# Can any observation lie more than 3 SDs from the mean if there are 10 observations constrained between 0 and 1?

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.

• 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. – Ray Oct 5 '16 at 20:24
• 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? – ROB Oct 5 '16 at 20:52
• 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 – Ray Oct 5 '16 at 20:54
• 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 $n-1$ – Glen_b Oct 5 '16 at 23:06

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.