# Assuming significant figures

1. Apr 14, 2013

### e-zero

I have a question which states that a car's speed is 110km/h. Should I assume this speed to be 3 significant figures or 4 significant figures? why??

2. Apr 14, 2013

### technician

the safest thing to do is stick to the number of figures given in the question, which is 3. If you feel that you must use 4 because calculations comu up like that then work with that but give 3 in your final answer.
Whatever you do...do not go beyond 4 !!!
The figures you do not know are, by definition, NOT SIGNIFICANT

3. Apr 14, 2013

### e-zero

Ok, I figured it would be 3 in this case, but I was taught that the last non-zero digit is considered the first uncertain digit, which would mean that 110km would have 2 significant figures not 3.

This is where I'm confused. Could you explain?

4. Apr 14, 2013

### technician

You have hit on a common problem !!!
It is difficult to know wheter the zero is significant or not.
If the speed was 115km/hr then there is no problem
I would say that in practice it is safe totake it to be 3 sig figs.
The problem is that these discussions about significant figures is not an exact science since they are used to convey accuracy of quantities. If greater accuracy is required then more significant figures are needed.
I am fairly certain you will get some other replies giving a variety of interpretations.

Hope this helps.
ps
if you think about it, giving a number to 3 sig figs implies that you know the value to better than 1%
when you write 115 you are saying that it is NOT 114 or 116, so you know the number to within +/- 1....about 1%
This is what makes it tricky deciding whether 110 is 3 or 2 sig figs. If it is only 2 then it means the value could be between 100 and 120 which is probably not intended !!! because it implies an accuracy of only +/-10%

Last edited: Apr 14, 2013
5. Apr 14, 2013

### mikeph

I'd say that if the speed is 110 +/- 5 then it should be written 1.1x10^2 km/h, then it's far simpler to derive implicit errors from stated significant figures.

6. Apr 14, 2013

### Staff: Mentor

I was always under impression that in most physics courses nobody cares about sig figs, they are treated seriously only by chemists (and even then not by all).

7. Apr 14, 2013

### technician

They are treated as a means of assessing and conveying accuracy but it would be laborious if every example had to have this error analysis applied.
They are very important in practical situations and in practical exams students are penalised if they cannot show an awareness of the significance of significant figures.
This usually shows up when pupils write down the number from a calculator without any regard as to how many figures should be shown.

8. Apr 14, 2013

### Staff: Mentor

I know what their application is, but I have seen on many occasions people being told to report the result with a "reasonable number of digits" - so obviously 1.23678923467854 doesn't make sense, but 1.24 or 1.237 is acceptable. Sig figs are about as wrong as easy to use, and treating them too religiously is a bad idea, as it gives students an unjustified feeling that they are doing something important.

9. Apr 15, 2013

### e-zero

Let me use an example to see opinion: ½(23.1 cm3 − 20.32 cm3 + 19.0 cm3)

Would you equate that to 3 significant digits or 1? I am suggesting 1 because the numerator & denominator of the fraction 1/2 are both composed of 1 significant digit.

10. Apr 15, 2013

### mikeph

I'd say common sense needs to be used, and the person doing the analysis needs to be aware of what terms are experimental data and what terms are exact. Without any further information (which we'd have if we did the experiment), I'd guess that 1/2 is an exact term and the other terms are to 3/4 s.f., so I'd quote the result to 3.

I agree there is some vagueness on what is acceptable, but there should definitely be clear pointers on what is unacceptable. I see all the time students measure results, round them, then average them (re-introducing false significant figures) and quoting the average to the same precision as the data. Unless you go through all the maths you'd never realise their result is ~10 standard deviations from the true mean. IMO that's very very bad practice. Rounding mid-calculation is obviously wrong, but a second error is quoting too many significant figures.

11. Apr 15, 2013

### Staff: Mentor

Are you suggesting that kinetic energy has never more that 1 sigfig, as in $\frac{mv^2}2$ 2 has only one sigfig?

12. Apr 15, 2013

### e-zero

No, I think it should be more, but just wanted to verify the ambiguity of sig figs.

13. Apr 15, 2013

### e-zero

14. Apr 15, 2013

### jbriggs444

1.4h pretty clearly.

The 130 km has either two or three significant figures depending on your guess as to whether the trailing zero is significant. The 95 has two significant figures. A quotient has as many significant figures as the less accurate of the dividend and the divisor. That's two either way.

However... That's a leading one on the 137 in the result. And that's a leading nine on the 95 in the input. That means that the relative error bounds on the input are tighter than "two significant figures" would normally suggest. And the relative error bounds on an output of 1.4 would be looser than "two significant figures" would normally suggest. So I would be quite willing to report 1.37 h if I were sure that the 130 km were accurate to +/- 0.5 km and that the 95 km/h were accurate to +/- 0.5 km/h. That way I wouldn't be throwing away accuracy in the name of significant figures.

15. Apr 15, 2013

### e-zero

When you say there is a leading 1 and a leading 9, what significance does that have on deciding on the number of significant figures??

16. Apr 15, 2013

### Staff: Mentor

To take a somewhat contrived example, consider two numbers, each with two significant figures:

(a) 11 - changing the last figure by ±1 is a change of about ±9%.

(b) 99 - changing the last figure by ±1 is a change of about ±1%.

Clearly and (a) and (b) have significantly different levels of precision, even though both have two sig figs.

Now consider the following number with three sig figs:

(c) 101 - changing the last figure by ±1 is a change of about ±1%.

Clearly (b) and (c) have about the same level of precision, even though one has two sig figs and the other has three.

17. Apr 15, 2013

### AlephZero

If you want to estimate the error in the answer, then do it properly.

"Counting significant figures" is such a crude method as to be no real practical use. If the answer happens to be 99, then to 2 significant figures you have an error of about 1%. If it happens to be 101, to 2 s.f. you have an error of about 10%. If you think that makes any sense, then carry on counting significant figures....

Edit: I started typing this before jtbell's post appeared.

18. Apr 15, 2013

### e-zero

I'm still a little confused. How about 65 * 1.96

What would you state that answer as?

19. Apr 15, 2013

### technician

101 is not given to 2 significant figures, it is 3. 101 means you know it is not 100 and it is not 102.