Experts of "significant digits" please enter...

[kingwinner]

I have long been confused with significant digits for questions that involve a combination of addition / subtraction & multiplication / division, in science classes I have lost marks frequently on this, I know the calculations but I lost marks on roundings...so frustrated :) I really need an expert on this to help me out...I would really appreciate because I am going to take Physics in which significant digits is particularly important!

 

Q1: The speed of P waves is 6.8 km/s and the speed of S waves is 4.1 km/s. How long would it take P waves and S waves to travel 100 km respectively?

 

This is simple!

t(P waves)=d/v

=100/6.8 (3 significant digits divided by 2 significant digits)

=14.7 sec (unround answer)

=15 sec (2 significant digits as a final answer to this question)

 

Similarily,

t(S waves)=d/v

=100/4.1 (3 significant digits divided by 2 significant digits)

=24.4 sec (unround answer)

=24 sec (2 significant digits as a final answer to this question)

 

Q2: What is the lag time between the arrival of P waves and S waves over a distance of 100km? (lag time is the time difference in arrival times of P waves and S waves)

 

I am starting to get confused...

Lag time=t(S waves) - t(P waves)

=(100/4.1) - (100/6.8)

=24.4 - 14.7 (use unrounded intermediate answers from above)

=9.7 or 9 or...? (problem starts to arise here...this calculation of lag time involve a combination of division and subtraction, some possible rounding methods popping off my head:

(i) (100/4.1) - (100/6.8) involve a combination of division and subtraction so it just follow the "rule of number of significant Digits in multiplication & division", i.e. the number of significant digits in an answer should equal the least number of significant digits in any one of the numbers of measurement, in this case, 2 significant digits, thus the final answer is 9.7 sec

(ii) 24.4 - 16.4 both have 1 deciminal place so the final answer should have 1 deciminal place as well, ie 9.7 sec

(iii) Follow "rule of number of significant Digits in multiplication & division" when doing the when doing multiplication/division and also follow "rule of number of significant Digits in addition & subtraction" when doing addition/subtraction, that is, rounding off every step:

=(100/4.1) - (100/6.8)

=24 (2 significant digits) - 15 (2 significant digits)

Now, doing subtraction and both have 0 deciminal place

Thus the final answer is 9 sec (0 deciminal place)

(iv) other...

 

Can someone point out the correct number of significant digits or deciminal places that should be in the final answer of Q2 and explain why?

 

Q3: Given the lag time for Austin is 150 sec, the distance from Austin to the location of the earthquake can be found using this formula:

distance = [lag time for Austin (s) x 100 km ] / [lag time for 100 km (s)]

The unrounded answer is 1546.3918 km, again I don't know how many significant digits I should keep in the final answer because I don't know the number of significant digits to the answer of Q2.

 

(*Also, in my class, we assume 100km has 3 significant digits, so no worry there...)

 

Thank you again for helping!

[Jay-qu]

use a calculator (if you can) leave you answer in the calculator never round halfway through a question and then continue on with that rounded number because it will be wrong - when you finish your calculations use the smallest no. of sig figs given in the question and then round it

 

so it should be 9.7, in the 3rd question the only new info you get given is the 150 so if you carry through any numbers from previous questions then it is acceptable to use your rounded answer and give it to the least amount of sig figs :) tell me if that made sense...

[kingwinner]

use a calculator (if you can) leave you answer in the calculator never round halfway through a question and then continue on with that rounded number because it will be wrong - when you finish your calculations use the smallest no. of sig figs given in the question and then round it

 

so it should be 9.7, in the 3rd question the only new info you get given is the 150 so if you carry through any numbers from previous questions then it is acceptable to use your rounded answer and give it to the least amount of sig figs :) tell me if that made sense...

But for Q2, it also involve a subtraction, and according to "rule of number of significant Digits in addition & subtraction", when doing subtraction, the final answer should have the same number of DECIMINAL PLACES as the given measurement with the least number of deciminal places. In this question, it also involve a division, so I don't know which rule of rounding to follow, deciminal places? significant digits?

