – flawr
    Commented Jan 5, 2017 at 16:15
  • $\begingroup$ @flawr Precisely. Any experimental measurement can be interpreted as the generation of a random number from a particular probability distribution (see frequentism). $\endgroup$
    – lemon
    Commented Jan 5, 2017 at 16:23
2
$\begingroup$

Normal distribution doesn't really matter. Having measured many X, Y's to compute real world Z's: we often don't know the parent distribution--we just propagate errors in quadrature as you have done. The biggest, or perhaps most common, mistakes occur when we fail to account for any correlations between X and Y--as real world system (or simulations of them) often get correlated.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.