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Glen_b
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Basil covered the essential issue (that the average distance from the mean is not the same thing as the standard deviation), but I think there's some additional points that should be added to that.

The mean of the absolute deviations (from the mean) - which is the average distance from the mean - will always be $\leq$ than the root-mean-square deviation from the mean (this, which is the standard deviation This follows from the triangle inequality, for example), or from Jensen's inequality.

In the case of data drawn from a normal distribution, in large samples the mean deviation is $\sqrt{2/\pi}$ times the standard deviation ... which is about 0.798, so that's your 80%.

Basil covered the essential issue (that the average distance from the mean is not the same thing as the standard deviation), but I think there's some additional points that should be added to that.

The mean of the absolute deviations (from the mean) - which is the average distance from the mean - will always be $\leq$ than the root-mean-square deviation from the mean (this follows from the triangle inequality, for example).

In the case of data drawn from a normal distribution, in large samples the mean deviation is $\sqrt{2/\pi}$ times the standard deviation ... which is about 0.798, so that's your 80%.

Basil covered the essential issue (that the average distance from the mean is not the same thing as the standard deviation), but I think there's some additional points that should be added to that.

The mean of the absolute deviations (from the mean) - which is the average distance from the mean - will always be $\leq$ than the root-mean-square deviation from the mean, which is the standard deviation This follows from the triangle inequality, for example, or from Jensen's inequality.

In the case of data drawn from a normal distribution, in large samples the mean deviation is $\sqrt{2/\pi}$ times the standard deviation ... which is about 0.798, so that's your 80%.

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Glen_b
  • 290.5k
  • 37
  • 652
  • 1.1k

Basil covered the essential issue (that the average distance from the mean is not the same thing as the standard deviation), but I think there's some additional points that should be added to that.

The mean of the absolute deviations (from the mean) - which is the average distance from the mean - will always be $\leq$ than the root-mean-square deviation from the mean (this follows from the triangle inequality, for example).

In the case of data drawn from a normal distribution, in large samples the mean deviation is $\sqrt{2/\pi}$ times the standard deviation ... which is about 0.798, so that's your 80%.

Basil covered the essential issue (that the average distance from the mean is not the same thing as the standard deviation), but I think there's some additional points that should be added to that.

The mean of the absolute deviations (from the mean) - which is the average distance from the mean - will always be $\leq$ than the root-mean-square deviation from the mean.

In the case of data drawn from a normal distribution, in large samples the mean deviation is $\sqrt{2/\pi}$ times the standard deviation ... which is about 0.798, so that's your 80%.

Basil covered the essential issue (that the average distance from the mean is not the same thing as the standard deviation), but I think there's some additional points that should be added to that.

The mean of the absolute deviations (from the mean) - which is the average distance from the mean - will always be $\leq$ than the root-mean-square deviation from the mean (this follows from the triangle inequality, for example).

In the case of data drawn from a normal distribution, in large samples the mean deviation is $\sqrt{2/\pi}$ times the standard deviation ... which is about 0.798, so that's your 80%.

Source Link
Glen_b
  • 290.5k
  • 37
  • 652
  • 1.1k

Basil covered the essential issue (that the average distance from the mean is not the same thing as the standard deviation), but I think there's some additional points that should be added to that.

The mean of the absolute deviations (from the mean) - which is the average distance from the mean - will always be $\leq$ than the root-mean-square deviation from the mean.

In the case of data drawn from a normal distribution, in large samples the mean deviation is $\sqrt{2/\pi}$ times the standard deviation ... which is about 0.798, so that's your 80%.