Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: [Phys-L] Sig Figs homework from my 7th grader



On 10/9/2013 12:30 PM, Paul Nord wrote:
Brian,

I'm not familiar with this method. What did you do there?
I made an arithmetic error.
My calculator reports that 11.145 - 0.005 - 0.0005 = 11.1395 and
11.145 + 0.005 + 0.0005 = 11.1505 which I could describe as
11.145 + - 0.0055 These represent extrema for the assumptions I gave.
I would have only added the errors in quadrature, not the values.

sigma = sqrt(0.005**2 + 0.0005**2) = 0.005025

I get 11.145 +- 0.005
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.
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


Hmmm....you offer two ranges: first given above
11.140 - 11.150
then 11.13 - 11.15
Meanwhile I offer (as corrected) 11.1395 - 11.1505

Given that my range represents extrema for given assumptions, only your first range occurs within my extrema. So I suppose one could say your" less generous" treatment carries a lower value less than my minimum. But if I simply accept your rounding, then this is just as acceptable.
The answer in the book is 11.1

Paul

This is evidently not wrong, implying a range 11.05 - 11.15 but could possibly be criticized as losing some available precision? I assert that my (corrected) treatment was not involved with rounding at all, but only with the discrete representation of continuous variables. On the other hand, I assert that the book answer makes an explicit rounding decision - which may have been your concern?

On Oct 9, 2013, at 10:26 AM, brian whatcott <betwys1@sbcglobal.net> wrote:
On 10/9/2013 8:49 AM, Paul Nord wrote:
I know the answer the teacher wants, but i find it troubling...

10.53 + 0.615 = ?

Share and enjoy,
Paul
_______________________________________________

Supposing I were troubled by the desire to best represent the last figure uncertainty, I might proceed in this way:
10.53 +-0.005
0.615 +-0.0005
==============
(DATA BELOW IN ERROR)
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