Solve Chebychev's Theorem: Mean 50, Standard Deviation 5

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SUMMARY

This discussion focuses on applying Chebychev's Theorem to a dataset with a mean of 50 and a standard deviation of 5, specifically analyzing the percentage of values within certain ranges. For part (a), it was established that at least 93.75% of the values fall between 10 and 30, calculated using the formula 1 - 1/k² where k equals 4. The user seeks clarification on part (b), which involves determining the percentage of values between 12 and 28, and questions the interpretation of "at least" in this context. The theorem asserts that a minimum of 1 - 1/k² of the distribution lies within k standard deviations of the mean.

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  • Understanding of Chebychev's Theorem
  • Basic knowledge of mean and standard deviation
  • Ability to perform algebraic calculations
  • Familiarity with statistical distributions
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  • Calculate the percentage of values using Chebychev's Theorem for different ranges
  • Explore the implications of "at least" in statistical contexts
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  • Study the differences between Chebychev's Theorem and the Empirical Rule
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Students, statisticians, and data analysts seeking to understand the application of Chebychev's Theorem in analyzing data distributions and calculating value ranges.

robasc
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Could you please explain and break the steps down in solving the answer to this question for me?



In a distribution of 200 values, the mean is 50 and the standard deviation is 5. Use Chebychev's theorem.



a. at least what percentage of the values will fall between 10 and 30?



50 - 30 = 20



k = 20 / 5 = 4



1 - 1 / k^2 = 1 - 1 / 4^2 = 1 - 1 / 16 = .0625 = 1 - .0625 = .9375 or
%93.75



I got this part right but now this is the part I am having trouble with:



b. At least what percentage of the values will fall between 12 and 28?





Also, does at least mean to subtract?
 
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Chebychev's theorem states that at least 1 - 1/k^2 of the distribution lies within k standard deviations of the mean. Personally I don't see how you can use it here, maybe there's a trick. For part (a) your answer of 93% is the minimum percentage of values that fall between 30 and 70, not between 10 and 30.
 
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