12.1 The Balancing Act

Last year we met two ways to describe the “middle” of a list of numbers:

This year we view the mean through a new lens and watch how it reacts when the data changes.

Pick any two numbers, say 6 and 8. Their mean is

\frac{6+8}{2} = 7.

Plot 6, 8 and their mean 7 on a number line. The mean lands exactly midway between them. Try the pair 11 and 12: the mean \frac{11+12}{2} = 11.5 again falls smack in the centre.

The Mean as a Balance Point Picture each data value as an equal-weight bead threaded onto a number line, and imagine resting that line on a single pivot. The mean is the pivot where the line balances. In other words, the total distance from the pivot to the values on its left equals the total distance to the values on its right.

For two numbers this is just the midpoint. But with three or more numbers, the mean is usually not the midpoint of the smallest and largest value, and yet the balancing property survives. Look at the collection 14,\ 14,\ 15,\ 21. Its mean is

\frac{14+14+15+21}{4} = \frac{64}{4} = 16.

Now check the distances from the pivot 16:

The two sides match exactly, even though 16 sits nowhere near the midpoint of 14 and 21.

13 14 15 16 17 21 mean = 16 left total = 5 right total = 5

Is there only one such centre?

Could some different value also balance the data? Suppose we shift the pivot to a point larger than 16. Then every left-side distance grows while every right-side distance shrinks, so the two totals stop matching. Slide the pivot below 16 and the reverse happens. Either way the balance breaks. So the balancing centre is one of a kind, there is exactly one mean.

How the mean responds to change

Math Talk Before reading on, make a guess for each:

  • If we insert a new value above the mean, does the mean rise or fall?
  • If we delete a value, when does the mean go up, go down, or stay put?
  • What changes if we insert (or delete) a value that is exactly equal to the mean?

With the balance picture in hand, the rules almost prove themselves:

  • Inserting a value above the mean loads the right side, so the pivot slides right, the mean rises. Inserting a value below the mean makes the mean fall.
  • Deleting a value above the mean unloads the right side, so the mean falls; deleting one below the mean makes the mean rise.
  • Inserting or deleting a value equal to the mean leaves the mean alone, it sits right on the pivot and tips nothing either way.

Unchanging mean: balancing the changes

Can we insert or remove two values and still keep the mean fixed? Yes, provided the two new values balance each other around the mean. If the mean is 16, for instance, adding 13 (which is 3 below) together with 19 (which is 3 above) keeps everything level.

More broadly, we can insert two values below the mean and one value above, and still leave the mean unmoved, as long as the total “below” distance equals the total “above” distance.

Try This Take the collection 14,\ 14,\ 15,\ 21 (mean 16). Find two new values you could add so the mean stays 16. Then find three new values that keep the mean at 16, using two below and one above. (Hint: the total below-distance must equal the total above-distance.)

Shifting every value (relatively unchanged)

What if we add a fixed number to every value? Consider

9,\ 4,\ 11,\ 14,\ 5,\ 7,\ 9,\ 8,\ 9,\ 9,\ 14.

Its mean comes out to 9. Now add 12 to each value:

21,\ 16,\ 23,\ 26,\ 17,\ 19,\ 21,\ 20,\ 21,\ 21,\ 26.

The new mean is 21, precisely 12 more. The entire dot plot just slid 12 units to the right, dragging the balance point along with it. The mean’s position relative to the data never moved.

Here is the algebra. Suppose the values are x_1, x_2, \dots, x_n with mean

\frac{x_1 + x_2 + \dots + x_n}{n} = a.

Adding a fixed number (say 3) to each value gives a new mean

\frac{(x_1+3) + (x_2+3) + \dots + (x_n+3)}{n} = \frac{(x_1+\dots+x_n) + 3n}{n} = a + 3.

So the mean increases by exactly the amount added to each value. Subtracting a fixed number works identically, the mean drops by that amount.

Stretching every value

What if every value is multiplied by a fixed number? Doubling each value doubles the mean. In general, multiplying each value by 4:

\frac{4x_1 + 4x_2 + \dots + 4x_n}{n} = \frac{(x_1+x_2+\dots+x_n) \times 4}{n} = 4 \times \frac{x_1+x_2+\dots+x_n}{n} = 4a.

So multiplying every value by a number multiplies the mean by the same number.

Two Handy Rules for the Mean If the mean of a data set is a, then:

  • adding c to every value gives a new mean of a + c (a shift);
  • multiplying every value by c gives a new mean of c\,a (a stretch).

These let you re-compute a mean on the spot, without redoing the whole sum.