 

Another question too! I was told that whole numbers have an infinite number of significant digits, so like Q1, "How long would it take P waves and S waves to travel 100 km respectively?", the 100km, would you consider this as having 3 significant digits or an infinite number of that? If the answer is an infinite number of significant digits, that doesn't make sense to me beucase, still, the "100km" is a measurement!

[Jay-qu]

according to "rule of number of significant Digits in addition & subtraction", when doing subtraction, the final answer should have the same number of DECIMINAL PLACES as the given measurement

 

never heard of such rule :) maybe it is taught differently in different countries. In my work I never quote an answer to less sig figs than the info given and only ever quote more (though i argued this with my teacher) when it changes your answer significantly.

For example a wave traveling at 340m/s has a wavelength of 2cm what is the frequency? quickly doing the math you see that the answer is 17,000Hz now when it comes to giving your final answer if you do it to 1 sig fig the answer is 2 x 10^4 - this increases the frequency by 3000Hz, thats a big change and hence you are intitiled to quote it as 1.7 x 10^4

 

you just have to use a bit of common sense, unless your been stupid you wont be penalised.

[Qfwfq]

A simple case of a pitfall is like:

 

r = b - a

 

Suppose a or b is an expression that you can't avoid a bit of approximation on when you calculate it and that the difference a - b is comparable to this approximation, or not much greater, then it can make a great difference to r. In such cases it is better to manipulate a - b so as to avoid the problem. A simple example of this could be when a is given and b is an algebaic sum of very large and small values, this suffers a bit of approximation and the value of a small addend may even be lost altogether:

 

b - a = B + s - a

 

In short, when summing and subtracting large and small values you need to watch how to group them, the associative property might not work when partial sums aren't of similar magnitudes. Even worse, if this r is a factor in a denominator and the approximations wrongly make it near zero this could give an overflow.

 

Even without minus signs, it may be better to first add the smallest terms with each other.

[cwes99_03]

To correct/ clarify the above.

 

You count the number of accurate digits when multiplying/dividing and you count the furthest digit to the right when adding/subtracting.

 

2.57-.07=2.50 not 3, no rounding necessary.

2.57-.5 =2.0 not 2.07, there is not a hundreths place on the 0.5 being subtracted.

 

20*55=1000 not 1100 because the zero on the 20 is not significant unless there is a designator saying it is significant (usually a bar above the zero that is significant).

 

20,"0"00 (I use the quotes to designate which is the lowest significant zero) has three significant digits and therefore any answer multiplied by it would have a maximum of 3 sig figs.

 

When adding/subtracting and multiplying/dividing, you simply apply these rules together.

 

* note * I too do not round any numbers until my calculation has been done, however you need to know how many significant numbers you should have once that is done. Multiple roundings is bad math and bad science.

 

2.3+3.57*3+5.99 = 2.3+"1"0.71+5.99 = 20 not 19 because only the number in the 10s position is significant (due to being multiplied by 3 which has only one sig fig).

 

Totally disregard the statement above made by Jay-qu. He obviously doesn't understand the point of significant figures is to show someone how accurate your data is. If he did understand then he would know why we choose to only show 20,000 Hz, and that is because his accuracy for measuring the wavelength is terribly inaccurate thus leading to low accuracy frequency calculations. If he knew his accuracy to be 2.0000 cm, then he would be able to justify an answer more accurate than 20,000Hz.

 

Oh and you have just scratched the surface. There is the whole how do you calculate the +/- accuracy of the calculation. That will make your head explode.

[Jay-qu]

Totally disregard the statement above made by Jay-qu. He obviously doesn't understand the point of significant figures is to show someone how accurate your data is. If he did understand then he would know why we choose to only show 20,000 Hz, and that is because his accuracy for measuring the wavelength is terribly inaccurate thus leading to low accuracy frequency calculations. If he knew his accuracy to be 2.0000 cm, then he would be able to justify an answer more accurate than 20,000Hz.

