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In Analytical Chemistry, 7th edition by Christian et al., in its Example 3.10 (screenshot below; see blue circle for the main matter), when it computed for the standard deviation after getting the average of the three measurements, the authors further divided the standard deviation by 3 to get its absolute uncertainty.

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However in its Example 3.12 (screenshot below; see blue circle for the main matter) it did not divide further the calculated standard deviation by 100 to get the absolute uncertainty, even though it's essentially the same operation as in Example 3.10. In addition, you can see in the price calculation near the end of the solution that he multiplied it too to the standard deviation to get the absolute uncertainty.

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Of the two, which is the correct action for the error when dividing or multiplying with an exact number? If they're both wrong, or if they're both right and I'm just misinterpreting the concept, what is, then, the proper way to deal with this situation?

Follow-up question:

On the other hand, what should I do with the error when adding/subtracting a measured number from an exact number or vice-versa?

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    $\begingroup$ Note that the first instance is a case of averaging (i.e. separate measurements), whereas the second case is making the percentage numbers that go into the calculation look nicer. This amounts to a (albeit exact) unit conversion, which probably should be treated differently. $\endgroup$ – TAR86 Nov 6 '17 at 5:58
  • $\begingroup$ 1/2 Actually, I think that in the case of the percent, the 100 really needs to be distributed to the absolute uncertainty. In the example, the (36.28 +- 0.04%)/100 is actually the conversion step of the percent to the actual factor. If the conversion factor, 100, was not applied to the absolute uncertainty, it would look like that the author was multiplying the mass to (0.3628 +- 0.04). By making the factor like this it would seem that the uncertain digit is actually the hundredths place, instead of the ten-thousandths place (0.3628 +- 0.0004) -- $\endgroup$ – Acnologia Nov 6 '17 at 10:27
  • $\begingroup$ 2/2 which would occur if we divided the absolute uncertainty too with 100. $\endgroup$ – Acnologia Nov 6 '17 at 10:27
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    $\begingroup$ A screenshot or picture of an exercise is not searchable. Please consider rewriting it, so that it can be of help for future visitors. $\endgroup$ – Martin - マーチン Nov 22 '17 at 6:33

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