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Thread: Normal distribution beams laid side by side

  1. #1
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    Normal distribution beams laid side by side

    Beams of mean width = 300.765mm and standard deveaition of 7.86. If 30 beams are laid side by side what width does support to beams need to be such that there is only a 1:1000 risk that the edge beams will not be fully supported? Assume normal distribution

    I have tried by using the standard normal tables to find Z relating to 1 beam and multplying it by 30 ie (P) = 0.001 therefore Z =3.9

    Z= X-mean/std. dev

    X = Z*std dev + mean = 3.9*7.86 + 300.765= 331.42mm

    Thus support must be 30*331.42= 9943mm wide

    However this does not seem correct. The probability of 30 of the units being 331 is likely to be even less than 1:1000. Can anyone help me to properly integrate the number of beams into the problem

    Thanks
    Last edited by mr fantastic; Mar 11th 2011 at 12:49 PM. Reason: Deleted irrelevant text.
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  2. #2
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    Quote Originally Posted by parrot77 View Post
    Beams of mean width = 300.765mm and standard deveaition of 7.86. If 30 beams are laid side by side what width does support to beams need to be such that there is only a 1:1000 risk that the edge beams will not be fully supported? Assume normal distribution

    I have tried by using the standard normal tables to find Z relating to 1 beam and multplying it by 30 ie (P) = 0.001 therefore Z =3.9

    Z= X-mean/std. dev

    X = Z*std dev + mean = 3.9*7.86 + 300.765= 331.42mm

    Thus support must be 30*331.42= 9943mm wide

    However this does not seem correct. The probability of 30 of the units being 331 is likely to be even less than 1:1000. Can anyone help me to properly integrate the number of beams into the problem

    Thanks
    The width of 30 beams laid side by side is normally distributed with mean $\displaystyle 30\times 300.765$ mm and SD $\displaystyle \sqrt{30}\times 7.86$ mm, and since the critical $\displaystyle $$z $value for a right tail probability on $\displaystyle $$0.001$ is $\displaystyle $$3.09$ you want:

    $\displaystyle 30\times 300.765+3.09\times \sqrt{30}\times 7.86\approx 9155.98$ mm

    CB
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  3. #3
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    Thanks for your help
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