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Homework Help: Measurement uncertainty and error

  1. Jun 30, 2013 #1
    Suppose we have [itex]x[/itex] plates of wood of thickness [itex] (0.0150 \pm 0.0002) \text{ m}[/itex] and we wish to stack them into a pile of height [itex] (1.000 \pm 0.001) \text{ m}[/itex]. The number of plates [itex]x[/itex] required is then

    [itex]x = \dfrac {1.000 \pm 0.001}{0.0150 \pm 0.0002} \, .[/itex]

    How many plates will certainly fit in the pile? Well, the minimum value of [itex]x[/itex] is the number of plates we with certainty can claim will fit in the stack of wood-plates. [itex]x[/itex] is minimal when the nominator is at its minimum and denominator at its maximum. Thus

    [itex]x_{\text{min.}} = \dfrac { 1.000 - 0.001 }{0.0150 + 0.0002} = \dfrac {0.999}{0.0152} \approx 65.7 \, .[/itex]

    So, since [itex]x \geq 65.7[/itex] we can conclude that the least integral value of [itex]x[/itex] is [itex]66[/itex]. Thus the number of wood-plates that we can with certainty claim will fit are [itex]66[/itex] plates. Problem is that my book claims [itex]65[/itex]; please tell me it is wrong? It just cannot be correct; if [itex]x = 65[/itex] then the length of the stack will no longer be as required in the premises.

    Note: I was unable to access the homework-section.
  2. jcsd
  3. Jul 1, 2013 #2

    Simon Bridge

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    What is the probability that 66 plates will be too high, vs 65 plates?
  4. Jul 1, 2013 #3


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    It depends on the exact wording of the question.
    65 could range from 0.9620 to 0.9880 so is certainly under height.
    66 could range from 0.9768 to 1.0032 so it may be over height.
    If the pile must fit below 0.999 then the answer must be 65.
  5. Jul 1, 2013 #4
    Hm, not quite sure how to calculate that; could you show?

    Well as I wrote in my previous post, we want the stack to be of height [itex](1.000 \pm 0.001) \text{ m}[/itex]. The smallest height of the stack is [itex]1.000-0.001 = 0.999 \text{ m}[/itex] and the greatest height of the stack is [itex]1.000 + 0.001 = 1.001 \text{ m}[/itex]. [itex]65[/itex] wood-plates of thickness [itex]0.0150 + 0.0002 = 0.0152 \text{ m} [/itex] (the denominator had to assume its smallest value) give a stack height of [itex]0.0152 \cdot 65 = 0.988 \text{ m}[/itex] but this is less than the smallest height, [itex]0.999 \text{ m}[/itex], of the stack so clearly [itex]65[/itex] is not the amount of plates we can guarantee will fit in the stack.
  6. Jul 1, 2013 #5


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    I believe you may have rephrased the question rather than quote it exactly as originally presented. The question and the answer are very closely related. I ask if the correct answer is given as 65, then how should the question be interpreted. Maybe Question; “If the height is restricted to 1.000 +/- .001 then how many 0.0150+/-.0002 could you certainly insert? Answer; 65.
    It all depends on the exact wording of the question.
    If you post the original question exactly as presented then it is probably resolvable.
  7. Jul 1, 2013 #6
    you cant have 65.7 plates...you cannot round up to 66. You must round down to 65
  8. Jul 1, 2013 #7
    The wording of the question was something in the lines of
    The equivalent question is clearly "what is the least amount of plates one can fit in the stack?", and you can see how I tackled this very question in my previous post. Are my reasonings incorrect or will I have to conclude that the book simply is incorrect when it says 65 (and not 66)?
  9. Jul 1, 2013 #8
    Hm, interesting point but I do not understand why we will have to round down? Could you please elaborate?
  10. Jul 1, 2013 #9
    You can only have an integer (whole number) of plates.
    65.7 calculated number of plates can, in my humble opinion, only ever be 65 real plates, not 66.
    Nearly 66 but not 66!
  11. Jul 1, 2013 #10
    Yes I understand that we can only have an integral number of wood-plates, but 65.7 is not an upper bound, rather it is a lower bound. Therefore it would not make sense to round down, since 65.7 is the smallest value [itex]x[/itex] – the number of wood-plates – can assume. The first and least integral value for [itex]x[/itex] is 66, not 65.
  12. Jul 1, 2013 #11

    Simon Bridge

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    Baluncore actually shows you how to do it by rule of thumb ... have you done no statistics? He shows that there are certainly 65 plates in the stack and there are probably 66 without having to do a hard calculation.

    The uncertainty ##\pm## values is not usually considered a hard limit - it is the standard deviation of a Gaussian (normal) distribution. Do you know how to find probabilities from the normal distribution?

    Your problem may be a little trickier than that -
    ... that's not good enough: you need to get the wording exactly.

    Note. 1 wood plate will certainly fit in the stack, so would 2.
    What is it asking for exactly.

    Your calculation of 65.7 says that the number of plates is less than 66 and more than 65 ... therefore...
  13. Jul 3, 2013 #12

    D H

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    No! It's not "least". Zero is the least number of plates that will fit. The equivalent statement is clearly what is the largest number of plates one can fit in the stack with zero chance of exceeding the lower limit on the stack height. You are looking for the greatest lower bound in the integers. That's 65, not 66.

