Find the percentage uncertainty of a wooden block

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SUMMARY

The discussion focuses on calculating the percentage uncertainty of a wooden block's volume and density using measurements from three different instruments. The measurements provided are: Instrument 1 (L=1.1 cm, W=1.1 cm, H=1.3 cm, V=1.573 cm³), Instrument 2 (L=1.13 cm, W=1.13 cm, H=1.31 cm, V=1.67 cm³), and Instrument 3 (L=1.12 cm, W=1.12 cm, H=1.29 cm, V=1.62 cm³). The inherent uncertainty in measurements is ±0.005 cm due to the precision of the instruments. The discussion emphasizes the need to combine fractional errors for volume and density calculations, suggesting the use of standard deviation for error estimation.

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agentlee
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Homework Statement


I have to find the percentage uncertainty of a wooden block for its volume and density. Using 3 different types of measurements for length, width, and height.

Let's say that this is my data (cm):
Instrument 1: L= 1.1 W= 1.1 H= 1.3 V=1.573
Instrument 2: L= 1.13 W= 1.13 H= 1.31 V=1.67
Instrument 3: L= 1.12 W= 1.12 H= 1.29 V=1.62

Average Mass could be 2.56g

Homework Equations



I was given ΔV/V and ΔP/P for the uncertaintiy

The Attempt at a Solution


I am not sure how to do this, without an acceptable given volume for the block, I am a bit lost.
 
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It looks like each measurement is to the nearest .01, so there's an inherent uncertainty of +/- .005 in those. On top of that, the measurements are not entirely consistent, suggesting a further source of error. You can apply standard techniques to estimate, say, one standard deviation on those. Your problem then is to combine into a volume error. If you have two numbers with known fractional errors, e.g. x1(1+δ1), x2(1+δ2), then for small errors the product is approximately x1x2(1+δ12). I.e. the fractional errors just add up. But since some may be plus and some minus, cancellation can occur. So treat the net fractional error as the sum of independent normally distributed random variables with zero mean. It will help if you have some basis for supposing the three dimensions have the same variance.
 
ΔV = V_measured - V_standard, V = V_standard
etcetera for density

You should have been given a standard, have to look one up, or incorrectly take an average of your results as a standard.

I've never used this method for errors, except in several physics labs, and it always seemed kind of hokey to me...
 

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