Thread: 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

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 mm and SD mm, and since the critical value for a right tail probability on is you want:

mm

CB

Thanks for your help