# Could uncertainty be that big?

1. Oct 20, 2012

### peripatein

Another question I have concerns a table's surface area.
If L=122.14±0.14[cm] and W=24.30±0.57[cm], I got that S=2968.00±69.70[〖cm〗^2], using ∆S=√((∂S/∂L)^2 〖∆L〗^2+(∂S/∂W)^2 〖∆W〗^2 ).
Would you kindly confirm this result? Is it plausible that ∆S would be nearly 70 cm^2??

2. Oct 20, 2012

### digfarenough

I do notice that if W had an error of 0.5 cm, then the error in the surface area would be around 0.5 cm * L = (0.5 cm)(122 cm) = 61 cm ^2 which is certainly close to 70 cm^2.

Others will surely have more insightful input!

[Edit. Just to be clear, though my argument is basically just half of the calculation you did, I'm trying to point out it seems pretty reasonable to me! A small error on the width times a long length can produce a sizable change in surface area.]

Last edited: Oct 20, 2012
3. Oct 20, 2012

### haruspex

What exactly do you mean by a ± error? In engineering terms, this usually means the actual limits of error and does not imply any particular distribution beyond that fact. In that model, the range for the area is min length * min width to max length * max width.
Your sum-of-squares approach effectively interprets the ± in the source data as meaning some (unstated) number of standard deviations.
In the numbers you quote, it happens that the (much) larger error range goes with the smaller dimension. As a result, your sum-of-squares calculation produces pretty much the same answer as above; the combination of width * error in length makes hardly any contribution.
If the ± in the source data represents hard limits but for the area you're more interested in standard deviation, you'll need to make some assumption about the source distributions.

4. Oct 21, 2012

### peripatein

Thank you very much for your replies!

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