# Thread: Normal distribution beams laid side by side

1. ## 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

2. Originally Posted by parrot77
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 $30\times 300.765$ mm and SD $\sqrt{30}\times 7.86$ mm, and since the critical $z$value for a right tail probability on $0.001$ is $3.09$ you want:

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

CB