The Mean Property of the Mean

Robert Muldoon (1921 - 1992), the erstwhile Prime Minister of New Zealand is reported to have said

New Zealanders who emigrate to Australia raise the IQ of both countries.

Might it really be possible?

Perhaps, not very surprisingly, the answer is Yes which is based on a simple property of the mean.

Let me mention in passing that the mean, or more accurately the arithmetic mean (also, the average) $M(a_{1}, a_{2}, \ldots, a_{N})\;$ of N numbers $a_{1}, a_{2}, \ldots, a_{N}\;$ is, by definition,

$\displaystyle M(a_{1}, a_{2}, \ldots, a_{N}) = \frac{a_{1} + a_{2} + \ldots + a_{N}}{N}.$

The fundamental property of the mean which is equivalent to the Pigeonhole Principle

$\min(a_{1}, a_{2}, \ldots, a_{N}) \le M(a_{1}, a_{2}, \ldots, a_{N}) \le \max(a_{1}, a_{2}, \ldots, a_{N}).$

In words, the minimum is at most and the maximum is at least the average.

Now, consider the original set of numbers $\{a_{1}, a_{2}, \ldots, a_{N}\}$ and augment it with an "extra" number $b.\;$ What is the relationship between $M(a_{1}, a_{2}, \ldots, a_{N})\;$ and

$\displaystyle M(a_{1}, a_{2}, \ldots, a_{N}, b) = \frac{a_{1} + a_{2} + \ldots + a_{N} + b}{N + 1}?$

I'll put a fat question mark between to numbers to indicate that this is the question of comparison. At this point we do not know whether one is greater or smaller or the two are equal. However, observe that regardless of which of the possibilities turns out correct, we still have the right to multiply both by the same positive number or add or subtract the same number from both without affecting as yet unknown relationship between the two. So,

$\displaystyle \frac{a_{1} + a_{2} + \ldots + a_{N}}{N}\;$ ? $\displaystyle \frac{a_{1} + a_{2} + \ldots + a_{N} + b}{N + 1}.$

This we multiply by the positive number $N(N + 1):$

$(N + 1)(a_{1} + a_{2} + \ldots + a_{N})\;$ ? $N(a_{1} + a_{2} + \ldots + a_{N} + b).$

Now subtract $N(a_{1} + a_{2} + \ldots + a_{N})$ from both sides:

$a_{1} + a_{2} + \ldots + a_{N}\;$ ? $Nb.$

And finally divide both sides by $N:$

$\displaystyle \frac{a_{1} + a_{2} + \ldots + a_{N}}{N}\;$ ? $b.$

But on the left we have the average of the original set: $M(a_{1}, a_{2}, \ldots, a_{N}).\;$ So, the meaning of the question mark at the beginning of the derivation is determined by its meaning in

$M(a_{1}, a_{2}, \ldots, a_{N})\;$ ? $b.$

If $b\;$ is greater than (less than, equal to) the average of a given set of numbers then adding it to the set will increase (decrease, not change) the average.

Similar result holds when an element is removed from a set. The removal increases, decreases, or does not change the average, according to whether the number being removed is less, greater, or equal to the average (of the whole set).

This could be used to explain Robert Muldoon's quip. Assume that, at the time, the set of individual IQs of the citizens of New Zealand was $Z,\;$ with the average $M(Z).\;$ Similarly, assume the set of Australian IQs was $A,\; with the average $M(A).\; If there was a New Zealander with IQ equal to c such that

$M(Z) \gt c \gt M(A),$

which at the very least assumes that $M(Z) \gt M(A),\;$ then removing $c\;$ from $Z\;$ and adding it to $A\;$ will certainly confirm Robert Muldoon's idea:

$M(Z - {c}) \gt M(Z)\;\text{and}\;M(A \cup {c}) \gt M(A).$

Related material
Read more...

  • The Means
  • Averages, Arithmetic and Harmonic Means
  • Expectation
  • The Size of a Class: Two Viewpoints
  • Averages of divisors of a given integer
  • Family Statistics: an Interactive Gadget
  • Averages in a sequence
  • Arithmetic and Geometric Means
  • Geometric Meaning of the Geometric Mean
  • A Mathematical Rabbit out of an Algebraic Hat
  • AM-GM Inequality
  • Harmonic Mean in Geometry
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