In the Middle – Weighted Average

In the post In the Middle we saw how 7th graders are very clear on how to find the center between two end points, if the end points are marked on paper with regular delineations, as on a quadrille pad or on a number line.  Their method, which I have unceremoniously dubbed “the two-handed approach”, was also seen to generalize well towards finding the median of a set of numbers.

In this post we will look at another generalization of the same basic idea.  What was that basic idea?  That you can get to the center by moving in towards the center from both ends, and that the place where you meet will have something to do with the center.

Typically, we’d offer that this only works if both sides move towards the middle at equal speed.  But what if the speeds are different?  If my daughter walks home from school, and I know the road she follows, I can pick her up by traversing that road in the opposite direction in my car.  Where we end up meeting depends on a number of different factors: the speed at which she walks, the speed at which I drive, and – very importantly – the difference between our starting times.  If I leave early enough, I’ll meet her at the parking lot right outside of school; if I leave late enough, I needn’t bother because she’ll be home already.  Inside of that range, all kinds of meeting places are possible.

Let’s simplify the meet-somewhere-in-the-middle scenario by stipulating that both parties leave at the same moment, both move towards each other, and each moves at a fixed speed.  The one thing we’re not insisting on this time, unlike the first post, is that both parties move at the same speed.  Can we make this scenario seem fair, in some way?  Maybe we could imagine that one of the parties is swimming, and that the other party is walking while carrying a baby.  Each expends the same number of calories per minute, but doesn’t cover the same distance as the other party even at the expenditure of the same number of calories.  Whatever the situation is that we want to draw attention to and want to equalize between the parties, we can always think of that as the cost for each.  We’ll say that each expends a different cost, and that we think it fair for each party to expend the same cost on the meeting.  This ‘cost’ doesn’t necessarily have to be expressed in dollars – it could be expressed in hours, or calories or sweat or steps, or something like that.

Now we have the problem stated in the way we want it: each party leaving from the end towards the other, moving at equal costs, though each party may have their own relationship between cost and miles.  Where will the parties meet?

Well, if you and I do this, and my miles cost .30 to your .70, we’ll meet at 70% of the way to your starting point, measured from mine.  If my miles cost .60 to your .40, we’ll meet at 40% of the way towards your starting point.  And if my miles cost the same as yours, .50 to your .50, we’ll meet at 50% of the way towards your starting point, or exactly half-way.  (Since the weights represent the cost per distance, the party with the largest weight ends up closest to his or her starting point.  This can be confusing for a while, until you are reconciled with the principle of equal cost for both parties.)

Below is a picture of the situation where my miles cost .60 to your .40, with me starting from the left, you from the right:

dataset-4The arrow indicates where we would meet.  This corresponds to the weighted average of your end point and mine, with my end point having a weight of 60% and your end point having the weight of 40%.

The amazing thing is that this reflects exactly the situation when the left end point represents the score on the final exam, weighted at 60%, and the right end point represents the score on the midterm exam, weighted at 40%!  This situation is shown below:

dataset-4-number-lineHow most of us would calculate a final score based on weighted scores is as follows: 40% of the midterm score 62 works out to 24.8 points towards the final score; 60% of the final exam score of 42 works out to 25.2 points towards the final score.  This gives us a final score of 24.8 + 25.2 = 50.  This matches our picture!  The two situations are essentially equivalent.  Can you work out any remaining doubt as to why this must be so?

Advertisements
This entry was posted in Uncategorized and tagged , , , , , , . Bookmark the permalink.

4 Responses to In the Middle – Weighted Average

  1. Pingback: In the Middle - Weighted Average, part II « Learning and Unlearning Math

  2. Pingback: In the Middle - Weighted Average, part III « Learning and Unlearning Math

  3. Pingback: In the Middle - Weighted Average, Part IV « Learning and Unlearning Math

  4. Pingback: In the Middle - the Series « Learning and Unlearning Math

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s