15.38 Which of the following five statements I to V about the operation of finding the average (arithmetic mean) are always true:
I. Finding the average is associative;
II. Finding the average is commutative;
III. Finding the average is distributive over addition;
IV. Addition is distributive over finding the average;
V. Finding the average has an identity element.

The ones that are always true are
(A) All.
(B) Only I and II.
(C) Only II and III.
(D) Only II and IV.
(E) Only II and V.
(24th American High School Mathematics Examination, 1973)
[Solution] Let * denote the operation of averaging two numbers. For any two numbers $a$ and $b$, we have
$$
a * b=\frac{1}{2}(a+b) \text {. }
$$

Therefore, statements II and N are the only correct ones, because in II,
$$
a * b=\frac{1}{2}(a+b)=\frac{1}{2}(b+a)=b * a \text {. }
$$

And in $\mathbb{N}$, $\quad a+(b * c)=a+\frac{1}{2}(b+c)$, while $(a+b) *(a+c)=\frac{1}{2}[a+b+a+c]=a+\frac{1}{2}(b+c)$.
On the other hand, in I, $a *(b * c)=\frac{1}{2} a+\frac{1}{4} b+\frac{1}{4} c$,
and $(a * b) * c=\frac{1}{4} a+\frac{1}{4} b+\frac{1}{2} c$,
which are not always equal, so * is not associative.
In III, $a *(b+c)=\frac{1}{2} a+\frac{1}{2} b+\frac{1}{2} c$,
while $(a * b)+(a * c)=a+\frac{1}{2} b+\frac{1}{2} c$, which are also not always equal.
In $\mathrm{V}$, since $a * e=\frac{1}{2}(a+e)$, there does not exist an identity element $e$ for each number $a$ such that $e * a=a$.
Hence, the answer is $(D)$.