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Topic: Propagation of error and sig. figs  (Read 8174 times)

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Offline stevensfreeman

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Propagation of error and sig. figs
« on: January 22, 2010, 12:14:37 AM »
I'm really embarrassed to ask this! I knew how to do it before but I forgot after 2 years. I remember learning it for analytical chemistry :(

So a basic problem:
What is x= a+b+c?
given that:
a= 100.7 +/- 0.05
b= 0.61 +/- 0.005
c=226 +/- 0.5

OK. My question is the following:
I understand the answer is
x = 327.31 +/- 0.5025186564...

HOWEVER, I have no idea how the sig figs go.

Do I treat each part "separately?" And by part, I mean the thing before the +/- and the thing after. If I do that, then the part before is 327 and the thing after is 0.5.. is my answer than 327 +/- 0.5??? It looks incorrect :(

Offline Ligander

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Re: Propagation of error and sig. figs
« Reply #1 on: January 22, 2010, 02:20:32 AM »
I understand the answer is
x = 327.31 +/- 0.5025186564...

How did you get this value?

Offline Golden_4_Life

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Re: Propagation of error and sig. figs
« Reply #2 on: January 22, 2010, 12:05:42 PM »
Your final analytical result will be reported as:  327.31 widgets.
Golden4Life

Offline stevensfreeman

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Re: Propagation of error and sig. figs
« Reply #3 on: January 22, 2010, 05:59:08 PM »
thanks for your responses but all you guys are wrong which is okay :)

uncertainty has only 1 sig fig so it's +/- 0.5
from that, u see only 1 decimal place

so the answer must have just 1 decimal place:
327.3


final answer is:
327.3 +/- 0.5

Offline MOTOBALL

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Re: Propagation of error and sig. figs
« Reply #4 on: January 30, 2010, 12:38:57 PM »
What is x= a+b+c?
given that:
a= 100.7 +/- 0.05
b= 0.61 +/- 0.005
c=226 +/- 0.5
 
(1) The least accurate measurement is b= 0.61, i.e 0.01 in 0.61, i.e 1 in 61.

(2) x = 100.7 + 0.61 + 226 = 327.31, but this implies accuracy of 0.01 in 327.31, i.e 1 in 32,731.  Must round to x = 327.
 Standard deviation is SD, where
(SD)2 = (0.5)2 + (0.05)2 + (0.005)2 = 0.252525

then SD = 0.502518656.

The accuracy of the uncertainty cannot be greater than the accuracy of the least accurate measurement.

Then SD =0.502 implies accuracy of 0.001 in 0.502, i.e 1 in 502

From (1) accuracy of least accurate is 1 in 61, so SD = 0.50 (2 sig fig)

x= 327 +/- 0.50

Please have a chem professor check the reasoning !!!

MOTOBALL

Offline stevensfreeman

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Re: Propagation of error and sig. figs
« Reply #5 on: January 30, 2010, 11:28:05 PM »
What is x= a+b+c?
given that:
a= 100.7 +/- 0.05
b= 0.61 +/- 0.005
c=226 +/- 0.5
 
(1) The least accurate measurement is b= 0.61, i.e 0.01 in 0.61, i.e 1 in 61.

(2) x = 100.7 + 0.61 + 226 = 327.31, but this implies accuracy of 0.01 in 327.31, i.e 1 in 32,731.  Must round to x = 327.
 Standard deviation is SD, where
(SD)2 = (0.5)2 + (0.05)2 + (0.005)2 = 0.252525

then SD = 0.502518656.

The accuracy of the uncertainty cannot be greater than the accuracy of the least accurate measurement.

Then SD =0.502 implies accuracy of 0.001 in 0.502, i.e 1 in 502

From (1) accuracy of least accurate is 1 in 61, so SD = 0.50 (2 sig fig)

x= 327 +/- 0.50

Please have a chem professor check the reasoning !!!

MOTOBALL


i do not know if you method is correct, but I will let you know. our professor does all the HW grading and refuses to delegate work to our TAs, so it takes her awhile to grade assignments. i'll keep in touch

Offline MOTOBALL

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Re: Propagation of error and sig. figs
« Reply #6 on: January 31, 2010, 02:39:30 PM »
Thanks for your response.

It's all very well me saying that "THIS IS HOW IT IS !!!" only if I HAVE got it right. I don't want to lead anybody astray here. It's always nice to be right, but if I'm wrong that is equally, or more, important for me to know.

Also, I tend to write it out line by line so that I can develop the reasoning through a problem for myself---it's not meant to be pedantic.

Please keep us updated.

Motoball

Offline ozols

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Re: Propagation of error and sig. figs
« Reply #7 on: February 14, 2010, 12:19:10 PM »
Calculation MOTOBALL gave in (2) is correct.
I somewhat disagree with the rounding. If you round 327.31 down to 327 it means you don't know anything after the point. So in this case to state the standard deviation as 0.50 (notice the zero!!!) is a bit absurd. So if you leave the standard deviation as 0.50 then the result should be 327.31 +/- 0.50.

c=226 +/- 0.5 Notice that it is given with only one significant figure after the point. This is the clue! Even though you know a and b more precisely, round down everything to 327.3 +/- 0.5 because the is the least significant number.

This appears to be a well crafted test question as you didn't even have to calculate anything to give an answer. A smart student will immediately notice that the others are at least an order of magnitude smaller and the overall uncertainty can be guessed from c alone (as calculation of MOTOBALL proves). A smart student would save a few minutes here for harder questions.

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