0
$\begingroup$

Let's say I a have a scale that can, for example, measure a mass of some object to precision of 0.1 g. If I do some measurements of different objects I can get results like this: 9.8 g, 9.9 g, 10.1 g, 10.2 g, 10.3 g...

My question is how many significant digits I have in my measurements. For the first 2 measurements I would say 2, but for the rest of them I would have 3. But nothing changed in my apparatus, so it shouldn't change.

Also if I did those measurements on the same object and would like to have a mean result, then would it have 2 or 3 significant figures?

Improve this question
$\endgroup$
5
  • $\begingroup$ If humans used hex instead of base 10 counting, the measured values would be 9.8, 9.9, A.1, A.2, and A.3. No one would truncate digits in this case. I have no doubt that the rules for significant figures never intended to interfere with which base was being used for reporting measurements. This means that 10.1, 10.2, and 10.3, should be reported, as that is what was actually measured. $\endgroup$
    – David White
    Mar 26, 2019 at 15:23
  • $\begingroup$ I agree that they should be reported. But what would be your conclusion to my last question in the post (if I take a mean value of those measurements, should the final result have 2 or 3 significant figures)? $\endgroup$
    – lodovico.arioso
    Mar 26, 2019 at 16:00
  • $\begingroup$ In my opinion, the average should be reported with one digit past the decimal point. $\endgroup$
    – David White
    Mar 26, 2019 at 16:35
  • $\begingroup$ More on significant figures. $\endgroup$
    – Qmechanic ♦
    Mar 27, 2019 at 8:43
  • $\begingroup$ I'm voting to close this question as off-topic because it's not about physics but about statistical uncertainty of measurement.. Plus there are no RULES about sig figs; there are guidelines that tedious people abuse. $\endgroup$
    – Bill N
    Jan 3, 2020 at 20:53

1 Answer 1

Reset to default
1
$\begingroup$

Rule of thumb is that a result takes the same number of significant figures as the input with the fewest significant figures.

You are correct in that for your measurements of

9.8 g, 9.9 g, 10.1 g, 10.2 g, 10.3 g

the values of 9.8 and 9.9 have two significant figures, and the values of 10.1, 10.2 and 10.3 have three significant figures. Although nothing changed in your measurement nor in your apparatus, this is still possible as we see here.

One way to check how many significant figures are in a value is to put it into scientific notation, i.e. use exponents. Thus 9.9 becomes $9.9\times 10^{0}$, which has two digits and hence significant figures, 10.1 becomes $1.01\times 10^1$, which has three digits and hence three significant figures, whereas 10.0 could become $1.00 \times 10^1$ with three significant figures, or $1.0 \times 10^1$ with two significant figures, or $1 \times 10^1$ with only one significant figure. In this you could preserve the level of precision by using $1.00 \times 10^1$.

Why? Because the zeroes are only significant if they have another digit on both ends (except when using scientific notation). This is why 0.1, 0.0001 and 1,000,000 all only have one significant figure - because the zeroes do not have other digits on both ends. On the other hand, 1.1, 1.0001 and 1,000,001 have two, five and seven significant figures, respectively, because the zeroes are surrounded by other digits.

Then, if you take the mean result: $\frac{9.8 + 9.9 + 10.1 + 10.2 + 10.3}{5}=10.06$, but of your input values there are some with two significant figures and some with three significant figures, so you would take two significant figures as the reliable precision of the result. This means that your result cannot be (accurately) stated to more than two significant figures.

The result of 10.06 has four significant figures, which breaks this "reliability rule", and rounding 10.06 to the next decimal place makes it 10.1, which is three significant figures and still breaks this "reliability rule", but rounding it further makes it 10, which is one significant figure. So, what to do in this case? I would call it 10.1 and err on the side of retaining the same number of decimal places, i.e. preserve the precision.

Given another data set containing e.g. {9.7, 9.7, 9.8. 9.9, 10.1}, for which the average would be 9.84, the result could be then rounded to 9.8 as two significant figures.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Thank you for your comment. I am aware of all these rules, but I couldn't find any rules about measurements like in this case in which measurements are in different orders of magnitude. Just some corrections: " $1.0 \times 10^1$ has only one significant figure" - This should be 2 significant figures. "10, which is one significant figure" - This should be 2 significant figures. But still, with these measurements if I take mean with 2 significant figures I would get 10 g, not 10.1 g. To me this seems a little weird, because I somehow lost this 0.1 g precision of my scale. $\endgroup$
    – lodovico.arioso
    Mar 26, 2019 at 16:22
  • $\begingroup$ Thanks, yes, $1.0\times 10^1$ has two s.f. I have corrected that., but 10 has only one s.f., just as 1300 has only two s.f. It seems to me that your best option is to preserve the level of precision and retain one decimal place. $\endgroup$
    – Mick
    Mar 27, 2019 at 7:55
  • 1
    $\begingroup$ I would calculate the mean and the standard deviation - either as 10.06+-0.19 or 10.1+-0.2. I prefer the first one since it provides information on how I'm rounding. $\endgroup$
    – Cinaed Simson
    Mar 27, 2019 at 10:01
  • $\begingroup$ If finding an average, you would add the numbers and keep the least significant decimal place. That's a new number. Then dividing by 4 you have options. More importantly, the person should find the standard deviation and report that alongside the average. People spend too much energy on perfecting their sig figs and ignore more important items. @CinaedSimson is absolutely right. $\endgroup$
    – Bill N
    Jan 3, 2020 at 20:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.