Uncertainty in Measurement and significnt figures in physics












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$begingroup$


(a)Find $sqrt{8}$ to four significant figures; then cube this number and round to three significant figures. (b) Find $sqrt{8}$ to three significant figures; then cube this number and round to three significant figures. (c)Which answer is more accurate? Explain.



In that question, the answer of (a) is $22.6$ and (b) is $22.7$. I can't think which answer is more accurate. So, please explain me. Thank you!






significant-figures







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  • $begingroup$
    Calculate it to 10 digits, then look to see.
    $endgroup$
    – John L Winters
    Dec 8 '18 at 5:29










  • $begingroup$
    I think uncertainty in Measurement is different thing.
    $endgroup$
    – tarit goswami
    Dec 8 '18 at 5:29










  • $begingroup$
    You can easily determine which is more accurate by pulling out a calculator and doing this without rounding. But on general principles, it should be obvious which would be more accurate: The two procedures differ only in how many figures you round $sqrt 8$ to in the first step. Which one is going to change the actual value less?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 14:34
















0












$begingroup$


(a)Find $sqrt{8}$ to four significant figures; then cube this number and round to three significant figures. (b) Find $sqrt{8}$ to three significant figures; then cube this number and round to three significant figures. (c)Which answer is more accurate? Explain.



In that question, the answer of (a) is $22.6$ and (b) is $22.7$. I can't think which answer is more accurate. So, please explain me. Thank you!






significant-figures







share|cite|improve this question











$endgroup$












  • $begingroup$
    Calculate it to 10 digits, then look to see.
    $endgroup$
    – John L Winters
    Dec 8 '18 at 5:29










  • $begingroup$
    I think uncertainty in Measurement is different thing.
    $endgroup$
    – tarit goswami
    Dec 8 '18 at 5:29










  • $begingroup$
    You can easily determine which is more accurate by pulling out a calculator and doing this without rounding. But on general principles, it should be obvious which would be more accurate: The two procedures differ only in how many figures you round $sqrt 8$ to in the first step. Which one is going to change the actual value less?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 14:34














0












0








0


1



$begingroup$


(a)Find $sqrt{8}$ to four significant figures; then cube this number and round to three significant figures. (b) Find $sqrt{8}$ to three significant figures; then cube this number and round to three significant figures. (c)Which answer is more accurate? Explain.



In that question, the answer of (a) is $22.6$ and (b) is $22.7$. I can't think which answer is more accurate. So, please explain me. Thank you!






significant-figures







share|cite|improve this question











$endgroup$




(a)Find $sqrt{8}$ to four significant figures; then cube this number and round to three significant figures. (b) Find $sqrt{8}$ to three significant figures; then cube this number and round to three significant figures. (c)Which answer is more accurate? Explain.



In that question, the answer of (a) is $22.6$ and (b) is $22.7$. I can't think which answer is more accurate. So, please explain me. Thank you!






significant-figures




significant-figures






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 6:52









tarit goswami

1,8511421




1,8511421










asked Dec 8 '18 at 5:17









Thinzar LinThinzar Lin

1




1












  • $begingroup$
    Calculate it to 10 digits, then look to see.
    $endgroup$
    – John L Winters
    Dec 8 '18 at 5:29










  • $begingroup$
    I think uncertainty in Measurement is different thing.
    $endgroup$
    – tarit goswami
    Dec 8 '18 at 5:29










  • $begingroup$
    You can easily determine which is more accurate by pulling out a calculator and doing this without rounding. But on general principles, it should be obvious which would be more accurate: The two procedures differ only in how many figures you round $sqrt 8$ to in the first step. Which one is going to change the actual value less?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 14:34


















  • $begingroup$
    Calculate it to 10 digits, then look to see.
    $endgroup$
    – John L Winters
    Dec 8 '18 at 5:29










  • $begingroup$
    I think uncertainty in Measurement is different thing.
    $endgroup$
    – tarit goswami
    Dec 8 '18 at 5:29










  • $begingroup$
    You can easily determine which is more accurate by pulling out a calculator and doing this without rounding. But on general principles, it should be obvious which would be more accurate: The two procedures differ only in how many figures you round $sqrt 8$ to in the first step. Which one is going to change the actual value less?
    $endgroup$
    – Paul Sinclair
    Dec 8 '18 at 14:34
















$begingroup$
Calculate it to 10 digits, then look to see.
$endgroup$
– John L Winters
Dec 8 '18 at 5:29




$begingroup$
Calculate it to 10 digits, then look to see.
$endgroup$
– John L Winters
Dec 8 '18 at 5:29












