Are there any limits on Standard Deviation of a data set with given $n$ and mean?

Question

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 ?

Answer 1

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 $$ \mathbb{E}[X^2] \leq \max_{x\in\Omega} \lvert x\rvert \cdot \mathbb{E}[\lvert X\rvert]. $$

Answer 2

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 $.

Answer 3

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 $50$.