Are there any limits on Standard Deviation of a data set with given $n$ and mean? Say a class of 200 students is graded out of 100 marks. The mean of the dataset is 50.
Can we put a maximum limit on Standard Deviation for the set ?
I thought of putting a number of people onto 100 and the rest to zero and came up with $\frac{(100a + 0*(200-a))}{(200)} = 50$, which gives a = 100, so if a 100 people got 100 and a 100 got zero the average would still be 50.
Now Standard Deviation = $\sqrt{\frac{\sum{(x_i - 50)}^2}{N}}$
which would equal $\sqrt{ \frac{100*50*50 + 100 *50 *50}{200}}$ = $50$
Is this the max value of standard deviation of this data set ?
How do I prove it ?
EDIT
I now tried it out for different mean values and see that all values are less than 50. 
Thus max value should be 50. But how do I prove it ?
 A: For $1 \leq i \leq 200$, let $x_i$ be the grade of the $i$th student.
Let $y_i = x_i - 50$.
Then $$\sqrt{\frac{\sum{(x_i - 50)}^2}{N}} = \sqrt{\frac{\sum y_i^2}{N}}.$$
But since $0 \leq x_i \leq 100$, it follows
that $|y_i| \leq 50$ and $y_i^2 \leq 2500.$
Therefore
$$ \frac{\sum y_i^2}{N} \leq \frac{N \cdot 2500}{N} = 2500,$$
and so
$$\sqrt{\frac{\sum{(x_i - 50)}^2}{N}} \leq \sqrt{2500} = 50.$$
This shows that the maximum cannot be greater than $50$. 
You have already shown that $50$ is achievable, so the maximum
is in fact equal to $50$.
A: For a non-negative random variable over (here) $[0,100]$, you can write $$\mathop{Var}[X] = \mathbb{E}[X^2]-\mathbb{E}[X]^2 \leq \max_{x\in[0,100]} x\cdot \mathbb{E}[X]-\mathbb{E}[X]^2 = 100\cdot 50 - 50^2 = 2500,$$ so the standard deviation is at most $\sqrt{2500} = 50$. Here, we used the fact that $X$ is non-negative to write 
$$
\mathbb{E}[X^2] \leq \max_{x\in[0,100]} x\cdot \mathbb{E}[X]
$$
since in general for domain $\Omega\subseteq\mathbb{R}$ you only get
$$
\mathbb{E}[X^2] \leq \max_{x\in\Omega} \lvert x\rvert \cdot \mathbb{E}[\lvert X\rvert].
$$
A: If all the $x_i$
satisfy
$0 \le x_i \le 2m$
where $m$ is the mean,
then 
$|x_i-m| \le m$
so that
the variance
is
$\begin{array}\\
v
&=\frac1{n}\sum (x_i-m)^2\\
&\le\frac1{n}\sum m^2\\
&=m^2
\end{array}
$
so the standard deviation
$\sigma
\le \sqrt{m^2}
= m
$.