$begingroup$
I think uncertainty in Measurement is different thing.
$endgroup$
– tarit goswami
Dec 8 '18 at 5:29




$begingroup$
I think uncertainty in Measurement is different thing.
$endgroup$
– tarit goswami
Dec 8 '18 at 5:29












$begingroup$
You can easily determine which is more accurate by pulling out a calculator and doing this without rounding. But on general principles, it should be obvious which would be more accurate: The two procedures differ only in how many figures you round $sqrt 8$ to in the first step. Which one is going to change the actual value less?
$endgroup$
– Paul Sinclair
Dec 8 '18 at 14:34




$begingroup$
You can easily determine which is more accurate by pulling out a calculator and doing this without rounding. But on general principles, it should be obvious which would be more accurate: The two procedures differ only in how many figures you round $sqrt 8$ to in the first step. Which one is going to change the actual value less?
$endgroup$
– Paul Sinclair
Dec 8 '18 at 14:34










1 Answer
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oldest

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$begingroup$

When working with significant figures, you should always round off at the end to avoid round-off error. For example, $2.5 * 2.5 = 6.25$ with two significant figures. If you round now ($6.3$) and then cube it, you get $250.047$ with two significant figures (rounds to $250$ or $2.5*10^2$). If you cube before rounding, you get $244.140625$ with two significant figures (rounds to $240$ or $2.4*10^2$). As you can see, rounding during the calculations versus rounding after the calculations often causes differences. The more accurate answer will be the one where the rounding comes at the end.



Let's test that theory. I claim that $(2.5^2)^3$ will yield a more accurate answer as to what the actual value of the data squared and then cubed will be than $round(2.5^2)^3$



The measured data $2.5$ means that the actual data might be anywhere between $2.45$ and $2.55$ so that it rounds to $2.5$.



Let's evaluate what the data squared and then cubed would be if the the actual measure of the data were the minimum that it could be given how accurate your measurement was.



$2.45^2 = 6.0025$
$6.0025^3 = 216.270112515625$



Remember that we are assuming that the exact measure of the data is 2.45. Therefore, it has infinite significant figures, and we don't round anywhere.



Now look at the maximum possible measure that the data could be.



$2.55^2 = 6.5025$
$6.5025^2 = 274.941996890625$



Now let's find the average of those two answers. Since our measured data is the average of the two extremes of the data, the answer should be the rounded average of the evaluations with the two extremes.



$(216.270112515625 + 274.941996890625)/2 = 245.606054703125$



Now let's solve using the measured data using significant figures.



Start with rounding during the calculations



$2.5 ^ 2 = 6.25$
$6.3 ^ 3 = 250.047$



That rounds to $2.5*10^2$ or $250$ with two significant figures.



Now do it with rounding only at the end



$2.5 ^ 2 = 6.25$
$6.25 ^ 3 = 244.140625$



That rounds to $2.4*10^2$ or $240$ with two significant figures. While the exact average value does round to the first answer, you can see that unrounded value the second time was much closer to the exact value. (244.14... was closer to 245.60... than 250.04... was). You can see how if we continued to do calculations on the data, the first method (with rounding during the calculations) would become more and more inaccurate because the calculations are being done with a number that is farther away from the exact average of the extremes of the data than the answer we got from rounding only at the end (and if we continued manipulating that number, 244.14..., we wouldn't round it, so it would indeed be closer to that average of 245.6...)



I hope that helps, that you were able to follow along, and that I didn't make any mistakes.






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    $begingroup$

    When working with significant figures, you should always round off at the end to avoid round-off error. For example, $2.5 * 2.5 = 6.25$ with two significant figures. If you round now ($6.3$) and then cube it, you get $250.047$ with two significant figures (rounds to $250$ or $2.5*10^2$). If you cube before rounding, you get $244.140625$ with two significant figures (rounds to $240$ or $2.4*10^2$). As you can see, rounding during the calculations versus rounding after the calculations often causes differences. The more accurate answer will be the one where the rounding comes at the end.



    Let's test that theory. I claim that $(2.5^2)^3$ will yield a more accurate answer as to what the actual value of the data squared and then cubed will be than $round(2.5^2)^3$



    The measured data $2.5$ means that the actual data might be anywhere between $2.45$ and $2.55$ so that it rounds to $2.5$.



    Let's evaluate what the data squared and then cubed would be if the the actual measure of the data were the minimum that it could be given how accurate your measurement was.



    $2.45^2 = 6.0025$
    $6.0025^3 = 216.270112515625$



    Remember that we are assuming that the exact measure of the data is 2.45. Therefore, it has infinite significant figures, and we don't round anywhere.



