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Math Help - Logic Puzzle, can any one solve this?

  1. #1
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    Angry Logic Puzzle, can any one solve this?

    Suppose someone tells you that the number of marbles in a jar is 100 with a margin of error of 5, and that 5 has a margin of error 2.

    What is the true margin of error of the number of marbles in the jar expressed as a single number?

    Hopefully this question should be hard enough to be a puzzle.
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  2. #2
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    Re: Logic Puzzle, can any one solve this?

    Seven, no?
    The margin of error could be as much as 7 and as little as 3, so we should account for a margin of error at greatest of seven.
    I'm not sure, maybe I don't have the best idea of what is meant by margin of error, but it makes sense to me.
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  3. #3
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    Re: Logic Puzzle, can any one solve this?

    I'm not sure if you can stack margin of errors like that.
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  4. #4
    Member agentmulder's Avatar
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    Re: Logic Puzzle, can any one solve this?

    Quote Originally Posted by Sneaky View Post
    Suppose someone tells you that the number of marbles in a jar is 100 with a margin of error of 5, and that 5 has a margin of error 2.

    What is the true margin of error of the number of marbles in the jar expressed as a single number?

    Hopefully this question should be hard enough to be a puzzle.
    Well, the margin of error on 100 could be 7 or 6 or 5 or 4 or 3

    If all margins of error are equally likely, would it not make sense to take the average margin of error?

    (7 + 6 + 5 + 4 + 3)/5 = 5

    My guess is margin of error stays 5, but I don't like it.

    If the largest margin of error could be 7, then shouldn't we consider that...just to be safe?

    As eulcer observed?

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  5. #5
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    Re: Logic Puzzle, can any one solve this?

    I think I figured it out.
    If the margin of error is 5 with a margin of error of 2, then, the number of marbles can be

    Code:
    93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 // error of 7
       94 95 96 97 98 99 100 101 102 103 104 105 106         // error of 6
          95 96 97 98 99 100 101 102 103 104 105                // error of 5
             96 97 98 99 100 101 102 103 104                        // error of 4
                97 98 99 100 101 102 103                               // error of 3
    Does anyone see where to go from here, how do I express the margin of error as a single number? Seems like you can have a margin of error but some parts of it are more likely of appearing.

    So there's 55 numbers.
    93: 1/55
    94: 2/55
    95: 3/55
    96: 4/55
    97: 5/55
    98: 5/55
    99: 5/55
    100: 5/55
    101: 5/55
    102: 5/55
    103: 5/55
    104: 4/55
    105: 3/55
    106: 2/55
    107: 1/55
    Last edited by Sneaky; October 9th 2011 at 04:46 PM.
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  6. #6
    Grand Panjandrum
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    Re: Logic Puzzle, can any one solve this?

    Quote Originally Posted by Sneaky View Post
    I think I figured it out.
    If the margin of error is 5 with a margin of error of 2, then, the number of marbles can be

    Code:
    93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 // error of 7
       94 95 96 97 98 99 100 101 102 103 104 105 106         // error of 6
          95 96 97 98 99 100 101 102 103 104 105                // error of 5
             96 97 98 99 100 101 102 103 104                        // error of 4
                97 98 99 100 101 102 103                               // error of 3
    Does anyone see where to go from here, how do I express the margin of error as a single number? Seems like you can have a margin of error but some parts of it are more likely of appearing.

    So there's 55 numbers.
    93: 1/55
    94: 2/55
    95: 3/55
    96: 4/55
    97: 5/55
    98: 5/55
    99: 5/55
    100: 5/55
    101: 5/55
    102: 5/55
    103: 5/55
    104: 4/55
    105: 3/55
    106: 2/55
    107: 1/55
    You should first decide what the term "margin of error" means, it is ambiguous. From what you post it seems you think is is the semi-range.

    You seem to assume that you have uniform distributions for your first two margin of error values, but you will not have a uniform distribution for the final result.

    CB
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