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Brian,I made an arithmetic error.
I'm not familiar with this method. What did you do there?
I would have only added the errors in quadrature, not the values.Is it more reasonable to algebraically add the supposed least significant place variability? Yes of course, but the RMS treatment does not represent extrema, and the RMS value makes a (plausible?) assumption about how the variability of each number may be distributed between extrema.
sigma = sqrt(0.005**2 + 0.0005**2) = 0.005025
I get 11.145 +- 0.005
Being less generous to the signifigance of the last digit, I would probably suggest:
sigma = sqrt(0.01**2 + 0.001**2) = 0.01005
and report: 11.14 +- 0.01
The answer in the book is 11.1
Paul
(DATA BELOW IN ERROR)On 10/9/2013 8:49 AM, Paul Nord wrote:
I know the answer the teacher wants, but i find it troubling...Supposing I were troubled by the desire to best represent the last figure uncertainty, I might proceed in this way:
10.53 + 0.615 = ?
Share and enjoy,
Paul
_______________________________________________
10.53 +-0.005
0.615 +-0.0005
==============
11.1450 +-0.0050 +-0.0005 = 11.1445 to 11.1505 or 11.1475 +- 0.0030
It is this last line that would prove worrying: why should the nominal value shift, when the last figure uncertainties are accounted for?
Brian Whatcott Altus OK