    Now look at the maximum possible measure that the data could be.



    $2.55^2 = 6.5025$
    $6.5025^2 = 274.941996890625$



    Now let's find the average of those two answers. Since our measured data is the average of the two extremes of the data, the answer should be the rounded average of the evaluations with the two extremes.



    $(216.270112515625 + 274.941996890625)/2 = 245.606054703125$



    Now let's solve using the measured data using significant figures.



    Start with rounding during the calculations



    $2.5 ^ 2 = 6.25$
    $6.3 ^ 3 = 250.047$



    That rounds to $2.5*10^2$ or $250$ with two significant figures.



    Now do it with rounding only at the end



    $2.5 ^ 2 = 6.25$
    $6.25 ^ 3 = 244.140625$



    That rounds to $2.4*10^2$ or $240$ with two significant figures. While the exact average value does round to the first answer, you can see that unrounded value the second time was much closer to the exact value. (244.14... was closer to 245.60... than 250.04... was). You can see how if we continued to do calculations on the data, the first method (with rounding during the calculations) would become more and more inaccurate because the calculations are being done with a number that is farther away from the exact average of the extremes of the data than the answer we got from rounding only at the end (and if we continued manipulating that number, 244.14..., we wouldn't round it, so it would indeed be closer to that average of 245.6...)



    I hope that helps, that you were able to follow along, and that I didn't make any mistakes.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      When working with significant figures, you should always round off at the end to avoid round-off error. For example, $2.5 * 2.5 = 6.25$ with two significant figures. If you round now ($6.3$) and then cube it, you get $250.047$ with two significant figures (rounds to $250$ or $2.5*10^2$). If you cube before rounding, you get $244.140625$ with two significant figures (rounds to $240$ or $2.4*10^2$). As you can see, rounding during the calculations versus rounding after the calculations often causes differences. The more accurate answer will be the one where the rounding comes at the end.



      Let's test that theory. I claim that $(2.5^2)^3$ will yield a more accurate answer as to what the actual value of the data squared and then cubed will be than $round(2.5^2)^3$



      The measured data $2.5$ means that the actual data might be anywhere between $2.45$ and $2.55$ so that it rounds to $2.5$.



      Let's evaluate what the data squared and then cubed would be if the the actual measure of the data were the minimum that it could be given how accurate your measurement was.



      $2.45^2 = 6.0025$
      $6.0025^3 = 216.270112515625$



      Remember that we are assuming that the exact measure of the data is 2.45. Therefore, it has infinite significant figures, and we don't round anywhere.



      Now look at the maximum possible measure that the data could be.



      $2.55^2 = 6.5025$
      $6.5025^2 = 274.941996890625$



      Now let's find the average of those two answers. Since our measured data is the average of the two extremes of the data, the answer should be the rounded average of the evaluations with the two extremes.



      $(216.270112515625 + 274.941996890625)/2 = 245.606054703125$



      Now let's solve using the measured data using significant figures.



      Start with rounding during the calculations



      $2.5 ^ 2 = 6.25$
      $6.3 ^ 3 = 250.047$



      That rounds to $2.5*10^2$ or $250$ with two significant figures.



      Now do it with rounding only at the end



      $2.5 ^ 2 = 6.25$
      $6.25 ^ 3 = 244.140625$



      That rounds to $2.4*10^2$ or $240$ with two significant figures. While the exact average value does round to the first answer, you can see that unrounded value the second time was much closer to the exact value. (244.14... was closer to 245.60... than 250.04... was). You can see how if we continued to do calculations on the data, the first method (with rounding during the calculations) would become more and more inaccurate because the calculations are being done with a number that is farther away from the exact average of the extremes of the data than the answer we got from rounding only at the end (and if we continued manipulating that number, 244.14..., we wouldn't round it, so it would indeed be closer to that average of 245.6...)



      I hope that helps, that you were able to follow along, and that I didn't make any mistakes.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        When working with significant figures, you should always round off at the end to avoid round-off error. For example, $2.5 * 2.5 = 6.25$ with two significant figures. If you round now ($6.3$) and then cube it, you get $250.047$ with two significant figures (rounds to $250$ or $2.5*10^2$). If you cube before rounding, you get $244.140625$ with two significant figures (rounds to $240$ or $2.4*10^2$). As you can see, rounding during the calculations versus rounding after the calculations often causes differences. The more accurate answer will be the one where the rounding comes at the end.



