Getting an average figure with ±

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To calculate an average value with uncertainties, it's essential to consider the precision of each measurement. The initial method proposed incorrectly combined the uncertainties, leading to an inaccurate result. A weighted mean approach is recommended, especially when measurements have different levels of precision. The discussion highlights the importance of understanding whether the measurements are statistically independent and how to properly calculate combined uncertainties. Clarifying the context of the measurements can further aid in determining the correct method for averaging.
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Homework Statement



Quick question, working on a lab

Just wondering how I could go about getting an average value for a series of figures with +- values.

For instance
1) 1.5 ± 0.02
2) 1.7 ± 0.84
3) 1.7 ± 0.08

The Attempt at a Solution



I tried enclosing the figures within parenthesis and then trying to find an average with wolfram alpha but no success.

I was just reading something on the net

would it be x "avg" ± Δx "avg

1.5 + 1.7 + 1.7 /3 = 1.63
(sqrt) 0.02^2 + 0.84 + 0.08 = 0.844

so we arrive at 1.63 ± 0.844

correct ?
 
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Monocerotis said:
would it be x "avg" ± Δx "avg

1.5 + 1.7 + 1.7 /3 = 1.63
(sqrt) 0.02^2 + 0.84 + 0.08 = 0.844

so we arrive at 1.63 ± 0.844

correct ?
Incorrect, and very much so. Assuming the measurements are statistically independent (are they? There's no telling from the original post) then

1. That 1.5 measurement is considerably more precise than either of the other two measurements. You should get something closer the 1.5 than 1.7 for your final estimate.

2. You combined error is larger than the largest error. It should be smaller than any of individual errors.Google "weighted mean".
 
k I've been reading and it seems as though what I'm dealing with is a common mean rather than a weighted mean

I'm dealing with slope figures of a v(t) graph and trying to find g avg +- delta g avg as to compare it with my theoretical results

the values I gave above were ambigous but the figures I'm actually dealing with from my perspective appear to have equal weight so

still confused as to what to do however because even the common mean will only give me a single value
 
What exactly are you measuring here? Without details it is a bit hard to help you.
 
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