# میانگین وزنی

سرفصل: بخش ریاضی / سرفصل: آمار / درس 3

## بخش ریاضی

14 سرفصل | 192 درس

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• زمان مطالعه 12 دقیقه
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## Weighted Averages

We did averages. The test sometimes will ask us, to combined average of two or more different groups. So here’s a sample problem. We’re not gonna do this sample problem right now, but just to give you an idea of this kind of question.

On a ferry, there are 50 cars and 10 trucks. The cars have an average mass of 1200 kilograms and the trucks, have an average mass of 3,000 kilograms. What is the average mass of all 60 vehicles on the ferry? So in other words, we have two different groups, that are unequal size. They have unequal averages, and we’d like to figure out the average of everyone combined together.

So that’s the basic gist of this problem type. There are a few different ways to approach problems such as this. The first approach would be to think in terms of sums. As we discussed in the first lesson of this module, think in terms of sums, can often simplify problems about averages. In that problem with vehicles, since all the masses are in thousands, we do the calculation with everything divided by 1000 for much simpler numbers.

So it just makes our life easier to make everything a thousand times smaller. So here’s the problem again. So, we’ll just work through this problem together. The car sum is 50, and instead of using 1200, I’m just gonna divide by a thousand. I’m gonna call that 1.2. 50 times 1.2.

Well that turns out to be 60. And then the truck sum, is gonna be 10 trucks times 3000 divided by 1000, so that’s 10 times three which is 30. That means that the total sum is 90. Well then, 90 divided by 60 vehicles. That’s nine divided by six, that’s three halves or 1.5.

And now we just multiply that by 1000, 1500 kilograms, that has to be the mass of all the vehicles together, all 60 vehicles on the ferry. Using sums is one way to approach weighted average problems. It’s one good trick to have up your sleeve. Another approach involves thinking about proportions, and this is very important to understand also.

Each group of individuals, is some proportion of the entire pool. We can simply multiply the average of each group by this proportion, find the sum of all those products, and that sum is the average of the entire set. So for example, symbolically, suppose there are three groups. Three different sized groups, and each, they have averages, A1, A2, and A3. Those are the averages of these three different groups.

They have three different averages. Furthermore, the first group is P1 of of the whole. That’s its proportion. The proportion of the whole. The second is P2 the third is P3, these three proportions should add up to one, because together they all add up to the entire group.

So what I’m saying is, we can multiply every average, times it’s corresponding proportion. And then add these products together. So the average of the whole equals A1 times P1 plus A2 times P2 times A3 plus P3. This is a very important idea.

So pause the video and then work on this problem. All right so here we don’t even have the number of employees, all we have are the percentages. So if you’re given percentages, it’s a very good indication, that you’re to solve the problem using proportional thinking.

So what we’re gonna do is, we’re going to take each one of those percentages, change it into a decimal and multiply it by its corresponding salary. And again, we’re gonna make these salaries much, much smaller. We’re gonna divide them all by 1,000. So, the total is gonna be, for the marketers, that’s 0.7, 17% is 0.7. And then their salary, we’re gonna count as 40 instead of 40,000.

The programmers are 20% which is 0.2. Their salary is gonna count as 80. And the managers 10% that’s 0.1 times their salary 120. So we’re just gonna multi, do all the multiplication there. We get 28 plus 16 plus 12, that adds up to 56. Remember we divided it by 1000 so this 56, it really should be 56,000.

That’s the average salary of all those employees at the company. We can use another kind of proportional reasoning. If the weighted average involves only two groups. So this is really complicated, but if you understand this, this is incredibly powerful and incredibly useful.

Suppose the two groups have proportions p1 and p2, where p1 plus p2 equals 1 and where p1, we’re just gonna pretend p1 is the bigger group. So p1 is larger than p2. The average of the whole collection, will be between the averages of the individual groups, and in fact, it will be closer to the average of the bigger group. That makes sense, the bigger group is gonna have more of an influence, so the average is gonna be closer to the bigger group.

