Solving Weighted-Average Problems
Of course!
Incorrect.
As you know, the average of a data set is given by the sum of all of the data points divided by the number of data points. Since the numbers of people that make $30,000 and $40,000 are different, the average income will not simply be $35,000. That is common mistake you will want to avoid on the GRE.
What is the total number of residents of Troy?
Incorrect.
Correct!
The total number of residents in Troy is equal to the number of residents who live on Xavier Avenue plus the number of residents who live on Yorick Avenue. So there are $$20 + 60 = 80$$ residents of Troy.
This means that you only need the sum of all of the incomes to find the average income of the residents of Troy. Unfortunately, you aren't given that information in the question. How might you be able to find the sum of all the incomes?
Incorrect. This assumes that the average income is $35,000, but that isn't the case.
Not quite. While this is technically correct, you don't know the overall average income of the residents of Troy. In fact, that is exactly what you are trying to ultimately solve for in this problem!
Indeed!
What is a way to express the sum of the incomes for the residents of Xavier Avenue?
Perfect!
Incorrect.
Incorrect.
This is just a rearrangement of the formula for average.
> $$\displaystyle \mbox{Average} = \frac{\mbox{Sum of values}}{\mbox{# of values}}$$
> $$\displaystyle \mbox{Average} \cdot \mbox{# of values}=\mbox{Sum of values}$$
Since you are told that there are 20 people that live on Xavier Avenue with an average income of $30,000, the sum of their incomes is $$(30{,}000)(20)=600{,}000$$.
Using the same logic for Yurick Avenue, you get a sum of $$(40{,}000)(60)=2{,}400{,}000$$.
So what is an expression for the average income of the residents of Troy?
Default step content.
Not so fast! While it might seem like the average should just be $35,000
The overall sum is just equal to the sum of the incomes on Xavier Avenue and the incomes on Yorick Avenue. The only problem is that you aren't given that information either.
That's right!
That's right!
To sum up:
[[summary]]
Incorrect.
On the GRE, it is possible that you will encounter a problem that can be solved using the concept of weighted averages. Here is an interesting problem based on this concept.
> _The small town of Troy has two streets: Xavier Avenue and Yorick Avenue. The residents of Xavier Avenue have an average annual income of $30,000, and the residents of Yorick Avenue have an average annual income of $40,000. Also, there are 20 people who live on Xavier Avenue and 60 people who live on Yorick Avenue. What is the average income of the residents of Troy?_
Before attempting to solve this problem, it is a good idea to define what it is you are looking for. Which of the expressions below gives you the average income of the residents of Troy?
In fact, both of the expressions are correct. It is usually easier to do the sum first and then divide, rather than the other way around, so the average income of the residents of Troy is
> $$\displaystyle\frac{600{,}000+ 2{,}400{,}000}{80} = \frac{3{,}000{,}000}{80} = 37{,}500$$.
One thing to notice is that the average is weighted more towards the avenue that had more residents. That is why this is called a **weighted average**.
20
60
80
Find the average of $30,000 and $40,000 and multiply it by 80
Find the overall average income of the residents of Troy and multiply it by 80
Find the sums of the incomes on Xavier and Yorick Avenues individually and then add them together
The average income of the residents times how many of them there are
The average income of the residents divided by how many of them there are
The number of residents divided by their average income
$$\displaystyle\frac{600{,}000}{80} + \frac{2{,}400{,}000}{80}$$
$$\displaystyle\frac{600{,}000+ 2{,}400{,}000}{80}$$
$35,000
$$\frac{\mbox{Number of residents}}{\mbox{Sum of all incomes}}$$
$$\frac{\mbox{Sum of all incomes}}{\mbox{Number of residents}}$$
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