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Math Help - [SOLVED] Probability distribution problem :(

  1. #1
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    Unhappy [SOLVED] Probability distribution problem :(

    An environmental artist is planning to
    construct a rectangle with 36 m of fencing as
    part of an outdoor installation. If the length
    of the rectangle is a randomly chosen
    integral number of metres, what is the
    expected area of this enclosure?

    No idea how to approach this question
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  2. #2
    Super Member Anonymous1's Avatar
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    Quote Originally Posted by ibetan View Post
    If the length
    of the rectangle is a randomly chosen
    integral number of metres
    Can you clarify what this means.
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  3. #3
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    Quote Originally Posted by Anonymous1 View Post
    Can you clarify what this means.
    I think it means the length of the side is an integer
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  4. #4
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    I'm going to take a quick stab at this, but hopefully you'll post the correct answer if/when you find it.

    Okay, we have 36m of fencing to use and the person wants to build a rectangle.

    So, the person can rectangles of the following dimensions:

    1m width by 17m length
    2m width by 16m length
    3m width by 15m length
    4m width by 14m length
    5m width by 13m length
    6m width by 12m length
    7m width by 11m length
    8m width by 10m length

    Those are all of the possible dimensions that can be used to create a rectangular enclose whose perimeter is 36m.

    We're looking for the expected area of the enclosure, which means you're looking for the average area from all of the possible enclosure sizes listed above. So we calculate the area for each of the dimensions listed above:
    1*17=17
    2*16=32
    3*15=45
    4*14=64
    5*13=65
    6*12=72
    7*11=77
    8*10=80

    So, the expected or average area would simply be:

    \frac{17+32+45+64+65+72+77+80}{8}=56.5m

    I'm fairly confident that this is the correct way of approaching the problem, but if it is wrong I'd love to see the proper way to do it. Thanks
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  5. #5
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    Quote Originally Posted by downthesun01 View Post
    I'm going to take a quick stab at this, but hopefully you'll post the correct answer if/when you find it.

    Okay, we have 36m of fencing to use and the person wants to build a rectangle.

    So, the person can rectangles of the following dimensions:

    1m width by 17m length
    2m width by 16m length
    3m width by 15m length
    4m width by 14m length
    5m width by 13m length
    6m width by 12m length
    7m width by 11m length
    8m width by 10m length

    Those are all of the possible dimensions that can be used to create a rectangular enclose whose perimeter is 36m.

    We're looking for the expected area of the enclosure, which means you're looking for the average area from all of the possible enclosure sizes listed above. So we calculate the area for each of the dimensions listed above:
    1*17=17
    2*16=32
    3*15=45
    4*14=64
    5*13=65
    6*12=72
    7*11=77
    8*10=80

    So, the expected or average area would simply be:

    \frac{17+32+45+64+65+72+77+80}{8}=56.5m

    I'm fairly confident that this is the correct way of approaching the problem, but if it is wrong I'd love to see the proper way to do it. Thanks
    Thank you! that is the correct answer from the solution
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