In the previous post, we looked at the weighted average of two numbers. We pictured the weighted average as where you meet if you move inwards from the two points, but at potentially different rates each. In this post we will look at the term “weight” and look at how we can picture a weighted average by using weights.
For this, let’s compare two different situations:
On top, we see two points, as in the previous post, and the arrow indicates their weighted average, where the left end point has a weight of 60% and the right end point has a weight of 40%. On the bottom, we see the weight expressed in a different way: each point is replicated some number of times appropriate to its weight. By using 6 copies of the left point, and 4 copies of the right point, we model their relative weight. In the bottom picture, we traded the complexity of having each end point having a different weight for the complexity of a lot of data points but all of the same weight. We ended up with 10 data points, and we could calculate the average of the 10 numbers. The average of 42, 42, 42, 42, 42, 42, 62, 62, 62, 62 can be obtained by adding up all 10 numbes and dividing by 10. We get 50 as our result!
From this, we may get the idea for a different way of picturing the top situation:
Interestingly, for the purpose of calculating the weighted average it doesn’t matter if we think of the weights as 6 and 4 or .6 and .4, nor does it matter if we think of the weights as pounds or tonnes. Only the relative weights matter.
In terms of weight, the location of the weighted average (50) is the place where you could put a fulcrum to keep the whole dumbbell thing in balance. Another way of saying this is that the center of weight is at 50, consistent with the weighted average.
We can also relate this picture back to our idea of traveling towards the center! Imagine we have 6 people at the left end point and 4 people at the right end point. If all people walk towards 50, and stop there, then each party will have walked the same amount in total. And if each person regularly dropped bread crumbs, like Hansel and Gretel, then there would be just as many bread crumbs to the left of 50 as there would be to the right of 50!
I will leave you with a puzzle. In the picture below, we have 3 points, each weighted. We can calculate the weighted average of these three points. We have 42 with weight 6, 57 with weight 2, and 62 with weight 2. The weighted average is no longer 50. Which of the various ways in which we could think about the weighted average of two numbers still applies to the weighted average of the three numbers?