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 is:

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EEJ MAIN Mathematics QUESTION #7493

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 is:

Correct Answer Explanation

Mean = 6 (from \(\sum x_i = 54\)). Variance = \(\frac{\sum x_i^2}{9} - 36 = \frac{360}{9} - 36 = 4\). SD = 2.