Defective Tube Estimation Using Central Limit Theorem

rhyno89
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Homework Statement



The cross sectional area of a tube is u = 12.5 and SD = .2. When the area is less than 12 or greater than 13 it won't work. They are shipped in boxes of 1000, determine how many per box will be defective

Homework Equations





The Attempt at a Solution



So I think I have the solution but am a little confused with the process. To start I normalized the variables into Z values. So I did (12-12.5)/.2 and (13-12.5)/.2. When it was all said and done I found it to equal .0124 x 1000 = aprx 12 defects.

My question is do I have to add any correction factor while normalizing like for the binomial approximation and do I have to change the mean. I know that some types of these problems require me to start by multiplying the mean by the number and doing a similar change to the standard deviation. I don't really need much help with the computation, just getting a little confused with all the different cases.

Thanks alot
 
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Nope - you followed the correct steps: use the information to find the chance an item will be defective, then estimate the number in your shipment that will be defective.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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