If I have 2 data points like

16.70 (+/-4.33)
54.70 (+/-16.30)

where 16.7 = mean and +/- 4.33 is the standard deviation, is it possible to calculate a range out of that? I would think that roughly, the min range here would be 16.7-4.33 and the max 54.7+16.30, so the range would be:

(12.37 - 71.00)

But I can't find a source to confirm or disconfirm this. All I find is examples how to calculate standard deviation from a range, but I am looking for the opposite. Is there a "standard" way to get a range from 2 means and 2 standard deviations?

  • 1
    $\begingroup$ Normally the "+/-" refers to the dispersion of measurement errors. This suggests that computing a range as you propose would mix up two completely different things: the range of your means (16.7 to 54.4) and the dispersions of their measurements. It is difficult to imagine a context in which that would be meaningful. Could you enlighten us concerning the intended interpretation or purpose of this calculation? $\endgroup$ – whuber Jul 31 '15 at 11:58
  • $\begingroup$ Thanks. My purpose was to make a statement like "the values are between min and max.". I was thinking that standard deviation means "at most, the calculated mean deviates this much". I understand now that this doesn't work. So, just to be sure: It would be correct to say "the range of the two means is (16.7-54-4)"? $\endgroup$ – poshtad Jul 31 '15 at 18:50
  • $\begingroup$ That would be correct--but a tiny bit misleading. It would be more informative (and simpler) to say it like it is: "The means were 16.7 and 54.7." In fact, is there any problem in reporting the full information, as in "The means were $16.7\pm 4.33$ and $54.7\pm 16.3$ (plus or minus one standard deviation)"? $\endgroup$ – whuber Jul 31 '15 at 18:53
  • $\begingroup$ Yes, this is just an example, I would like to "merge" many more of those datapoints, perhaps dozens or hundreds. Giving the highest and the lowest value ever measured should literally give an idea, not a precise result, about the range of data, even if the values are means from an unknown number of single measurements. $\endgroup$ – poshtad Jul 31 '15 at 19:05

Assume the population from which these data points are taken is normal (mean $\mu$, variance $\sigma^2$). I think your question is ill-posed because there is no population parameter that you can call "the" range here, and there is no use in trying to estimate it. Think of it this way: if the number of data points becomes very large the sample standard deviation will approach to $\sigma$, but the range of the sample will converge to $+\infty$. So it is useful to talk about standard deviation of the population, but not of its range. The "range" is a statistic, a random quantity of which the distribution keeps changing with the sample size (shifting to larger and larger values). The standard deviation $\sigma$ on the other hand is a fixed population parameter that you can estimate from any given sample.

This is the reason why you will find methods of estimating $\sigma$ from the range (28.00 in your case) of a sample, but not the other way around. Usually these methods suppose the population to be normal. If not, you need to apply methods from order statistics (Tippet integrals...).

In quality engineering for example, for Shewhart control charts, it is still widely customary to use the sample range to estimate $\sigma$, even if this is somewhat less optimal than using the sample standard deviation directly.

  • $\begingroup$ While I don't quite understand why the range would converge to +∞, I see your point with the standard deviation being a population parameter, thanks. Could you explain why σ is 28.00? What did you calculate there? $\endgroup$ – poshtad Jul 31 '15 at 18:55
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    $\begingroup$ No, I meant to say that the range in your example of two data points is 28.00, not $\sigma$. If you keep drawing random numbers from a normal distribution, the difference between the smallest and the largest number you have observed will keep increasing (to $\infty$). $\endgroup$ – StijnDeVuyst Jul 31 '15 at 20:16

If all you know is the population (i.e., true, not sample) mean and standard deviation, and you neither know nor assume anything else about the distribution of the random variable, then here's what you can conclude about the range of the random variable.

  1. If standard deviation = 0, you can conclude that the range is [mean,mean], i.e., the random variable equals its mean, with probability one.

  2. If standard deviation > 0, the range could be as big as $[-\infty, \infty]$, even though it might be tighter. For instance, think of a Normal random variable with even a tiny, but positive, standard deviation.


You should be able to use the range distribution for this: (Wikipedia).

You can create a CDF given your two distributions and find the range(t) that gives f(t)=.025 and .975. That should give you your lower and upper bounds at p=.95.


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