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Significant figures

Discussion in 'Mathematics' started by lizzywilmot, Aug 5, 2012.

  1. lizzywilmot

    lizzywilmot New commenter

    Are there any websites that will help me work out a signficant numbers problem I have?
  2. lizzywilmot

    lizzywilmot New commenter

    Are there any websites that will help me work out a signficant numbers problem I have?
  3. 3 years on this forum and no reply.
    Have you tried google
  4. lizzywilmot

    lizzywilmot New commenter

    I onlyposted this a few minutes ago [​IMG]

    I have tried google, all I get is the rules for significant figures and a few examples that are not relevant to what I need. Oh well, more searching on google.
  5. What sig fig problem do you have?
  6. What is the significant figures problem you need answers to?
  7. harsh-but-fair

    harsh-but-fair Star commenter

    Welcome to TES, Perses.
  8. lizzywilmot

    lizzywilmot New commenter

    Calculate the upper and lower bounds for the length marked x cm on the rectangle. The area and length are both accurate to 2 s.f.
    There is a drawing of a rectangle with one side marked 10 cm and the other side marked x cm. The area is given as 100 cm^2.

    Thanks elainem.
  9. lizzywilmot

    lizzywilmot New commenter

    And thank you ProfCalculus
  10. That is a very very very stupid question..
    Sorry, but it it really is a very stupid question.
    I'm not normally this forthright, but I get really
    upset about silliness.
  11. Which bit of the question are you unsure about? I don't want to tell you things you probably know already but willing to help if I can.
  12. lizzywilmot

    lizzywilmot New commenter

    Am I right in thinking the boundaries for 10 will be 9.5 and 10.5?

    And the boundaries for 100 will 95 and 105?
  13. adamcreen

    adamcreen Occasional commenter

    yes and yes, but only if it's 10 to the nearest unit, and 100 to the nearest ten.

    the limits are plus/minus HALF the degree of accuracy.

    so 24 to the nearest 2 would have limits of 23 and 25, for example.

    but that's not significant figures, it's limits of accuracy.
  14. lizzywilmot

    lizzywilmot New commenter

    Thanks Adam, that is most helpful.[​IMG]
    Yes, I know it was a limits of accuracy question but it was the 2 sf in the problem I wanted to check I was right on.[​IMG]
  15. frustum

    frustum Star commenter

    There's a kind of discontinuity with significant figures as the number of digits changes. Maybe your question has been set by someone who wanted to exploit this to make you think. Or maybe it's just random numbers, computer-generated, in which case it's a bad one to have come up (and I suspect the program will have calculated using 9.5/10.5).
    9.5 is 9.5 to 2s.f., so if we're thinking of the lower bound as the lowest number which would get rounded to 10 when rounded to 2s.f., I make it 9.95. 10.5 is fine as the upper bound.

  16. lizzywilmot

    lizzywilmot New commenter

    The answers are given as lower bound = 9.5 cm and upper bound = 10.6 (1dp).
    Using your figures I get lower bound 95 / 9.95 = 9.54 cm so that agrees with the answer given.
    But I can't get their answer for the upper bound, can you?

    I've just tried this using 100.5 and 99.5 for the boundaries of 100 to 2 sf. and it gives their answers. Have they got it wrong?

  17. Absolutely, and those are the numbers you should be working with. Consider 1 s.f., an example would be 1, then it would be a fairly inaccurate measurement, with a possible range between 0.5 and 1.5. For 2 s.f., an example would be 10, and the accuracy changes by a factor of 10, and its range would be + or - 0.5, therefore values of 10.5 or 9.5. The area was also to 2 s.f., therefore the first two digits are significant, therefore you can only expect accuracy to the nearest ten, plus or minus 5, giving a range from 95 to 105.
    So, the largest possible value for the unknown side would be the largest possible area divided by the smallest possible side, i.e. 105 / 9.5 = 11.0526cm, but the answer must also be given to the nearest 2 s.f., which would be 11cm. For the smallest value for the unknown side, it would use the smallest possible area divided by the largest possible side, i.e. 95 / 10.5 = 9.0476cm, and making this to 2 s.f. would give 9.0cm.
    No, this is not a stupid question because in the real world (e.g. Engineering) tolerances and accuracies are very relevant.

  18. Nazard

    Nazard New commenter

    Alex - you might want to revisit this.

    The upper and lower bound, given to significant figures, of a power of 10 is not symmetrical. For example, the upper bound of 100 given to 2sf is 105 - that is not a problem. Round off 104.999... to 2sf and you get 100.

    The lower bound is not 95, though. Round off 95 to 2sf and you get ... 95 ! The lower bound is actually 99.5

    I hope that the original question was designed to exploit this.
  19. lizzywilmot

    lizzywilmot New commenter

    Did I misunderstand the question then? What you say Nazard makes sense, but I thought it was the 100 rounded to 2sf so why do we have to round the 95 to 2sf too?
  20. I am thinking from a practical point of view: the 100 to me signifies some sort of measurement, in this case an area. The tolerance is given by significant figures, so in this case the two most significant figures indicate the tolerance. Taking the original value of 100 and tolerance as 2sf, then the reading may be out by + or - 5, hence 105 or 95.
    Had the original value been 99.99999 then the 2sf tolerance would have had a different meaning, i.e. by + or - 0.5. So a measurement of 99 with a tolerance given by 2sf accuracy, the reading could be 99.5 or 98.5.
    This would provide a more meaningful problem, so if my understanding is incorrect, then the problem may well be a silly one with little real world relevance.

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