We are doing a lab test where one of the criteria for the material being tested to "pass" is:

no individual specimen shall deviate more than 3 standard deviations from the mean for the 10 specimens

I can’t for the life of me create a set of data where 1 specimen is outside of 3 standard deviations. The data points we are dealing with is “percent mass loss” so by definition it is constrained to zero to one.

I am beginning to think it might not be possible to have a set of data that fails the test for this criteria.

Am I correct in saying:

An outlier has such a strong influence on the standard deviation that the "mean+3*stdev maximum" always is greater than the outlier itself when the size of the dataset is 10.

  • $\begingroup$ I strongly suspect this has something to do with Chebyshev's inequality (probability of being more than 3 SDs away from the mean is at most 1/9) or its proof: surely you've tried the example of nine 0 and one of any other value, where the mean plus 9 SD's gives precisely the 10th value. Note that if we had a sample size of 11, ten 0 and one 1 would work. $\endgroup$
    – Ray
    Oct 5, 2016 at 20:24
  • $\begingroup$ Thank you for your input. Looking at the case of ANY 9 values (with zero variance) and any single other value. It seems that the mean+3*stdev will always(?) equal the 10th value? $\endgroup$
    – ROB
    Oct 5, 2016 at 20:52
  • $\begingroup$ Of course; standard deviation 'scales' and is invariant to shifts, and all you're doing if you fix 9 values and let the 10th value be free is essentially stretching a line segment with 2 endpoints, one corresponding to your 9 points and the other to the 10th $\endgroup$
    – Ray
    Oct 5, 2016 at 20:54
  • 1
    $\begingroup$ With 10 observations you can't get more than $9/\sqrt{10}=2.846$ standard deviations from the mean. See the analysis here for example (among others). [In the notation of that analysis, $n^2=8$] ... to get it to be $3$ sd's away at $n=10$ you'd have to use the $n$-divisor for standard deviation, rather than the usual $n-1$ $\endgroup$
    – Glen_b
    Oct 5, 2016 at 23:06

1 Answer 1


I believe you are correct. I think the general sketch of the proof comes from Cantelli's Lemma (which is related to Chebyshev's Inequality).

Note that in our specific case, we get $$ P((X-E[X])/\sigma \geq k) \leq 1/(1+k^2)=1/10 \qquad\text{for k=3 in our case}$$ If we want a strict inequality for the difference, e.g. $k=3+\epsilon$ standard deviations for $\epsilon>0$, then the right hand side is strictly less than 1/10. But we have 10 observations, so each observation must have at least probability 1/10; hence we have a contradiction.

  • $\begingroup$ Thank you again. I'm satisfied. Now to query the test standard's governing body... $\endgroup$
    – ROB
    Oct 5, 2016 at 21:13
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    $\begingroup$ Does probability less or equal to 1/10 means that it is impossible?! $\endgroup$
    – Tim
    Oct 5, 2016 at 21:36
  • $\begingroup$ With an empirical distribution with 10 mass points, the only possible probability smaller than 1/10 is 0, so while the event might not be impossible, it cannot be observed with positive probability. $\endgroup$
    – Ray
    Oct 5, 2016 at 21:53

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