# standard dev

• Apr 23rd 2006, 02:44 AM
miavaro
standard dev
"Hai"
I hope Somebody help me
2.0±0.2 - 1.0±0.2
:rolleyes:
(All related to weight")
Thank you !
• Apr 23rd 2006, 03:18 AM
CaptainBlack
Quote:

Originally Posted by miavaro
"Hai"
I hope Somebody help me
2.0±0.2 - 1.0±0.2
:rolleyes:
(All related to weight")
Thank you !

I'm not sure that anyone will be able to tell what your question is
from this but one interpretation (not the most likely, but this is the
interval arithmetic solution) will give:

$\displaystyle (2 \pm 0.2) - (1 \pm 0.2)=1 \pm 0.4$

RonL
• Apr 23rd 2006, 07:16 AM
topsquark
Quote:

Originally Posted by CaptainBlack
I'm not sure that anyone will be able to tell what your question is
from this but one interpretation (not the most likely) )will give:

$\displaystyle (2 \pm 0.2) - (1 \pm 0.2)=1 \pm 0.4$

RonL

I would say that equally likely to CaptainBlack's answer would be the answers:

1) $\displaystyle 1.0 \pm 0.2$ assuming the standard error in the measuring instrument is 0.2

2) $\displaystyle 1.0 \pm 0.3$ using the propagation of errors technique.
(Define a function f = f(a, b), where we have measurements $\displaystyle a = \bar{a}+\delta a$ and $\displaystyle b=\bar{b}+ \delta b$. Then $\displaystyle \bar{f} = f(\bar{a}, \bar{b})$ and $\displaystyle \delta f = \sqrt{ \left ( \frac{\partial f}{\partial a} \delta a \right ) ^2 + \left ( \frac{\partial f}{\partial b} \delta b \right ) ^2}$. )

Lacking any other information, my best guess is that the propagation of errors answer (2) is the answer, though there are many statistical reasons for choosing either of the other two. It depends entirely on how the original error estimates were made.

-Dan
• Apr 24th 2006, 04:04 PM
miavaro
thank you