Uncertainty in Measurement and significnt figures in physics












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












$begingroup$


(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






share|cite|improve this question















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
1






active

oldest

votes


















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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030720%2funcertainty-in-measurement-and-significnt-figures-in-physics%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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












      $begingroup$

      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




        186






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3030720%2funcertainty-in-measurement-and-significnt-figures-in-physics%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

            Image
            2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
            Read more

            ComboBox Display Member on multiple fields

            Image
            2 I'm trying to set the .DisplayMember property of a ComboBox in C#, but I want to bind it to multiple columns in the .DataSouce . My SQL looks like this: SELECT PersNbr, PersFirstName, PersMiddleName, PersLastName FROM Pers WHERE PersNbr = :persNbr; I'm saving this query in a DataTable so each column selected has it's own column in the Datatable . I want to make the .DisplayMember a combination of PersFirstName + PersMiddleName + PersLastName so their full name appears like this: comboBox.DisplayMemeber = "PersFirstName" + "PersMiddleName" + "PersLastName" I know I can just to this on the query: SELECT PersNbr, (PersFirstName || PersMiddleName || PersLastName) PersName and then just do this: comboBox.DisplayMember = "PersName"; but I
            Read more

            Is it possible to collect Nectar points via Trainline?

            Image
            2 Since I am not eligible for any Railcard, the only available discount/bonus scheme is Nectar Card for me. GWR, SWR, LNER and Virgin Trains allow to collect Nectar points and I have already linked a card to my GWR account. A friend suggested me to use Trainline App, because it can search for bus tickets either. I had a look at the application, but could not see any option for Nectar. Is there a way to collect Nectar points when I buy tickets via Trainline? uk trains loyalty-programs trainline share | improve this question asked Dec 2 '18 at 15:50 ahmedus 3,224 5 19 49
            Read more