Calculating Defect Rate Probability with BINOMDIST

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SUMMARY

The discussion centers on calculating the probability of finding no more than two defective closures in a sample of 1000 vaccine vials using the Microsoft Excel BINOMDIST function. The defect rate is established at 0.08%, with parameters set as number_s = 2, trials = 1000, and probability_s = 0.0008. It is confirmed that setting cumulative = TRUE directly provides the probability of two or fewer defects, negating the need to subtract from 1. The calculated probability is 0.9526, indicating a high likelihood of meeting the defect criteria.

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  • Understanding of Microsoft Excel functions, specifically BINOMDIST
  • Knowledge of probability concepts, including cumulative distribution
  • Familiarity with defect rate calculations and statistical significance
  • Basic grasp of Poisson processes and chi-square estimations
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  • Study the application of Poisson distribution in quality control scenarios
  • Explore chi-square tests for goodness of fit and their applications
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kristymassi
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it seems me true but prof. said its not..

You are asked to calculate the probability that a sample of 1000 vaccine vials from a filling line contains no more than two (2) defective closures. From prior data, you know that the particular filling line in question has a defect rate of 0.08%. Given this information, you use the Microsoft Excel BINOMDIST function with number_s =2, trials = 1000, and probability_s = 0.0008. In order to yield the desired probability, you should set the value of cumulative = TRUE and subtract the result of the function from 1.000.

is that true?

what is your idea?
 
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kristymassi said:
it seems me true but prof. said its not..

You are asked to calculate the probability that a sample of 1000 vaccine vials from a filling line contains no more than two (2) defective closures. From prior data, you know that the particular filling line in question has a defect rate of 0.08%. Given this information, you use the Microsoft Excel BINOMDIST function with number_s =2, trials = 1000, and probability_s = 0.0008. In order to yield the desired probability, you should set the value of cumulative = TRUE and subtract the result of the function from 1.000.

is that true?

what is your idea?

You gave an observed rate of 8 defects per 1000 but no variance. If you can assume a Poisson process then the mean and variance is 8 and the SD=sqrt (8). From this you can estimate the probability of 2 defects per thousand.

EDIT: I'm confused about your denominator. Is it 1000 or 10000? In either case the above works as long as your denominator doesn't change. You can also use a chi- square estimate where E=8, O=2 and Chi sq=(E-O)^2/E. You can get the probability from Chi Square tables.
 
Last edited:
SW VandeCarr, thank you very much

When we set the value of cumulative to TRUE, we already getthe probability of 2 or less
than 2 (that is no more than 2) defectives. So we need not subtract the answer from 1.
(The probability answer is 0.9526.)
 

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