It's common to see people report variability in their realized sample through the $\bar{x} \pm 2s$ range, where $\bar{x}$ is the realized mean and $s$ the realized standard deviation estimate in their sample.

Beyond informing about the realized sample, I am guessing $\bar{x} \pm 2s$ is often reported in the hope that it provides information about $\mu \pm 2\sigma$.

But unless the sample is very large, $\bar{x}$ and $s$ will be imperfect estimates of $\mu$ and $\sigma$, and $\bar{x} \pm 2s$ will therefore be an inaccurate estimate of $\mu \pm 2\sigma$.

Whether or not it's a good idea to try and estimate $\mu \pm 2\sigma$, this leads me to wonder about standard practices to do so, and to assess the accuracy of the estimated range (including the relationship between accuracy and sample size $n$).

I guess you could construct standard CIs around both $\mu$ and $\sigma$ and somehow "combine" these two CI looking at worse case scenarios to get a "maximum" CI for $\mu \pm 2\sigma$ you're equally confident in (say by taking the highest value in the CI for $\sigma$ plus/minus the highest/lowest value in the CI for $\mu$).

But that seems super hacky and not well-founded theoretically (among other things, I suspect there are issues with $s$ appearing in the CI formula for both $\mu$ and $\sigma$?).

If that's indeed not a good idea, what would be a better approach to estimating $\mu \pm 2\sigma$ with confidence? In the classical CI sense, and under some reasonable distributional assumptions, is there an observable random interval $[\underline{F}(X_1, \dots, X_n; \alpha), \bar{F}(X_1, \dots, X_n; \alpha)]$ that is guaranteed to include the whole $\mu \pm 2\sigma$ range $(1-\alpha)$ percent of the time (in a similar way that $(1-\alpha)$ percent of the time, the range $\bar{X} \pm z_{\alpha/2} (s/\sqrt{n})$ includes $\mu$ itself)?

That is, I am looking for a family of statistics $\underline{F}(X_1, \dots, X_n; \alpha)$ and $\bar{F}(X_1, \dots, X_n; \alpha)$ such that:

$$P\big(\underline{F}(X_1, \dots, X_n; \alpha) \leq \mu - 2\sigma < \mu + 2\sigma \leq \bar{F}(X_1, \dots, X_n; \alpha) \big) = 1-\alpha.$$

  • $\begingroup$ You are right that the sample mean and standard deviation $\bar{x}$ and $s$ are estimates of the population mean $\mu$ and standard deviation $\sigma$. But you seem to confuse the standard deviation $s$ with the standard error of the mean $s/\sqrt{n}$. These are not the most intuitive of concepts; see here. $\endgroup$
    – dipetkov
    Apr 24, 2022 at 11:02
  • $\begingroup$ @dipetkov: When I wrote "is there an observable random interval similar to 𝑋¯±π‘§π›Ό/2(𝑠/π‘›βˆš) that includes the whole πœ‡±2𝜎 range (1βˆ’π›Ό) percent of the time?", what I meant was: "is there an observable random interval that includes the whole πœ‡±2𝜎 range (1βˆ’π›Ό) percent of the time ? (in the same way that similar to 𝑋¯±π‘§π›Ό/2(𝑠/π‘›βˆš) includes πœ‡ (1βˆ’π›Ό) percent of the time)". Maybe that's where the impression of confusion came from? I'll make a correction to the question anyways. $\endgroup$
    – FZS
    Apr 24, 2022 at 23:20
  • 1
    $\begingroup$ Re "I am guessing x¯±2s is often reported in the hope that it provides information about ΞΌ±2Οƒ." Not usually. You describe a tolerance interval. This has important applications, especially in quality control. Those are computed using special formulas that account for the uncertainty in $\mu\pm2\sigma.$ Please search our site for more information. stats.stackexchange.com/questions/26702 would be a good starting point. $\endgroup$
    – whuber
    Apr 25, 2022 at 13:11
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    $\begingroup$ @whuber: Thanks a lot for mentioning the notion of "tolerance interval" which I wasn't familiar with. Most likely the keyword I was missing in my search. Answers are still welcome, but I will look into it myself and hopefully be able to answer my own answer on that basis soon. $\endgroup$
    – FZS
    Apr 25, 2022 at 14:12

1 Answer 1


As @whuber wrote in the comments, what I am describing is a "Tolerance Interval" (see e.g., Prediction and Tolerance Intervals).

If I understand things correctly so far, one connection with the $\bar{x} + 2s$ rule of thumb is that, with normal data, as the sample-size $n$ tends to infinity, $\bar{x} + 1.96s$ obviously gets closer and closer to including 95% of the population (as $n$ tends to infinity, this is true, I believe, for any level of confidence, be it 95%, 99%, 99.9%,...).

However, for finite $n$, intervals including 95% of the population with, say, 95% confidence are larger than $\bar{x} + 1.96s$ (e.g., the interval is around $\bar{x} + 2.75s$ when $n=20$, where calculations are from https://statpages.info/tolintvl.html with underlying statistical assumption --- in particular, normal data --- described in https://www.itl.nist.gov/div898/handbook/prc/section2/prc263.htm).


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