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On average, how far each data point deviates from the mean
Mean square deviation
σ2=Σ(x−xˉ)2n\sigma^2 = \frac{\Sigma (x - \bar{x})^2}{n}σ2=nΣ(x−xˉ)2
σ=σ2\sigma = \sqrt{\sigma^2}σ=σ2 (std dev is square root of variance)
σ2=x2‾−(xˉ)2\sigma^2 = \overline{x^2} - (\bar{x})^2σ2=x2−(xˉ)2
Mean of all squared data
Mean of all data then squared
All data
No, it ignores outliers
σ\sigmaσ (standard deviation)
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