Measurement uncertainty and error

In summary: So the equivalent question is, what is the largest integral number of plates that will fit in the stack with zero chance of exceeding the lower limit?What is the probability that 66 plates will be too high, vs 65 plates?In summary, the conversation discusses the calculation of the number of wood-plates that can fit in a stack of a certain height, taking into account the uncertainty in the dimensions. The minimum number of plates that can certainly fit in the stack is 66, while the maximum number that may fit is 65. The probability of 66 plates being too high is dependent on the exact wording of the question and the probability of 65 plates being too high is
  • #1
eroxore
23
0
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.
 
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  • #2
What is the probability that 66 plates will be too high, vs 65 plates?
 
  • #3
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.
 
  • #4
Simon Bridge said:
What is the probability that 66 plates will be too high, vs 65 plates?

Hm, not quite sure how to calculate that; could you show?

Baluncore said:
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.

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.
 
  • #5
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.
 
  • #6
you can't have 65.7 plates...you cannot round up to 66. You must round down to 65
 
  • #7
The wording of the question was something in the lines of
We have wood-plates of thickness (0.0150 ± 0.0002) [m] and form a stack with these of height (1.000 ± 0.001) [m]. How many wood-plates will certainly fit in this stack?
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)?
 
  • #8
technician said:
you can't have 65.7 plates...you cannot round up to 66. You must round down to 65

Hm, interesting point but I do not understand why we will have to round down? Could you please elaborate?
 
  • #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!
 
  • #10
technician said:
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!

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.
 
  • #11
eroxore said:
Hm, not quite sure how to calculate that; could you show?
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 -
The wording of the question was something in the lines of...
... 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...
 
  • #12
eroxore said:
The wording of the question was something in the lines of "How many wood-plates will certainly fit in this stack?" The equivalent question is clearly "what is the least amount of plates one can fit in the stack?"
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.
 
  • #13
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?).
 
  • #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:
A couple of plates of thickness (1.50 ± 0.02) cm [= (0.0150 ± 0.0002) m] are to be piled up in a stack, which is (1.000 ± 0.001) m high.
a) How many plates can maximally fit?
b) How many plates will most certainly fit?

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.
 
  • #15
66.7...which clearly cannot be rounded down to 65.
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.
 
  • #16
Simon Bridge said:
1000/15=66.67 ... so you've just got the expected number of plates.

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).

Simon Bridge said:
You have to pick a whole number ... probably less than 67.

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.

Simon Bridge said:
The number of plates has to stack to a height between 999mm and 1001mm (if I follow your thinking)...

Yes, according to the problem's statement.

Simon Bridge said:
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.

Why is choosing the least integral value for [itex]x[/itex] in the established interval an incorrect approach?
 
  • #17
Why is choosing the least integral value for x in the established interval an incorrect approach?
If you follow the suggestion
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 aught to see why for yourself.
 
  • #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.
 
  • #19
MrAnchovy said:
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"
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.
 
  • #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.

MrAnchovy said:
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.

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?
 
  • #21
eroxore said:
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?

I used a different interpretion of the problem; instead of:
The number of plates has to stack to a height between 999mm and 1001mm
I read the requirement as:
The number of plates has to stack to a height that is less than a height of between 999mm and 1001mm
I can justify this interpretation in three different ways:
  1. it is the first interpretation that occurred to me, although I appreciate that I was not reading the problem in its original language;
  2. it appears to be what the questioner intended, as it has the answer given by the questioner;
  3. if the problem requires a pile that is between 0.999m and 1.001m then there is no solution so this is clearly not what the questioner intended.

You need to get used to the fact that sometimes problems are not stated clearly; sometimes this is deliberate (in order to get you to think harder), but often it is simply poor question setting. Sometimes questions (or answers) are actually incorrect - nobody is perfect, including examiners/authors/proof readers.
 
  • #22
eroxore said:
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).
The trouble with this is that this sort of problem requires exact wording ... the math has specialized language to reflect this. We have to interpret the problem or it cannot be completed.
There is no way to tell what a reasonable interpretation would be for your class, especially considering the words have been translated from an unfamiliar language - we are not there, we do not know the people involved. You have to extrapolate that from the discussion... sorry.

@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.
You did fine, you just didn't finish. I think it was the decision to round off that was not appropriate to the situation. If you round up you risk being too high and if you round down you may be too low - that is the nature of rounding off. In the problem description, you'd accept "too low" but reject "too high" - therefore you truncate instead of rounding.

The figure was arrived at by just working out the maximum conceivable, then working the possible range that this number of plates would give. Just like I said ... start at 67, work out the range, you'll see it's too high so go to 66, work out the range, could 66 plates possibly end up too high?

I didn't do the math myself - the idea is that you do it.
It is not unheard of for a model answer to be wrong after all ;)
 
  • #23
Δ
eroxore said:
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?
The phrase "fit in" carries implications that differ from those I'd associate with "comprise". To ensure an answer that agrees with the textbook, perhaps the question should be cast along the lines: "The shelves in a storage cupboard offer a vertical stacking space of 1.000 ± 0.001m. How many plates of thickness 0.0150 ± 0.0002m can you guarantee will fit in this space?"
 
  • #24
Having to test different number of plates in order to see whether the yielded stack-height satisfies the given interval becomes a very tedious task, undeniably so. In the key of the book, they simply form the fraction [itex]\frac {0.999}{0.0152}[/itex] (i.e. they compute the minimal value for [itex]x[/itex] by minimizing the nominator and maximizing the denominator), which essentially is what I did. But seemingly, they round down the quotient 65.7 to 65 (and not 66).

Now, building on my approach in post #16, we had that [itex]x \geqslant 65.7[/itex] (unless I did something wrong in that post, which has not been pointed out yet). Since [itex]x[/itex] not ought to be smaller than 65.7, what justifies rounding the quotient down to 65?
 
  • #25
It's the largest whole number smaller than 65.7.

You understand that the number of plates has to be an integer ... therefore the number must be 65 or 66.
The purpose of getting you to test each candidate number (there's only two - hardly tedious) was to try to get you to understand that it has to be less than 65.7 and not greater than 65.7.

You should repeat the exercise with different dimensions ... see if you can find a situation where rounding up will give the right answer.
 
  • #26
Simon Bridge said:
It's the largest whole number smaller than 65.7.

You understand that the number of plates has to be an integer ... therefore the number must be 65 or 66.
The purpose of getting you to test each candidate number (there's only two - hardly tedious) was to try to get you to understand that it has to be less than 65.7 and not greater than 65.7.

You should repeat the exercise with different dimensions ... see if you can find a situation where rounding up will give the right answer.

All right thanks, I will have to ponder the question a bit more for myself.
 

What is measurement uncertainty?

Measurement uncertainty refers to the potential error or variability in a measurement. It is the difference between the true value of a quantity and the value obtained from a measurement.

What are sources of measurement uncertainty?

Sources of measurement uncertainty can include instrumental limitations, human error, environmental conditions, and the inherent variability of the quantity being measured.

How is measurement uncertainty calculated?

Measurement uncertainty is typically calculated using statistical methods, such as standard deviation or confidence intervals. It involves analyzing the potential sources of error and their contributions to the overall uncertainty.

Why is measurement uncertainty important?

Measurement uncertainty is important because it provides a measure of the reliability and accuracy of a measurement. It allows scientists to understand the limitations and potential errors in their data, and make informed decisions based on that information.

How can measurement uncertainty be reduced?

Measurement uncertainty can be reduced by using more precise instruments, improving measurement techniques, and minimizing sources of error. It can also be reduced by repeating measurements multiple times and taking the average to reduce random error.

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