 

Excuse me... you may have misread my post! I DO understand the point of sig figs and accuracy that is why I put in brackets the fact that I had argued the point with my teacher. In actual fact I got this question on an exam with the data as 2cm and I quoted my answer as 2x10^4Hz and I got marked WRONG! like I said i disagreed with my teacher but he says that you cant just change the answer by 3000Hz because of one sig fig he said it was just stupid - which I found quite insulting, but hey thats learning :)

[Garak]

I have long been confused with significant digits for questions that involve a combination of addition / subtraction & multiplication / division, in science classes I have lost marks frequently on this, I know the calculations but I lost marks on roundings...so frustrated :) I really need an expert on this to help me out...I would really appreciate because I am going to take Physics in which significant digits is particularly important!

 

Q1: The speed of P waves is 6.8 km/s and the speed of S waves is 4.1 km/s. How long would it take P waves and S waves to travel 100 km respectively?

 

This is simple!

t(P waves)=d/v

=100/6.8 (3 significant digits divided by 2 significant digits)

=14.7 sec (unround answer)

=15 sec (2 significant digits as a final answer to this question)

 

Similarily,

t(S waves)=d/v

=100/4.1 (3 significant digits divided by 2 significant digits)

=24.4 sec (unround answer)

=24 sec (2 significant digits as a final answer to this question)

 

Q2: What is the lag time between the arrival of P waves and S waves over a distance of 100km? (lag time is the time difference in arrival times of P waves and S waves)

 

I am starting to get confused...

Lag time=t(S waves) - t(P waves)

=(100/4.1) - (100/6.8)

=24.4 - 14.7 (use unrounded intermediate answers from above)

=9.7 or 9 or...? (problem starts to arise here...this calculation of lag time involve a combination of division and subtraction, some possible rounding methods popping off my head:

(i) (100/4.1) - (100/6.8) involve a combination of division and subtraction so it just follow the "rule of number of significant Digits in multiplication & division", i.e. the number of significant digits in an answer should equal the least number of significant digits in any one of the numbers of measurement, in this case, 2 significant digits, thus the final answer is 9.7 sec

(ii) 24.4 - 16.4 both have 1 deciminal place so the final answer should have 1 deciminal place as well, ie 9.7 sec

(iii) Follow "rule of number of significant Digits in multiplication & division" when doing the when doing multiplication/division and also follow "rule of number of significant Digits in addition & subtraction" when doing addition/subtraction, that is, rounding off every step:

=(100/4.1) - (100/6.8)

=24 (2 significant digits) - 15 (2 significant digits)

Now, doing subtraction and both have 0 deciminal place

Thus the final answer is 9 sec (0 deciminal place)

(iv) other...

 

Can someone point out the correct number of significant digits or deciminal places that should be in the final answer of Q2 and explain why?

 

Q3: Given the lag time for Austin is 150 sec, the distance from Austin to the location of the earthquake can be found using this formula:

distance = [lag time for Austin (s) x 100 km ] / [lag time for 100 km (s)]

The unrounded answer is 1546.3918 km, again I don't know how many significant digits I should keep in the final answer because I don't know the number of significant digits to the answer of Q2.

 

Thank you again for helping!

 

 

For your first two questions, I don't like that 100 km bit. It's ambiguous. Is it 104 km? Is it just over 95 km? Is it supposed to represent 100 km exactly, arbitrarily? Many textbooks make the same stupid mistake. They write about avoiding ambiguity, but then use 100 when they mean 1.00 X 10^2. To me, 100 km has one significant figure. 100.0 has four. 100. or 1.00 X 10^2 has three.

 

As for the subtraction, the closest thing I've seen to a "firm" rule about that was a chemistry professor stating that in such a calculation is you did, round your multiplication and division by appropriate significant figures first, then do your addition or subtraction to the least number of decimal places.

 

We in physics don't always agree with how chemistry people do things, mathematically speaking. When I took that same class, it seemed the grad students marking assignments interpreted the rules however they wanted, so I avoided it by doing it the physics way. Manipulate the integrated rate law in such a way to isolate your constant, make it look like a real bastard in such a way that no one but a physicist or a mathematician will touch it, and the chem grad student just compares it to the final answer and puts a check mark, since he can't find where a significant figure error might be! :)

[kingwinner]

For your first two questions, I don't like that 100 km bit. It's ambiguous. Is it 104 km? Is it just over 95 km? Is it supposed to represent 100 km exactly, arbitrarily? Many textbooks make the same stupid mistake. They write about avoiding ambiguity, but then use 100 when they mean 1.00 X 10^2. To me, 100 km has one significant figure. 100.0 has four. 100. or 1.00 X 10^2 has three.

