# Can mean plus one standard deviation exceed maximum value?

I have mean 74.10 and standard deviation 33.44 for a sample that has minimum 0 and maximum 94.33.

My professor asks me how can mean plus one standard deviation exceed the maximum.

I showed her many examples about this, but she doesn't understand. I need some reference to show her. It could be any chapter or paragraph from a statistics book that talks particularly about this.

• Why do you want to add (or subtract) one standard deviation from the mean? The SD is a measure of the spread of the data. Did you want the standard error of the mean instead perhaps? – Gavin Simpson Nov 17 '14 at 22:38
• I don't want to add or subtract, the one that wants this is my professor. That is the way she understands the standart deviation – Boyun Omuru Nov 17 '14 at 23:22
• An interesting example is the sample (0.01,0.02,0.98,0.99). Both the mean plus the standard deviation and the mean minus the standard deviation lie outside [0,1]. – Glen_b Nov 18 '14 at 9:59
• Maybe she's just thinking of a Normal distribution? – user765195 Nov 27 '14 at 7:00

Certainly the mean plus one sd can exceed the largest observation.

Consider the sample 1, 5, 5, 5 -

it has mean 4 and standard deviation 2, so the mean + sd is 6, one more than the sample maximum. Here's the calculation in R:

> x=c(1,5,5,5)
> mean(x)+sd(x)
[1] 6

It's a common occurrence. It tends to happen when there's a bunch of high values and a tail off to the left (i.e. when there's strong left skewness and a peak near the maximum).

--

The same possibility applies to probability distributions, not just samples - the population mean plus the population sd can easily exceed the maximum possible value.

Here's an example of a $\text{beta}(10,\frac{1}{2})$ density, which has a maximum possible value of 1:

In this case, we can look at the Wikipedia page for the beta distribution, which states that the mean is:

$\operatorname{E}[X] = \frac{\alpha}{\alpha+\beta}\!$

and the variance is:

$\operatorname{var}[X] = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\!$

(Though we needn't rely on Wikipedia, since they're pretty easy to derive.)

So for $\alpha=10$ and $\beta=\frac{1}{2}$ we have mean$\approx 0.9523$ and sd$\approx 0.0628$, so mean+sd$\approx 1.0152$, more than the possible maximum of 1.

That is, it's easily possible to have a value of mean+sd that cannot be observed as a data value.

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For any situation where the mode was at the maximum, the Pearson mode skewness need only be $<\,-1$ for mean+sd to exceed the maximum. It can take any value, positive or negative, so we can see it's easily possible.

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A closely related issue is often seen with confidence intervals for a binomial proportion, where a commonly used interval, the normal approximation interval can produce limits outside $[0,1]$.

For example, consider a 95.4% normal approximation interval for the population proportion of successes in Bernoulli trials (outcomes are 1 or 0 representing success and failure events respectively), where 3 of 4 observations are "$1$" and one observation is "$0$".

Then the upper limit for the interval is $\hat p + 2 \times \sqrt{\frac{1}{4}\hat p \left(1 - \hat p \right)} = \hat p + \sqrt{\hat p (1 - \hat p )} = 0.75 + 0.433=1.183$

This is just the sample mean + the usual estimate of the sd for the binomial ... and produces an impossible value.

The usual sample sd for 0,1,1,1 is 0.5 rather than 0.433 (they differ because the binomial ML estimate of the standard deviation $\hat p(1-\hat p)$ corresponds to dividing the variance by $n$ rather than $n-1$). But it makes no difference - in either case, mean + sd exceeds the largest possible proportion.

This fact - that a normal approximation interval for the binomial can produce "impossible values" is often noted in books and papers. However, you're not dealing with binomial data. Nevertheless the problem - that mean + some number of standard deviations is not a possible value - is analogous.

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In your case, the unusual "0" value in your sample is making the sd large more than it pulls the mean down, which is why the mean+sd is high.

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(The question would instead be - by what reasoning would it be impossible? -- because without knowing why anyone would think there's a problem at all, what do we address?)

Logically of course, one demonstrates it's possible by giving an example where it happens. You've done that already. In the absence of a stated reason why it should be otherwise, what are you to do?

If an example isn't sufficient, what proof would be acceptable?

