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Weighted Averages II (Advanced)


In the first weighted average video we discussed two straightforward methods that could be used to solve these problem. So that first video you could say that was kind of a lesson designed for everyone. Here's what you need to know to solve weighted average problems and they're not really too bad if you understand these basic techniques. In this video I'm gonna show you an advanced technique.

If you are good at math and a good visual thinker, this could be a time saver. You absolutely do not need to know this. And so what I'll say is, if math is a struggle to you please do yourself a favor, don't even watch this video. It will just confuse you. Don't worry about it, you don't need it at all.

This is a video for people that are really good at math and it can show you a time saver. And if you're curious, by all means watch it. But keep in mind, don't be frustrated if this is something that is not natural to you because it is something quite advanced. So just to summarize, in the first lesson we talked about two basic methods.

We can find the sum of each subsection and add the sums, this works if we have the actual counts of the sums. If rather than have counts, we have proportions or percents for every subsection, we can multiply each subsection's average by those same subsections proportions and add those. And we saw this formula in the first video.

So again, those are the very straightforward methods. We can use another kind of proportional reasoning if the weighted average involves only two groups. Suppose the two groups have proportions p1 and p2. So those are two proportions that add to 1 and we're gonna say for the sake of argument, that p1 is the bigger group.

So p1, group one is bigger than group two. If we think of the averages on a number line, the average of the whole collection will be somewhere between the average of group one and the average of group two. And it'll be closer to the average of group one because one is the bigger group, it would have to be closer to it. So it would look something like this.

So again, group one is the bigger group and so the average is gonna be closer to it. In fact, the proportion of those distances on the number line will be the same as the proportions of the size of the two groups. Of course the average of the bigger group will be closer to the total average so the bigger group will have the smaller distance.

So again, one is the bigger group so it has d1 the smaller distance but if we take the ratio of those distances and the ratio of the sizes of the groups and of course we have to flip them over. Big goes to small, small goes with big, that sort of thing, but we can set them equal and it's the same ratio. Now typically we don't set up that ratio but we use that ratio to solve problems.

So here's a problem, we have this in the first video. On a ferry there are 50 cars and 10 trucks. The cars have an average mass of 1200 kg, the trucks have an average mass of 3000 kg. What is the average mass of all of them? So the weighted average of all 60 things all together.

Well, the proportion, cars to trucks, is 5 to 1, or 5. Another way to say this is cars are 5 parts, trucks are 1 part, and so all together, in all vehicles, that's 6 parts. So cars are five-sixths of all the vehicles and trucks are one-sixth of all the vehicles. And so the proportion between the two groups is 5 to 1, well that's gonna have to be the same proportion on the number line.

We have these two averages, 1,200 and 3,000, think of them as their locations on the number line, and we're gonna have a five to one ratio in the breakdown of the distances from the weighted average to those two individual averages. So let's think about that. That distance is 1800.

Well, just as the whole group of vehicles is divided into 6 parts, we have to divide that into 6 parts also. So how do we divide 1800 into 6 parts? Well, we divide it by 6. 1800 divided by 6 is 300. So 300 is one part and so that means the larger group, cars, are one part of the distance or 300 away from the average.

And trucks which is the smaller group they're five parts away. And so the easiest thing to do just start from the cars, which are 1,200. And one part away from that is plus 300. So we move up 300, 1200 + 300 = 1500. We could also start from trucks, 3,000, move down 5 parts, well 5 times 300 is 1500.

3000 minus 1500 is, again, 1500. So this is the same answer we found the first time. Here's a harder practice problem. Pause the video and then I'll talk. So at Didymus Corporation, there are two classes of employees: silver and gold. We have the average salary is $56,000 higher than the average of the silver.

So we have the difference between the gold and silver. Notice we don't actually have the actual number, what is the average salary of the gold employee? We don't have that number, we don't have average number of the silver, all we have is the difference between them. Then we're told the size of the two groups.

And we wanna know, the average salary for the company is how much higher than the average salary for the silver employees? Well let's think about this. On a number line, somewhere there'd be an average for the silver. Somewhere higher, there would be an average for the gold. The difference between these is 56,000.

And we'd expect the average for the company, of course, it would have to be between them because gold is a slightly bigger group, it would have to be slightly closer to the gold side. And so we're trying to ask for this distance. How much higher is that company average above the average for the silver employees.

So let's think about the size of the groups first, let's set up that ratio cuz we know those numbers, 120 to 160. 120 is 3 times 40. 160 is 4 times 40. Divide by 40 we get a ratio of 3 to 4. So silver employee are 3 parts, gold employee are 4 parts there are seven part in total.

So now let's think distances on the number line, from the average of silver up to the average of the total is 4 parts of the distance. From the average of the total up to the average of the gold is 3 parts of the distance. And again, there are seven parts in total. Well we know that those seven parts all together make up a difference on the number line of 56,000.

So 7 parts = $56K. Divide both sides by 7, we get 1 part = $8K. Well, now we know how big a part is on the distances, well, from the silver to the gold is 4 of those parts. Well, what is 4 times 8,000?

That's 32,000. So the total average for the company is $32,000 higher than the average for silver. If there are only two groups, the distances from the two group averages to the total average are in a ratio that is the reciprocal of the ratio of the proportions of the sizes of the groups.

If you intuitively understand this approach, it can be a time-saving trick on advanced weighted average problems.

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