Mean vs Median: Balancing Point of Population Density Graphs

  • Thread starter Thread starter MeatyDumplings
  • Start date Start date
  • Tags Tags
    Statistics
Click For Summary

Homework Help Overview

The discussion revolves around understanding the concepts of mean and median in the context of population density graphs. The original poster seeks clarification on the definition and properties of the "balancing point" of a graph, as well as how to calculate its location.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the definitions of mean and median, questioning the meaning of the "balancing point" and its calculation. There are inquiries about the axes of the graph and the representation of probability density.

Discussion Status

The discussion is active, with participants providing insights into the properties of the mean as a balancing point in relation to area under the density curve. Some participants are attempting to clarify the relationship between the axes and the density function.

Contextual Notes

There is a mention of homework posting guidelines, indicating that the original poster is cautious about the appropriateness of the forum section for their question.

MeatyDumplings
Messages
3
Reaction score
0
in population density graphs, the median is known to be the line where it splits the graph into two equal sections. And the mean, according to the textbook, is the 'balancing point" of the graph. What exactly is this "balancing point"? what are its properties and how do i calculate the coordinate/location?
thanks in advance


sorry if this is in the wrong section, the Statistics section says not to post homework probelms there...
 
Physics news on Phys.org
What's on the x-axis?

The mean is the flat line drawn such that the (rectangular) area to the left of it equals the area to the left of the density curve.
 
Last edited:
What do the axes represent?
 
Assuming that x (taking on values from 0 to b) is some random variable and y (taking on values of 0 to a) is the probability density, you can think of "probability density" as if it were a real "weight density". The "balancing point" would be where the total weight (integral of density) is the same on both sides. If the total weight were one, that would mean that the weight on each side of the balancing point would be 1/2. You are looking for x such that
[tex]\int_0^x f(t)dt= \int_x^b f(t)dt= \frac{1}{2}[/tex].
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 2 ·
Replies
2
Views
17K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 309 ·
11
Replies
309
Views
17K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
10K
Replies
1
Views
2K