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idku

  • one year ago

a question about statistics.

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  1. idku
    • one year ago
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    The list X of ten numbers has a mean of 20, and population SD, \(\color{black}{{\LARGE \sigma}_X}\) equal to 5. THEN, Someone comes and claims that \(40\) is the maximum on this list X. (My task is:) `Show why this is impossible (i.e. why 40 can't be the maximum on the list). `

  2. idku
    • one year ago
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    \(\color{black}{\sqrt{\dfrac{(40-20)^2+R}{10}}=6}\) where R is the sum of all remaining component of the Standard deviation. R=(\(X_2\) - 20)\(^2\)+(\(X_3\) - 20)\(^2\)+(\(X_4\) - 20)\(^2\)+(\(X_5\) - 20)\(^2\)+(\(X_6\) - 20)\(^2\)+(\(X_7\) - 20)\(^2\) where it is obvious that \(R\ge0\), thus: \(\color{black}{\sqrt{\dfrac{(20)^2+R}{10}}=6}\) \(\color{black}{\sqrt{\dfrac{400+R}{10}}=6}\) \(\color{black}{\sqrt{\dfrac{400+R}{10}}>6}\) knowing the restriction for R

  3. idku
    • one year ago
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    oh my SD is 5, not 6. But then it certainly works.

  4. idku
    • one year ago
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    done I guess:)

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