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I have two sensors giving me each one measurement. Out of these I calculate the average value of the measurements, and want to express it upto the right amount of significant figures. To do that, I follow the rules to determine the significant figures in combined operations (in this case, Addition and Division). $$ AVG = \frac{S_1+S_2}{2}, $$ where $S_1$ and $S_2$ are the measurements provided by sensors 1 and 2. Considering that the number $2$ in the denominator has infinite precision, the result average will have the same amount of Significant Figures as the numerator.

Now let's say I have the following measurements, each with it's corresponding Average (expressed with the right amount of Significant Figures):

  • $S_1 = 500$, $S_2 = 1001$ $\,\,\,\,\,\,\Rightarrow\,\,\,\,\,\,$ $AVG = 750.5$
  • $S_1 = 70$, $S_2 = 80$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow\,\,\,\,\,\,$ $AVG = 75.0$
  • $S_1 = 238$, $S_2 = 237$ $\,\,\,\,\,\,\,\,\Rightarrow\,\,\,\,\,\,$ $AVG = 238$

From a formula/rules point of view, I do understand why sometimes the result is showing a decimal digit and why sometimes it is not.

I find it difficult to understand why this is correct. If someone would just see the $AVG$ values, without knowing where they come from, it would seem as if sometimes the resolution of the measurement is equal to $1$ and sometimes equal to $0.1$.

Could someone explain verbally what it's happening under the hood when applying the significant figures rules. Why does it seem there's more resolution in some cases compared to others?

Miguel
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  • What's your question? – md2perpe Jul 24 '23 at 16:05
  • Sorry, I thought it was clear. If someone can explain verbally and more in detail what the rules are implying in this case. How is it that it seems there's more resolution on some cases compared to others. – Miguel Jul 24 '23 at 19:01
  • Why do you use one decimal on $75.0$? Is it because $150$ (the sum of $70$ and $80$) has three digits so you think that $150/2$ also should have it? – md2perpe Jul 24 '23 at 22:14

1 Answers1

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There's more resolution in some cases as compared to others because bigger inputs have more significant figures. Scientific notation helps to show this.

  • $500 + 1001 = 1501 = 1.501 * 10^3$. That's 4 significant figures. $1501 \div 2.\bar0 = 750.5 = 7.505 * 10^2$. That's also 4 significant figures.

  • $70 + 80 = 150 = 1.50 * 10^2$. That's 3 significant figures. $150 \div 2.\bar0 = 75.0 = 7.50 * 10^1$. That's also 3 significant figures.

  • $238 + 237 = 475 = 4.75 * 10^2$. That's 3 significant figures. $475 \div 2.\bar0 = 237.5 \approx 238 = 2.38 * 10^2$. That's 3 significant figures.

From the perspective of addition, all integers have the same precision. But from the perspective of multiplication, an integer has the precision of its number of digits, so when you divide a big number, you'll end up with more precision than when you divide a small number.

mrfoogles
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