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The definition of the range has been given as the difference between the largest and the smallest value of the distribution. This, in my perception, tells us the discrepancy between the smallest and largest value and does not measure the dispersion from the central value.

So, my question is how does that help us in finding the dispersion of the observation from the central value of the same? In other words, why do we consider range as a measure of dispersion?

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3 Answers 3

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In statistics, dispersion (also called variability, scatter, or spread) is the extent to which a distribution is stretched or squeezed.

(Wikipedia)

It does not say “from the central value”. Dispersion is about how the data is spread out, obviously, if it has a big range it is very spread out.

Some dispersion statistics like variance or MAD measure the spread relative to the central value, but others like range or IQR do not.

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    $\begingroup$ Variance and so also SD can be defined without reference to the mean as a summary of pairwise squared differences. At some point you're taking a mean, but that is different. $\endgroup$
    – Nick Cox
    Commented May 15 at 9:15
  • $\begingroup$ Fisher and many others have used the term scale (and also location), which is a distraction for any group accustomed to their own meanings for these words (e.g. geographers). $\endgroup$
    – Nick Cox
    Commented May 15 at 9:20
  • $\begingroup$ See also stats.stackexchange.com/questions/225734/… $\endgroup$
    – Nick Cox
    Commented May 16 at 21:25
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The range is twice the maximum absolute deviation from the mid-range, a tenable 'central value'; so on the face of it ought to pass muster as a measure of dispersion around that.

Formal criteria that measures of location & dispersion ought to meet have been proposed in papers by Bickel & Lehmann—see my answers to Is mode a measure of central tendency? & Is it really appropriate to say "standard deviation" is variation or dispersion. The mid-range & range meet these respective criteria.

Note B. & L. appear to side with you in defining dispersion around a measure of location (they distinguish it from spread throughout a distribution). As @NickCox points out, a location measure may not appear explicitly in a particular definition of a dispersion measure; what matters is that the ordering of distributions by their range, say, can't contradict the partial ordering based on probabilities of absolute deviations from the mid-range, whereas it can contradict that based on probabilities of absolute deviations from the mean.


† The mid-range $\mu(X)$ is shift equivariant, $$\mu(X + b) = \frac{\min(X+b) + \max(X+b)}{2}= \frac{(\min(X)+b) + (\max(X) + b)}{2} = \mu(X) + b$$ scale equivariant, $$\mu(aX) = \frac{\min(aX) + \max(aX)}{2} = \frac{a\min(X) + a\max(X)}{2} =a\mu (X)\;\forall a > 0$$ & reflection equivariant $$\mu(-X) = \frac{\min(-X) + \max(-X)}{2} = \frac{-\max(X) - \min(X)}{2} =-\mu (X)$$ Stochastic dominance of $X$ over $Y$ implies the mid-range of $X$ is no less than that of $Y$:

$$\begin{align} \mu(X) -\mu(Y) &= \frac{\min(X)+\max(X)}{2} - \frac{\max(Y)+\min(Y)}{2}\\ &=\frac{\min(X)-\min(Y)}{2} + \frac{\max(X)-\max(Y)}{2} \end{align} $$

but for $X$ to dominate $Y$, both $\min(X)\geq\min(Y)$ & $\max(X)\geq\max(Y)$; therefore $\mu(X) \geq\mu(Y)$

The range $\sigma$ is shift-invariant,

$$\sigma(X + b) = \max(X+b) - \min(X+b) = (\max(X) + b) - (\min(X) +b)= \sigma (X)$$

scale-equivariant,

$$\sigma(aX) = \max(aX) - \min(aX) = a\max(X) - a\min(X) = a\sigma (X)\;\forall a > 0$$

& reflection-invariant:

$$\sigma(-X) = \max(-X) - \min(-X) = -\min(X) + \max(X)= \sigma (X)$$

Stochastic dominance of the absolute deviation of $X$ from the mid-range of $X$ over the absolute deviation of $Y$ from the mid-range of $Y$ implies the range of $X$ is no less than that of $Y$:

$$\begin{align} \sigma(X) - \sigma(Y) &= \max(X) - \min(X) - (\max(Y)-\min(Y))\\ &= \max(X) - (2\mu(X) - \max(X)) - (\max(Y) - (2\mu(Y) - \max(Y)))\\ &= 2(\max(X) - \mu(X) - (\max(Y)-\mu(Y))) \end{align} $$

But for $|X-\mu(X)|$ to dominate $|Y-\mu(Y)|$, $\max(X) - \mu(X) \geq \max(Y)-\mu(Y)$; therefore $\sigma(X) \geq \sigma(Y)$

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There is potentially semantic ambiguity in play here: in some fields one uses dispersion in a narrow sense, meaning standard deviation. However, in general case standard deviation is only one of various possible measures of dispersion (some other measures of statistical dispersion are cited in Wikipedia article, already quoted in the other answer.)

This, in my perception, tells us the discrepancy between the smallest and largest value and does not measure the dispersion from the central value.

One obvious advantage of range in comparison to standard deviation is that it is not dependent on how one estimates the central value (using mean, median, mode or something else), which is important when we do not know the underlying distribution or when it is contaminated with outliers.

Remark
Speaking of semantics dispersion here really means statistical dispersion. Indeed, the term is routinely used in physics, biology, chemistry, and some other fields in completely different senses, totally unrelated to statistical properties. Another example of a seemingly obvious term that in fact causes much cross-field confusion is frequency.

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  • $\begingroup$ I've never seen dispersion used to mean SD in a statistical context. I've often seen it used to refer to spatial data. $\endgroup$
    – Nick Cox
    Commented May 15 at 9:17
  • $\begingroup$ @NickCox many members of this community have background in fields other than statistics. I believe that I was taught to use dispersion and sd synonimously (in probabilistic context), although it was too long ago to give a specific reference (and it likely wouldn't be in English.) Still, most of the time dispersion meant for me $\partial k/\partial \omega$ (which may get back to sd in a case of a Gaussian wave packet) :-) $\endgroup$
    – Roger V.
    Commented May 15 at 9:24
  • $\begingroup$ Indeed. I am not a statistician myself. $\endgroup$
    – Nick Cox
    Commented May 15 at 9:25

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