(a) The two numbers must be of the form

x

and

2 - x

, where

x \geq 0

and

2 - x \geq 0

. The standard deviation is

σ = \sqrt{\frac{(x - 1)^{2} + (2 - x - 1)^{2}}{2}} = | x - 1 | .

0 \leq x \leq 2

we see that

σ \leq 1

and that we have

σ = 1

when

a = 0

and

b = 2

(or

a = 2

and

b = 0

(b) The three numbers $x$ , $y$ and $z$ must satisfy $x + y + z = 3$ (because their mean is $1$ ). Thus the point $(x, y, z)$ in 3-space must lie on the plane $x + y + z = 3$ , and it is in the first octant of 3-space (as $x$ and $y$ and $z$ are non-negative). This means that $(x, y, z)$ lies on the triangle $T$ (in 3-space) which has vertices $(3, 0, 0)$ , $(0, 3, 0)$ , $(0, 0, 3)$ . Note that the point $(1, 1, 1)$ is the centroid of $T$ .

Now

σ^{2} = \frac{(x - 1)^{2} + (y - 1)^{2} + (z - 1)^{2}}{3},

and we want to maximize

σ

or, equivalently,

\sqrt{3} σ

, which is the distance from

(x, y, z)

T

to the centroid

(1, 1, 1)

T

. Clearly, this maximum occurs at each vertex; for example at

(x, y, z) = (3, 0, 0)

. Thus

σ \leq \sqrt{\frac{(3 - 1)^{2} + (0 - 1)^{2} + (0 - 1)^{2}}{3}} = \sqrt{2} .

(c) If the $n$ numbers are $n, 0, 0, \dots, 0$ (which have mean $1$ ), then the standard deviation is

$σ = \sqrt{\frac{(n - 1)^{2} + (0 - 1)^{2} + \dots + (0 - 1)^{2}}{n}} = \sqrt{\frac{(n - 1)^{2} + (n - 1)}{n}} = \sqrt{n - 1} .$