Think about this all on a number line. We’re gonna have A1 somewhere on the number line. We’re gonna have A2 somewhere on the number line. Somewhere between them we’re goinna have a total. The total average of the group, of course this is gonna be closer to A1 and A2 and suppose we look at those distances so D1 is the distance on the number line from A1 to the total average.

D2 is the distance from A2 to the total average. All right? Here’s the interesting part. The ratio of those distances, is inversely proportional to the ratio of the two portions. In other words, p1 over p2, the ratio of the two proportions, equals d2 over d1.

And this makes sense because the one that has a larger size. A larger proportion is gonna be closer to the average so it’s going to have a smaller distance, a smaller separation from the average. All right, well this is interesting. How on earth are we gonna use this. So let’s first of all talk this through very carefully, using this problem.

We actually solved this problem earlier, but we’re gonna solve this same problem another way right now. So we have 50 cars, 10 trucks, so the cars are the bigger group. The cars have an average of 1200, the trucks have an average of 3000, we know that the total average has to be closer to 1200 than to 3000, and we want the total average.

Well the ratio of cars to trucks, clearly is five to one, or five. So another way to say this is that, cars account for five parts. The trucks account for one part. And so the total is six parts. And so, the proportions, in other words, cars are five six of the vehicles, on this ferry and trucks are one sixth of the vehicles on this ferry.

So those are the proportions. And that ratio five six to one six, that’s just a five to one ratio. Now we’re gonna look at the distance between the two averages. 3000 minus 1200 is 1800. So that’s gonna be 1800. That’s the real estate, on the number line between those two averages.

And that’s gonna have to be divided into six parts, because our proportion is in six parts. So we divide that into six parts and so one part, would be 1800 divided by 6 which is 300. And so that means, that one part is 1800. One part is 300, five parts is 1500.

And so that, that tells us how that 1800 is split up. So that means that we start at the mass of a car, which is the larger group. We only have to go 300, we have to go one part away from that up to the average or if we started at trucks 3,000 we’d have to go five parts down from there, 1,500 down from there, to get to the average. Well it’s easier to up from 1,200 just by one part.

The distances is from the, from the cars mean to the total mean must be one of these parts, 300. So we just add 1200 plus 300, that gives us the total mean at 1500. Notice we could have gotten there also by starting at 3000, going down five parts. 3000 minus 1500, also lands at 1500. So that is a very different way to approach this problem.

And notice also, this is the exact same answer we found the first time, using a very different approach. So two very different ways. And as always, if you can solve a problem in two different ways, you really understand it well.

Here’s another practice problem, pause the video and then we’ll talk about this. Okay, at Didymus Corporation, there are just two classes of employees, silver and gold. The average salary of gold employees is \$56000 higher than that of silver employees. Guess it’s good to be a gold employee here.

If there are 120 gold employees and 160, sorry 120 silver employees and 160 gold employees, then the average salary for the company, the total average salary, is how much higher, than the average salary for the silver employees, so the average salary for the lower employees. Okay. First thing we’re gonna do is figure out this ratio.

120 to 160, that’s 12 over 16, which is 3 over 4. So the distance from the silver to the total is gonna be four parts, it’s gonna the longer distance, the slightly longer distance, because the silvers are a slightly smaller group. The distance from the total, up to the average of the gold, that’s only gonna be three parts, because that’s the, the larger group, it’s gonna have a smaller distance.

Together that’s seven parts, and that’s gonna have to equal \$56,000. We’ll divide, and we get one part is 8 or 8000 and so now, since we’re starting out from silver, we have to go up four parts. Well 4 times 8 is 32 or \$32000. So the total is \$32000 higher than the average for the silver employee. One way to approach weighted averages is to find the sum, we can also find the proportion of each group and multiple each group average, by that group’s proportion, and simply add those up.

And 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. And this final way of thinking about it, admittedly, it’s very hard to put it into words in a way that’s clear and it can be very confusing when you first encounter it, but if you can learn to think about it this way, it will give you a powerful advantage on the test.

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