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EEJ MAIN Mathematics QUESTION #7445
Question 1
If \(\sum_{i=1}^{9}(x_i-5) = 9\) and \(\sum_{i=1}^{9}(x_i-5)^2 = 45\), then the standard deviation of the 9 items \(x_1, x_2, \ldots, x_9\) is:
  • 2✔️
  • 3
  • 9
  • 4
Correct Answer Explanation
We have \(\sum(x_i-5) = 9\), so \(\sum x_i - 45 = 9\), giving \(\sum x_i = 54\). Mean = \(\frac{54}{9} = 6\). Also, \(\sum(x_i-5)^2 = 45\). Expanding: \(\sum x_i^2 - 10\sum x_i + 225 = 45\), so \(\sum x_i^2 = 45 + 540 - 225 = 360\). Variance = \(\frac{\sum x_i^2}{n} - (\bar{x})^2 = \frac{360}{9} - 36 = 40 - 36 = 4\). Standard deviation = \(\sqrt{4} = 2\).