My intuition is that the standard deviation is: a measure of spread of the data.
You have a good point that whether it is wide, or tight depends on what our underlying assumption is for the distribution of the data.
Caveat: A measure of spread is most helpful when the distribution of your data is symmetric around the mean and has a variance relatively close to that of the Normal distribution.
(This means that it is approximately Normal.)
In the case where data is approximately Normal, the standard deviation has a canonical interpretation:
- Region: Sample mean +/- 1 standard deviation, contains roughly 68% of the data
- Region: Sample mean +/- 2 standard deviation, contains roughly 95% of the data
- Region: Sample mean +/- 3 standard deviation, contains roughly 99% of the data
(see first graphic in Wiki)
This means that if we know the population mean is 5 and the standard deviation is 2.83 and we assume the distribution is approximately Normal, I would tell you that I am reasonably certain that if we make (a great) many observations, only 5% will be smaller than 0.4 = 5 - 2*2.3 or bigger than 9.6 = 5 + 2*2.3.
Notice what is the impact of standard deviation on our confidence interval? (the more spread, the more uncertainty)
Furthermore, in the general case where the data is not even approximately normal, but still symmetrical, you know that there exist some $\alpha$ for which:
- Region: Sample mean +/- $\alpha$ standard deviation, contains roughly 95% of the data
You can either learn the $\alpha$ from a sub-sample, or assume $\alpha=2$ and this gives you often a good rule of thumb for calculating in your head what future observations to expect, or which of the new observations can be considered as outliers. (keep the caveat in mind though!)
I don't see how you are supposed to interpret it. Does 2.83 mean the values are spread very wide or are they all tightly clustered around the mean...
I guess every question asking "wide or tight", should also contain: "in relation to what?". One suggestion might be to use a well-known distribution as reference. Depending on the context it might be useful to think about: "Is it much wider, or tighter than a Normal/Poisson?".
EDIT:
Based on a useful hint in the comments, one more aspect about standard deviation as a distance measure.
Yet another intuition of the usefulness of the standard deviation $s_N$ is that it is a distance measure between the sample data $x_1,… , x_N$ and its mean $\bar{x}$:
$s_N = \sqrt{\frac{1}{N} \sum_{i=1}^N (x_i - \overline{x})^2}$
As a comparison, the mean squared error (MSE), one of the most popular error measures in statistics, is defined as:
$\operatorname{MSE}=\frac{1}{n}\sum_{i=1}^n(\hat{Y_i} - Y_i)^2$
The questions can be raised why the above distance function? Why squared distances, and not absolute distances for example? And why are we taking the square root?
Having quadratic distance, or error, functions has the advantage that we can both differentiate and easily minimise them. As far as the square root is concerned, it adds towards interpretability as it converts the error back to the scale of our observed data.
Robustness to outliers is a double-edged sword: Sometimes we want to estimate things in a way that is robust to outliers, which means that we do not mind getting large outliers. At other times we want to avoid large outliers, so we want to estimate things in a way that is not robust to outliers. Similarly, with measures of spread, sometimes we want something that is robust to outliers, so that large outliers do not increase the measure. At other times we want our measure of spread to reflect the presence of large outliers by manifesting in a larger value.
In decision-theory, issues like this are dealt with by specifying a penalty/loss function which penalises you for your error in estimation of a quantity. Two common loss functions are absolute-error loss and squared-error loss (shown in the following plots, taken from this answer by Jean-Paul).
Absolute-error loss penalises you according to the absolute deviation of your estimate from the true value. This form of loss function leads to estimation using medians. This form of loss function is robust to outliers in the sense that outliers contribute a penalty that is proportionate to their size. Measures of spread in this context reflect the expected loss of a particular estimate of central location, with the expected loss being a weighted sum of absolute deviations from the estimated central location.
Squared-error loss penalises you according to the squared deviation of your estimate from the true value. This form of loss function leads to estimation using means. This form of loss function is sensitive to outliers in the sense that outliers contribute a penalty that is proportionate to their squared deviation - this magnifies the effect of large outliers. Measures of spread in this context reflect the expected loss of a particular estimate of central location, with the expected loss being a weighted sum of squared deviations from the estimated central location.
In regard to the choice between median absolute deviation and standard deviation these same considerations apply. The former measure is a measure of spread that represents expected absolute-error loss, and is more robust to outliers. In this case, outliers do not manifest in large increases in the measure of spread. The latter is a measure of spread that represents expected squared-error loss, and is more sensitive to outliers. In this case, the outliers will manifest in large increases in the measure of spread.
Best Answer
The two easiest approaches:
One way to do it would be to count the standard deviation on $x$s and $y$s separately, to get you a box shape.
If you prefer the circle shape, then simply given the mean $(\bar{x},\bar{y})$ and the points $(x_i,y_i)$ compute the distance $r_i = \sqrt{(\bar{x} - x_i)^2 + (\bar{y} - y_i)^2}$ and then calculate the standard deviation of the $r_i$'s around $0$. The resulting standard deviation (or if you divide by $\sqrt{N}$ then standard error) will be the radius of the circle.