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Sets A & B have the same number of terms and range of SET A > range of SET B. Is the standard deviation of set A greater than the standard deviation of set B ?

For, E.g.

SET B : $(0, 0, 0, 0)$; Range of B is $0$ in this case. $SD_{B}=0$

SET A : $(0, 0, 0, 4)$; Range of A is $4$ in this case. $SD_{A}=\sqrt{3}$

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What have you tried? Could you add the meaning of "range" in this context? Did you e.g. work out any small examples? In general I suspect the answer is "no" but it might depend on the definition of "range". –  Lord_Farin Apr 22 '13 at 8:00
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1 Answer

The conjecture is false. An example is worked out below, but let me first pay attention to why it intuitively fails.

Namely, one or two "extreme" measurements can greatly increase the range. However, if the sample space is reasonably large, the effect on the standard deviation will be marginal. Conversely, any sample where the data points are all distributed close to two very separated values (e.g. $\pm 1$) might not have a huge range, but will have a considerable standard deviation because all measurements are a certain distance from the mean.


The above is exemplified in the following choice of samples $A$ and $B$ of size $2n$:

  • Let $A$ consist of $2n-2$ zeroes, and one of $\pm n$ each.
  • Let $B$ consist of $n$ values $1-n$ and $n$ values $n-1$.

Then we may check that both samples have mean $0$, and that the standard deviations are:

$$\sigma_A = \sqrt{\frac{n^2 + n^2}{2n}} = \sqrt n; \qquad \sigma_B = \sqrt{\frac{2n (n-1)^2}{2n}} = n-1$$

while the range of $A$ is $r_A = 2n$ and the range of $B$ is $r_B = 2n-2$. For $n \ge 3$ however we see that $\sigma_B > \sigma_A$.


I hope the above provides you with some insight in the matter.

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