 

As for the subtraction, the closest thing I've seen to a "firm" rule about that was a chemistry professor stating that in such a calculation is you did, round your multiplication and division by appropriate significant figures first, then do your addition or subtraction to the least number of decimal places.

 

We in physics don't always agree with how chemistry people do things, mathematically speaking. When I took that same class, it seemed the grad students marking assignments interpreted the rules however they wanted, so I avoided it by doing it the physics way. Manipulate the integrated rate law in such a way to isolate your constant, make it look like a real bastard in such a way that no one but a physicist or a mathematician will touch it, and the chem grad student just compares it to the final answer and puts a check mark, since he can't find where a significant figure error might be! :)

Do you mean this way? In the physics way how would you do? Can you show me for Q2? (this question is actually from my Earth Science course)

"(iii) Follow "rule of number of significant Digits in multiplication & division" when doing the when doing multiplication/division and also follow "rule of number of significant Digits in addition & subtraction" when doing addition/subtraction, that is, rounding off every step:

=(100/4.1) - (100/6.8)

=24 (2 significant digits) - 15 (2 significant digits)

Now, doing subtraction and both have 0 deciminal place

Thus the final answer is 9 sec (0 deciminal place)"?

But if you don't round to the correct number of significant digits after doing the 2 divisions, how would you be able to know how many deciminal places(because of subtraction) the final answer should have?

 

*Also, in my class, we assume 100km has 3 significant digits, so no worry there...

[cwes99_03]

Excuse me for saying you were wrong Jay-qu. Might I say, you were correct, and your professor was wrong for marking you wrong. The point of sig figs is to show accuracy. Had the professor stated in the directions to ignore sig figs when making the calculation then 17,000 would be fine. Unfortunately, not all professors are willing to correct their own mistakes, but instead want you to believe that it is not the world which has the laws, but the professor themselves that make up the laws (obviously this would result in anarchy.)

[Jay-qu]

;) its ok, also not the first time I have tried to proove my teachers wrong ;)

[cwes99_03]

To put an answer to that ? I would say, you do it on a calculator first, therefore not rounding anything until your answer is figured out, then you sit down and figure out each individual part to know how many sig figs your final answer should have, then round your final answer to the correct number of sig figs

[Garak]

Do you mean this way? In the physics way how would you do? Can you show me for Q2? (this question is actually from my Earth Science course)

"(iii) Follow "rule of number of significant Digits in multiplication & division" when doing the when doing multiplication/division and also follow "rule of number of significant Digits in addition & subtraction" when doing addition/subtraction, that is, rounding off every step:

=(100/4.1) - (100/6.8)

=24 (2 significant digits) - 15 (2 significant digits)

Now, doing subtraction and both have 0 deciminal place

Thus the final answer is 9 sec (0 deciminal place)"?

But if you don't round to the correct number of significant digits after doing the 2 divisions, how would you be able to know how many deciminal places(because of subtraction) the final answer should have?

 

*Also, in my class, we assume 100km has 3 significant digits, so no worry there...

 

Okay. The physics way for Q2 doesn't offer much of an advantage due to the simplicity of the calculation. However, to follow the generally accepted rules for significant figures, if there indeed are any, I would proceed as follows.:

 

(100/4.1) ~ 24.39 keeping an extra decimal place.

(100/6.8) ~ 14.71 keeping an extra decimal place.

 

If this were straight division, there would be only two significant figures remaining due to the denominators having the fewest significant figures. However, subtraction gets involved.

 

Round to significant figures before the subtraction, leaving you with 24 and 14 respectively. Then 24-14 is 10, with two signficant figures, which I would express as 1.0*10^1.

 

To clarify things for you, say, for example, that the above problem were as follows:

 

(100.00/4.1000)-(100.0/6.80) with the numbers having five, five, four, and three significant figures, respectively.

(100.00/4.1000) ~ 24.390 to five significant figures.

(100.0/6.80) ~ 14.7 to three significant figures.

 

When you do the subtraction, you go to the least decimal places, not to significant figures.

