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Uncertainty in Calculations

  1. Dec 8, 2009 #1
    Ok, not really a homework question, just something I am confused about. Hopefully it belongs here.

    Alright this is in my lab book:

    The overriding consideration for significant figures in computations is that a result should not be quoted with a precision higher than the absolute uncertainty associated with it.

    For example, it makes little sense to write I = (1.56250 +/- 0.01288) x 10^-2 kg.m^2, because the result is already uncertain in the third significant digit. It would be more appropriate to write I = (1.563 +/- 0.013) x 10^-2 kg.m^2 or, even better, I = (1.56 +/- 0.01) x 10^-2 kg.m^2.

    I guess I am just confused why its "already uncertain in the third significant digit". Can anyone explain.. like how is this known.
  2. jcsd
  3. Dec 8, 2009 #2


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    Homework Helper

    The result 1.56250 +/- 0.01288 actually means "between 1.57538 and 1.54962." (I got those two numbers by computing 1.56250 + 0.01288 and 1.56250 - 0.01288.) That range tells you that the first digit of the actual result is definitely 1, and the second digit is definitely 5. The third digit could be 4, 5, 6, or 7. That's what it means to say "uncertain in the third significant digit," that the given range allows several possibilities for that digit.

    Moving on, the fourth digit could be anything, technically, but if the third digit is 7 we know the fourth is 0, 1, 2, 3, 4, or 5, whereas if the third digit is 4 we know the fourth is 9. So it's slightly more likely to be one of those than 6, 7, or 8, but that information may or may not be important depending on how precise you are. Beyond that, the fifth and further digits have a practically equal chance to be any digit, 0 through 9, so there's not much point in knowing them.
  4. Dec 8, 2009 #3
    Awesome.. Thanks a lot! That was helpful.
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