Interval for a normal distribution

USN2ENG
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Homework Statement



A machine fills cereal boxes, normally distributed, with standard deviation of .1 oz. What amount setting should the machine be set to if only 1% of the boxes can have less than 16oz of cereal?

Homework Equations





The Attempt at a Solution



I am thinking that I just manipulate the confidence interval for a normal distribution. So x - 2.33(σ) = 16, with 2.33 being the z-score for 99%. My amount would then be 16.233. Am I way off? Thanks
 
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USN2ENG said:

Homework Statement



A machine fills cereal boxes, normally distributed, with standard deviation of .1 oz. What amount setting should the machine be set to if only 1% of the boxes can have less than 16oz of cereal?

Homework Equations


The Attempt at a Solution



I am thinking that I just manipulate the confidence interval for a normal distribution. So x - 2.33(σ) = 16, with 2.33 being the z-score for 99%. My amount would then be 16.233. Am I way off? Thanks

Your reasoning and your answer are correct.
 
Last edited:
Thanks, Dr. Vickson
 
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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