There's really no point simply pointing to a statement in a book, since any book may make a statement in error - I see them all the time. One must rely on direct demonstration that it's possible, either a proof in algebra (one could be constructed from the beta example above for example*) or by numerical example (which you have already given), which anyone can examine the truth of for themselves.

* whuber gives the precise conditions for the beta case in comments.

• +1 The Beta example is a nice idea. In fact, provided $0\lt\beta\lt 1$ and $\alpha \gt \beta(1+\beta)/(1-\beta)$, any Beta$(\alpha,\beta)$ distribution will have mean+sd exceeding $1$. – whuber Nov 17 '14 at 23:11
• Let me explain further. I am looking for accuracy percentage of particular appliance used for correction of teeth. And this appliance performed accuracy percentage for 7 tooth as follow: %76,19, %77,41, %94,33, %91,06, %0, %87,77, %91,96. My professor adds one standart deviation to mean and states that the result can not exceed maximum value even %100 because %100 is the maximum accuracy percentage that appliancek can perform. – Boyun Omuru Nov 17 '14 at 23:32
• She's right that a percentage > 100% makes no sense in your situation. The problem is actually the unstated premise that adding one sd to the mean should make sense in this context, when it doesn't. That's where I believe your difficulty originates. If we understood where the premise came from, it might lead to a better resolution. It's possible that the simple fact is stated in a book somewhere (it's a trivial observation, though, so it's possible it isn't, either), but I doubt it will ever be put in a way that will satisfy her, because her false premise is the source of the problem. – Glen_b Nov 18 '14 at 0:06
• Indeed - my minor point is that this curiosity is a result of what standard deviations represent for strongly non-symmetric distributions rather than a result of taking a sample. But in general, I think your answer is excellent – Henry Nov 18 '14 at 10:00
• @tomka I have attempted to help many students in a similar position. I eventually learned the (possibly unsurprising) rule of thumb that it's effectively impossible to teach a supervisor anything through the medium of their student. – Glen_b Nov 6 '16 at 22:59

Per Chebyshev's inequality, less than k -2 points can be more than k standard deviations away. So, for k=1 that means less than 100% of your samples can be more than one standard deviation away.

It's more interesting to look at the low bound. Your professor should be more surprised there are points which are about 2.5 standard deviations below mean. But we now know that only about 1/6th of your samples can be 0.

The essence of the problem may be that your distribution is not a normal distribution which a standard deviation assumes. Your distribution is likely left skewed, so you need to transform your set into a normal distribution first by picking a suitable transform function, this process is called transformation to normality. One such function candidate in your case might be a mirrored log transform. Once your set satisfies a normality test you may then take the standard deviation. Then to use your 1$\sigma$ or 2$\sigma$ values you must transform them back into your original data space using the inverse of your transform function. I'm thinking this is what your professor was hinting at.

• This is a nice contribution. I'm not sure that the SD really "assumes" a normal distribution, though. – gung Mar 25 '15 at 18:08
• "Distribution fitting" and finding a transformation to normality are distinct procedures with different aims. – whuber Mar 25 '15 at 18:45

In general for the Bernoulli random variable $X$, that takes the value $1$ with probability $0<p<1$ and the value $0$ with probability $1-p$, we have

$$E(X) = p,\;\; SE(X) = \sqrt {p(1-p)}$$

And we want

$$E(X)+ SE(X) > 1 \Rightarrow p +\sqrt {p(1-p)} >1$$

$$\Rightarrow \sqrt {p(1-p)} > (1-p)$$

Square both sides to obtain

$$p(1-p) > (1-p)^2 \Rightarrow p > 1-p \Rightarrow p > \frac 12$$

In words, for any Bernoulli random variable with $p>1/2$ the theoretical expression $E(X)+ SE(X) > \max X$ holds.

So for example, for any i.i.d. sample drawn from a Bernoulli with, say, $p=0.7$, in most cases the sample mean plus the sample standard deviation will exceed the value $1$, which will be the maximum value observed (bar the case of an all-zeros sample!).

For other distributions we always have the opposite direction in the inequality, e.g. for a Uniform $U(a,b)$, it is always the case that $E(U)+ SE(U) < \max U=b$.
Therefore, no general rule exists.