Interval for a normal distribution

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SUMMARY

The discussion centers on determining the appropriate machine setting for filling cereal boxes to ensure that only 1% of the boxes contain less than 16 oz of cereal. The standard deviation of the filling process is 0.1 oz. The correct calculation involves using the z-score of 2.33, which corresponds to the 99% confidence level. The resulting machine setting should be 16.233 oz to meet the specified requirement.

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  • Familiarity with z-scores and confidence intervals
  • Basic knowledge of statistical calculations
  • Ability to manipulate equations involving standard deviation
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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
 

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