24.390-14.7=9.7

 

The arithmetic answer is actually 9.69 and 14.7 has three significant figures and 24.390 has five. However, 14.7 has only one decimal place, and in subtraction or addition, the answer must have the same decimal precision as the operator with the fewest decimal places, which, in this case, is 14.7, giving us one decimal place.

 

I hope this clears things up somewhat.

[kingwinner]

"(100/4.1) ~ 24.39 keeping an extra decimal place.

(100/6.8) ~ 14.71 keeping an extra decimal place.

 

If this were straight division, there would be only two significant figures remaining due to the denominators having the fewest significant figures. However, subtraction gets involved.

 

Round to significant figures before the subtraction, leaving you with 24 and 14 respectively. Then 24-14 is 10, with two signficant figures, which I would express as 1.0*10^1."

 

Hello,

 

Wouldn't that be 24 and 15 respectively?

[cwes99_03]

Actually, I believe he was referring to the idea that you wouldn't round the answer until you had subtracted the 14.71 from the 24.39. When you do that you get 9.68, which rounds to 10. If you round your numbers early then you will get a rounded answer of 9 not 10, which we find to be wrong.

However, if you are doing this for a class assignment, I might suggest reading your textbook to determine what answer the teacher is going to be looking for. Then, if you are bold enough, stick with the answer we have supplied you and leave notes in the margins explaining why you chose not to round early.

 

Otherwise, yes 14.71 would round to 15, but instead he reduced 14.71 by the .39 that would be lost if you rounded the 24 down, which makes the 14.71 into 14.32, which rounds to 14.

 

Cheers

[kingwinner]

Actually, I believe he was referring to the idea that you wouldn't round the answer until you had subtracted the 14.71 from the 24.39. When you do that you get 9.68, which rounds to 10. If you round your numbers early then you will get a rounded answer of 9 not 10, which we find to be wrong.

However, if you are doing this for a class assignment, I might suggest reading your textbook to determine what answer the teacher is going to be looking for. Then, if you are bold enough, stick with the answer we have supplied you and leave notes in the margins explaining why you chose not to round early.

 

Otherwise, yes 14.71 would round to 15, but instead he reduced 14.71 by the .39 that would be lost if you rounded the 24 down, which makes the 14.71 into 14.32, which rounds to 14.

 

Cheers

My science text books are terrible, when they round their final answers to the correct number of significant digits, it is either completely wrong or they don't even bother to round it! (especially for complicated cases like involving both subtraction and division) That's way I can draw into conclusions onto what to do...

[kingwinner]

How about doing it this way?

Q2: Lag time=t(S waves) - t(P waves)

=(100/4.1) - (100/6.8)

=24.39 (only 2 sig fig so no decimal places) -14.71 (only 2 sig fig so no decimal places)

=9.68 (0 decimal places)

=10 (0 decimal places so round up to 10) ?

 

This way avoid rounding intermediate answers, but for this way I have to keep track of significant digits & sometimes deciminal places (when doing subtraction/addition) for every step, this is driving me crazy and sometimes the formula is just really complicated, like a= [d * (b+c) - d * (abc+e) ] + f, if I would have to do it the above way, it would take like forever...

 

Sometimes I do have strict teachers caring about significant digits, like a question that worth 2 marks, they take off 0.5 marks off even if you got everything right, crazy....this is what makes me to make this long post because I don't want the same thing to happen again in future physics and chemistry or other science courses...

 

Alternatively, is it acceptable to do it this way?

Lag time=(100/4.1) - (100/6.8)

=9.68446155

=10 (round to 2 significant digits because the least number of significant digits in the measured values is 2 significant digits, and don't care about the subtraction problem? *Although this answer "10" is still the same as my above method, conceptially they are different, with the above being rounded to 0 deciminal places and this one being rounded to 2 significant digits, and sometimes doing one way would make a difference compared to the other)

 

This follows only the "number of significant digits of multiplication & division" when doing a question involving a combination, say, division & subtraction, really saves a lot of time, but is this the accepted way in high schools and universities? (By the way, I remember I have one teacher in grade 10 saying this same rounding rule like this and this ignores the addition/subtraction problem and only uses the multiplication rules)