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Homework Help: Error propagation and significant digits

  1. Jan 20, 2016 #1
    Moderator's note: Thread moved to homework section. Thus no template.

    I have an exercise in which I have to calculate the Area from the following measurements:
    L = 22.1 ± 0.1 cm
    W = 7.3 ± 0.1 cm

    Of course, A = W * L = 161.33 but since I have a measurement with just 2 significant digits the results is limited to 2,
    A = 160 (or 1.6 x 102 cm).

    Now I apply error propagation:
    computing the terms, it gives σA = 2.327 cm

    Now I write my answer:
    160 ± 2 cm

    My question is, can I have a error in a decimal place with no significant digits in that decimal place in my answer?
    Should it be
    161 ± 2 cm?
  2. jcsd
  3. Jan 21, 2016 #2


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    It should be 161 ± 2 cm, sure. The 1 is not "significant" (you are not sure about it), but it is the best estimate, and you certainly know that 161 is a much better estimate than 164.9 or 155.

    Significant figures are a bad hand-waving way to estimate uncertainties, error propagation is a much better tool.
  4. Jan 21, 2016 #3


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    With regard to the number of digits in the principal answer, since you will be specifying an error range you can show as many as you like. The usual limit is because when you do not specify a range the number of digits you show implies the level of accuracy.
    Indeed, consider that your final answer has a top end of 160+2 = 162, whereas your more detailed calculation says it could easily reach 163.

    As for the error range calculation, I'm sure you have followed what you were taught, but I would like to point out an issue with it. In many experiments, a source of error is from reading off a scale. The error distribution is uniform, +/- half the granularity of the scale markings. The error propagation formula you quote assumes the input error ranges represent some unstated but consistent number of standard deviations, and churns out the corresponding number of standard deviations for the answer. But the product or sum of two uniformly distributed random variables will not have a uniform distribution, so applying the formula will not represent the same number of deviations.

    Finally, does it make sense to round down the 2.327? At the least, I would figure out the upper and lower limits based on the unrounded numbers, giving 161.33-2.327=159, and 161.33+2.327=163.66. To encompass that, take the midpoint etc.: 161.3+/-2.3. But that's just what makes sense to me. What you have been taught may be different. If you insist on showing fewer digits, it becomes, say, 161+/-4 in order to encompass the true range, which seems silly.
  5. Jan 22, 2016 #4
    Thank you, mfb and haruspex.

    I'll keep my answer 161.3 ± 2.3 cm. My confusion originated with the two rules about significant digits: rest/sum and division/multiply. I was using this exercise to understand this and use it in further problems.
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