5.1 The Balancing Act — Mean and Median Recap

5.1 The Balancing Act — Mean and Median Recap

Last year, we learnt about the mean^? and median^?. Recall: the mean of some data is the sum of all values divided by the number of values in the data. The median is the middle value when the data is sorted.

We shall now try to understand the mean and median from a different perspective and see how the mean behaves with changing data.

Mean as the Balancing Point

We saw earlier that the arithmetic mean is a measure of central tendency and represents the "centre" in the case of 2 numbers. Can there be more than one "centre"? In other words, is there any other value such that the sum of the distances to the values lower than it and the values higher than it will still be equal?

Consider the data 2, 4, 6, 8. Mean \(=\frac{2+4+6+8}{4}=5\). Distances on LHS of 5: \((5-2)+(5-4)=3+1=4\). Distances on RHS of 5: \((6-5)+(8-5)=1+3=4\). LHS = RHS ✓.

This is exactly halfway between the two numbers. Let us see if this holds for more than 2 numbers.

Consider 2, 4, 6. Mean = 4. LHS: (4-2)=2; RHS: (6-4)=2. Balanced. Consider 1, 3, 4, 5, 7. Mean = 4. LHS: (4-1)+(4-3)=3+1=4; RHS: (5-4)+(7-4)=1+3=4. Balanced again.

Can there be more than one such 'centre'? In other words, is there any other value such that the sum of the distances to the values lower than it and the values higher than it will still be equal?

Answer: No. There is only one centre. If we shift from the mean by even a tiny amount, the LHS total goes up while RHS total goes down, breaking the balance.

What Happens When a New Value is Included?

Will including a new value in the data increase or decrease the mean?

When a new value greater than the mean is included, the mean increases to maintain the balance between the sum of distances on the LHS and RHS. For example, start with 8, 10, 12, 14, 16 (mean = 12). Include 20 (which is greater than 12). New mean becomes \(\frac{8+10+12+14+16+20}{6}=\frac{80}{6}\approx 13.33\). The mean shifted right.

Similarly, if a value smaller than the mean is included, the new mean will be less than before.

Unchanging Mean!

We saw earlier how the mean varies when a value is included or removed. Explore if it is possible to include or remove 2 values such that the mean is unchanged.

You may use the following data to experiment with. Mean = 9 for a given collection. Adding two values that average to 9 (for example 6 and 12, or 7 and 11) keeps the mean at 9.

Can we include 2 values less than the mean and 1 value greater than the mean, such that the mean still remains the same? One of the possibilities: include 2, 4 (both less than 9) and 21 (greater than 9). Sum of 2+4+21 = 27 = 3×9, so the mean stays 9.

Try to include 2 values greater than the mean and 1 value less than the mean, so that the mean stays the same. Example for mean 9: include 11 and 14 (both > 9) and 2 (< 9). Sum = 27 = 3 × 9 ✓.

Relatively Unchanged!

We saw what happens to the mean when values are included or removed from the collection. What happens to the mean if every value in the collection increases by some fixed number? Consider the data 5, 10, 13, 4, 6, 7, 8, 6, 5. Calculate its mean: sum = 64, mean = \(\frac{64}{9}\approx 7.11\).

Now, consider this data with every value increased by 10: 15, 20, 23, 14, 16, 17, 18, 16, 15. What is its mean? Is there a quicker way to find this? Sum increases by \(9\times 10=90\), so new mean = \(7.11+10=17.11\).

That is, the new mean is \(k\) more than the previous mean.

Try to explain, using algebra, what the mean becomes when a fixed number is subtracted from every value in the collection (say, you have just lost!).

Scaling Property of Mean

What happens to the average if every value in the collection is doubled? The following is an example with the data we saw earlier: mean 7.11 becomes 14.22.

We can see that the average has doubled. Let us prove this using algebra. Suppose there are \(n\) values represented by \(x_1, x_2, \ldots, x_n\). Their average is \(\frac{x_1+x_2+\cdots+x_n}{n}=\bar{x}\).

When a fixed number, for example 5, is multiplied to every value in the collection, the new average becomes: \[\frac{5x_1+5x_2+\cdots+5x_n}{n}=\frac{5(x_1+x_2+\cdots+x_n)}{n}=5\bar{x}.\]

That is, the new average is 5 times more than the previous average.