What is the Probability of Insect Contamination in Multiple Chocolate Bars?

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SUMMARY

The discussion focuses on calculating the probability of insect contamination in chocolate bars using the Poisson distribution. The mean number of insect fragments in 225-gram chocolate bars is established at 14.4, with specific attention to a brand's contamination level of 1.82 for 28.35-gram bars. Participants debate the interpretation of probabilities for consuming one or more contaminated bars and the definition of "typical" contamination levels, with one participant calculating a probability of 0.000463 for contamination exceeding twice the mean.

PREREQUISITES
  • Understanding of Poisson distribution and its applications
  • Familiarity with EXCEL's POISSON.DIST function
  • Basic probability concepts, including P(A happens) and P(A does not happen)
  • Knowledge of statistical terminology related to mean and contamination levels
NEXT STEPS
  • Study the application of Poisson distribution in real-world scenarios
  • Learn advanced statistical analysis techniques using EXCEL
  • Research definitions and implications of "typical" in statistical contexts
  • Explore the impact of sample size on probability calculations in contamination studies
USEFUL FOR

Statisticians, food safety analysts, and anyone involved in quality control or risk assessment in food production will benefit from this discussion.

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Homework Statement


Data from www.centralhudsonlab.com determined
the mean number of insect fragments in 225-gram chocolate
bars was 14.4, but three brands had insect contamination
more than twice the average. Assume
the number of fragments (contaminants) follows a Poisson
distribution.
1)If you consume seven 28.35-gram (one-ounce) bars this
week from a brand at the mean contamination level, what
is the probability that you consume one or more insect
fragments in more than one bar?
2)Is the probability of a test result more than twice the mean
of 14.4 unusual, or can it be considered typical variation?
Explain.


Homework Equations


Using EXCEL's POISSON.DIST Function.

The Attempt at a Solution



#1 is really confusing me. I find the wording to be very obscure.
For a poisson distribution we can scale the mean to match the 28.35g bars.
Thus our mean of interest is (28.35/225)*14.4=1.82

My instructor told me that the answer is simply the (probability of consuming one or more insect fragments in ONE bar)^7.
Does this make sense to you guys? Because it doesn't to me.
To me the above is calculating the probability of consuming one or more insect fragments in SEVEN bars, rather than "more than one bar."


#2 seems subjective. More than twice the mean is 28.8. I don't understand what they mean by "typical." Seeing that the problem statement said that 3 brands had contamination more than twice the average, I would assume it's typical, then again it depends on what you define typical as. Any suggestions>
 
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For number 1, I would use the fact that P(A happens) = 1-P(A does not happen).

Number 2 is terribly worded. I think they mean something like "find the probability that a tested bar has at least twice as many insect fragments as the mean, and see if that number is really really small". The fact that you don't know how many brands were tested though makes this less than solid logic, but I assume this is what they are going for.
 
Office_Shredder said:
For number 1, I would use the fact that P(A happens) = 1-P(A does not happen).

Number 2 is terribly worded. I think they mean something like "find the probability that a tested bar has at least twice as many insect fragments as the mean, and see if that number is really really small". The fact that you don't know how many brands were tested though makes this less than solid logic, but I assume this is what they are going for.

Thanks for the reply.

Regarding #1, yes, that is what I used, except my (1-P(A does not happen)) is equal to the probability that one is consuming one or more insect fragments in ONE bar. I am just confused on how to apply this to ONE OR MORE bars.

The probability I calculated for #2 is 0.000463. I consider this to be really small, but I really don't understand what is the meaning of typical for this case.
 

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