        Let's test that theory. I claim that $(2.5^2)^3$ will yield a more accurate answer as to what the actual value of the data squared and then cubed will be than $round(2.5^2)^3$



        The measured data $2.5$ means that the actual data might be anywhere between $2.45$ and $2.55$ so that it rounds to $2.5$.



        Let's evaluate what the data squared and then cubed would be if the the actual measure of the data were the minimum that it could be given how accurate your measurement was.



        $2.45^2 = 6.0025$
        $6.0025^3 = 216.270112515625$



        Remember that we are assuming that the exact measure of the data is 2.45. Therefore, it has infinite significant figures, and we don't round anywhere.



        Now look at the maximum possible measure that the data could be.



        $2.55^2 = 6.5025$
        $6.5025^2 = 274.941996890625$



        Now let's find the average of those two answers. Since our measured data is the average of the two extremes of the data, the answer should be the rounded average of the evaluations with the two extremes.



        $(216.270112515625 + 274.941996890625)/2 = 245.606054703125$



        Now let's solve using the measured data using significant figures.



        Start with rounding during the calculations



        $2.5 ^ 2 = 6.25$
        $6.3 ^ 3 = 250.047$



        That rounds to $2.5*10^2$ or $250$ with two significant figures.



        Now do it with rounding only at the end



        $2.5 ^ 2 = 6.25$
        $6.25 ^ 3 = 244.140625$



        That rounds to $2.4*10^2$ or $240$ with two significant figures. While the exact average value does round to the first answer, you can see that unrounded value the second time was much closer to the exact value. (244.14... was closer to 245.60... than 250.04... was). You can see how if we continued to do calculations on the data, the first method (with rounding during the calculations) would become more and more inaccurate because the calculations are being done with a number that is farther away from the exact average of the extremes of the data than the answer we got from rounding only at the end (and if we continued manipulating that number, 244.14..., we wouldn't round it, so it would indeed be closer to that average of 245.6...)



        I hope that helps, that you were able to follow along, and that I didn't make any mistakes.






        share|cite|improve this answer









        $endgroup$



        When working with significant figures, you should always round off at the end to avoid round-off error. For example, $2.5 * 2.5 = 6.25$ with two significant figures. If you round now ($6.3$) and then cube it, you get $250.047$ with two significant figures (rounds to $250$ or $2.5*10^2$). If you cube before rounding, you get $244.140625$ with two significant figures (rounds to $240$ or $2.4*10^2$). As you can see, rounding during the calculations versus rounding after the calculations often causes differences. The more accurate answer will be the one where the rounding comes at the end.



        Let's test that theory. I claim that $(2.5^2)^3$ will yield a more accurate answer as to what the actual value of the data squared and then cubed will be than $round(2.5^2)^3$



        The measured data $2.5$ means that the actual data might be anywhere between $2.45$ and $2.55$ so that it rounds to $2.5$.



        Let's evaluate what the data squared and then cubed would be if the the actual measure of the data were the minimum that it could be given how accurate your measurement was.



        $2.45^2 = 6.0025$
        $6.0025^3 = 216.270112515625$



        Remember that we are assuming that the exact measure of the data is 2.45. Therefore, it has infinite significant figures, and we don't round anywhere.



        Now look at the maximum possible measure that the data could be.



        $2.55^2 = 6.5025$
        $6.5025^2 = 274.941996890625$



        Now let's find the average of those two answers. Since our measured data is the average of the two extremes of the data, the answer should be the rounded average of the evaluations with the two extremes.



        $(216.270112515625 + 274.941996890625)/2 = 245.606054703125$



        Now let's solve using the measured data using significant figures.



        Start with rounding during the calculations



        $2.5 ^ 2 = 6.25$
        $6.3 ^ 3 = 250.047$



        That rounds to $2.5*10^2$ or $250$ with two significant figures.



        Now do it with rounding only at the end



        $2.5 ^ 2 = 6.25$
        $6.25 ^ 3 = 244.140625$



        That rounds to $2.4*10^2$ or $240$ with two significant figures. While the exact average value does round to the first answer, you can see that unrounded value the second time was much closer to the exact value. (244.14... was closer to 245.60... than 250.04... was). You can see how if we continued to do calculations on the data, the first method (with rounding during the calculations) would become more and more inaccurate because the calculations are being done with a number that is farther away from the exact average of the extremes of the data than the answer we got from rounding only at the end (and if we continued manipulating that number, 244.14..., we wouldn't round it, so it would indeed be closer to that average of 245.6...)



        I hope that helps, that you were able to follow along, and that I didn't make any mistakes.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 13 '18 at 1:46









        ElliotThomasElliotThomas

        186




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