Ruth from Manchester High School for Girls sent us her work on this problem. Well done, Ruth!

(a) When there are $2$ numbers,
their total is $2$. The possibilities for the values of the two
numbers can be represented as a straight line $x+y=2$ on a graph.
The mean is the point $(1,1)$ in the centre of the line. The
standard deviation ($\sigma$) is greatest when the distance between
the values and the mean is largest. The endpoints are furthest from
the centre so $\sigma$ is largest at the point $(2,0)$ when it is
equal to $1$.

(b) When there are $3$ values,
their total is $3$. This is represented by a plane. As the numbers
are non-negative the only values for the numbers are in a triangle
with the corners at $(0,0,3)$, $(0,3,0)$ and $(3,0,0)$. These are
the points furthest away from the mean which is at $(1,1,1)$ so are
where $\sigma$ is greatest and it is $\sqrt{2}$.

(c) When there are $n$ values, the
total is $n$. The region where this is true and they are
non-negative is a $n-1$-dimensional shape with $n$ corners which
have coordinates $(0,0,...,0,n)$ each with the $n$ in a different
position and $n-1$ $0$s. These are furthest from the mean (which is
at $(1,1,...,1,1)$) so are where the value of $\sigma$ is greatest.
$\sigma$ is the square root of the difference between the square of
mean and the mean of the squares. The mean is $1$ so the mean
squared is $1$. The squares are $n^{2}$, 0, 0, ... so the total of
the squares is $n^{2}$ and their mean is $n$. This makes $\sigma=
\sqrt{n-1}$.

In fact, as is explained in the
notes , we'd have to do some geometry for (c) to prove that the
points are indeed the furthest from the mean. But for this question
all we are asked to do is to show that the standard deviation can take a certain value, and Ruth
has done this.