    A simpler way to look at it: If every plate in the stack was 0.2 mm higher than nominal, a stack of 66 plates would be 1.0032 meters tall. That exceeds the limit. The answer cannot be 66.
  14. Jul 3, 2013 #13

    Simon Bridge

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    It depends how the course is handling it a bit though doesn't it?
    ... is the quoted uncertainty to be treated as a margin for error (so any stack between 999mm and 1001mm is acceptable, for eg?) or a statistical uncertainty (say, in the equipment used to measure the height?).
  15. Jul 4, 2013 #14
    Thanks for the replies. I am beginning to view the problem much differently, but let me give you my rendition of this problem. The problem, in its verbatim form, reads:
    This is how I approach it: Let [itex]x[/itex] be the number of plates in the stack; this number will have a maximum and minimum value, so we can write [itex]x_{\text{min}} \leqslant x \leqslant x_{\text{max}}[/itex]. [itex]x[/itex] can be calculated by forming the fraction where the nominator is the height of the stack and the denominator is the thickness of each plate. I.e.

    [itex]x = \dfrac {1.000 \pm 0.001}{0.0150 \pm 0.0002} \, .[/itex]

    We will calculate the value which lies right in the middle of the interval [itex]x_{\text{min}} \leqslant x \leqslant x_{\text{max}}[/itex] by ignoring the error bounds. Thus

    [itex]x = \dfrac {1.000}{0.0150} \approx 66.7 \, .[/itex]

    The maximum value for [itex]x[/itex] can be obtained by letting the nominator assume its largest value and the denominator its smallest value. Thus

    [itex]x_{\text{max}} = \dfrac {1.000 + 0.001}{0.0150 - 0.0002} \approx 67.6 \, .[/itex]

    Similarly, the minimum value for [itex]x[/itex] can be obtained by letting the nominator assume its smallest value and the denominator its largest value. Thus

    [itex]x_{\text{min}} = \dfrac {1.000 - 0.001}{0.0150 + 0.0002} \approx 65.7 \, .[/itex]

    The difference divided by two between these two extreme values is 0.95 (in order to find the error bounds), so we can write

    [itex]x = 66.7 \pm 0.95 \, .[/itex]

    Now we can conclude that the smallest value for [itex]x[/itex] is 66.7 - 0.95 = 65.75, which clearly cannot be rounded down to 65. Where is the fault in the argument above?

    Note: The book I have is extremely basic and fundamental, so it surely does not have to get too advanced.
  16. Jul 4, 2013 #15

    Simon Bridge

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    1000/15=66.67 ... so you've just got the expected number of plates.

    You have to pick a whole number ... probably less than 67.
    The number of plates has to stack to a height between 999mm and 1001mm (if I follow your thinking)...

    Examine this in terms of the ranges you have got.
    If you stacked 67 plates, what are the maximum and minimum height the stack could be?
    Do the same for fewer and fewer plates until you get a number which is certainly going to be in the final pile.

    You don't have to round off - just truncate.
  17. Jul 4, 2013 #16
    Yes and then we establish an interval for [itex]x[/itex] from which we ought to be able to state how many plates certainly will fit (the least value of [itex]x[/itex] in that interval).

    Yes an integer indeed, but what do you mean with "probably less than 67"? 67 is the least integral value for [itex]x[/itex] in the established interval.

    Yes, according to the problem's statement.

    Why is choosing the least integral value for [itex]x[/itex] in the established interval an incorrect approach?
  18. Jul 4, 2013 #17

    Simon Bridge

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    If you follow the suggestion
    ... you aught to see why for yourself.
  19. Jul 4, 2013 #18
    This is a problem of linguistics, not mathematics: how do we interpret the words "most certainly fit"?

    The question starts with the words "A couple of plates"; "a couple" means "two", whereas the number of plates that is stacked is certainly a lot more than two. Immediately we are aware that the language of the question is not reliable.

    One possible interpretation is "what is the upper bound on the number of plates whose maximum possible aggregate thickness is less than or equal to the lower bound on the allowable height of the pile" - the answer to this is clearly 65, and we can infer that this is what the question setter intended.

    66 plates could have an aggregate thickness between 0.9768m and 1.0032m; I am not sure what interpretion would have the answer 66.
  20. Jul 4, 2013 #19


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    I agree. The height of the stack of plates cannot be guaranteed to be in the given range, so there is definitely something wrong with that interpretation of the question. A reasonable bet is that the range for the stack height represents an uncertainty in the available space. Therefore we're looking for the largest number of plates that will definitely fall short of the lower bound of the range: 65.
  21. Jul 6, 2013 #20
    Note: The language cannot be examined that meticulously due to the fact that I have had it translated. Simply consider it as a very simple problem presented to someone who has never taken physics before i.e. it is not that precise and complicated as some of you are making it (though I appreciate the effort, it presents new perspectives).

    @Simon Bridge: Ok, I realize 65 plates yield a stack-height that suits the interval given in the problem, whereas 66 plates exceed it. But then, what exactly was wrong in my approach (presented in post #16)? Clearly I must have reasoned incorrectly but in my eyes it is correct.

    How and why exactly did you manage to arrive at 65 and not 66, unlike me? I need to know what went wrong in my argument in post #16, only then will I be able to comprehend; what do you say?
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