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Mean as the balancing point

AP.STATS:
UNC‑1 (EU)
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UNC‑1.I (LO)
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UNC‑1.I.2 (EK)
CCSS.Math: ,
Explore how we can think of the mean as the balancing point of a data distribution.
You know how to find the mean by adding up and dividing. In this article, we'll think about the mean as the balancing point. Let's get started!

Part 1: Find the mean

Find the mean of left brace, 5, comma, 7, right brace.
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Find the mean of left brace, 5, comma, 6, comma, 7, right brace.
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Interesting! In the first two problems, the data was "balanced" around the number six. Try the next one without finding the total or dividing. Instead, think about how the numbers are balanced around the mean.
Find the mean of left brace, 1, comma, 3, comma, 5, right brace.
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Notice how the 1 and 5 were "balanced" on either side of the 3:
Find the mean of left brace, 4, comma, 7, comma, 10, right brace.
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Can you see how the data points are always balanced around the mean? Let's try one more!
Find the mean of left brace, 2, comma, 3, comma, 5, comma, 6, right brace.
  • Your answer should be
  • an integer, like 6
  • a simplified proper fraction, like 3, slash, 5
  • a simplified improper fraction, like 7, slash, 4
  • a mixed number, like 1, space, 3, slash, 4
  • an exact decimal, like 0, point, 75
  • a multiple of pi, like 12, space, start text, p, i, end text or 2, slash, 3, space, start text, p, i, end text

Part 2: A new way of thinking about the mean

You might have noticed in Part 1 that it's possible to find the mean without finding the total or dividing for some simple data sets.
Key idea: We can think of the mean as the balancing point , which is a fancy way of saying that the total distance from the mean to the data points below the mean is equal to the total distance from the mean to the data points above the mean.

Example

In Part 1, you found the mean of left brace, 2, comma, 3, comma, 5, comma, 6, right brace to be start color #e07d10, 4, end color #e07d10. We can see that the total distance from the mean to the data points below the mean is equal to the total distance from the mean to the data points above the mean because start color #e84d39, 1, end color #e84d39, plus, start color #e84d39, 2, end color #e84d39, equals, start color #1fab54, 1, end color #1fab54, plus, start color #1fab54, 2, end color #1fab54:

Reflection questions

What is the total distance start color #e84d39, start text, b, e, l, o, w, end text, end color #e84d39 the mean in this example?
Choose 1 answer:
Choose 1 answer:

What is the total distance start color #1fab54, start text, a, b, o, v, e, end text, end color #1fab54 the mean in this example?
Choose 1 answer:
Choose 1 answer:

Part 3: Is the mean always the balancing point?

Yes! It is always true that the total distance below the mean is equal to the total distance above the mean. It just happens to be easier to see in some data sets than others.
For example let's consider the data set left brace, 2, comma, 3, comma, 6, comma, 9, right brace.
Here's how we can calculate the mean:
start fraction, 2, plus, 3, plus, 6, plus, 9, divided by, 4, end fraction, equals, start color #e07d10, 5, end color #e07d10
And we can see that the total distance below the mean is equal to the total distance above the mean because start color #e84d39, 2, end color #e84d39, plus, start color #e84d39, 3, end color #e84d39, equals, start color #1fab54, 1, end color #1fab54, plus, start color #1fab54, 4, end color #1fab54:

Part 4: Practice

Problem 1

Which of the lines represents the mean of the data points shown below?
Choose 1 answer:
Choose 1 answer:

Problem 2

Which of the lines represents the mean of the data points shown below?
Choose 1 answer:
Choose 1 answer:

Challenge problem

The mean of four data points is 5. Three of the four data points and the mean are shown in the diagram below.
Choose the fourth data point.
Choose 1 answer:
Choose 1 answer:

Want to join the conversation?

  • leaf blue style avatar for user Judith Aragon
    I do not get where the mean is located on a number line. Can someone help me i'm stuck
    (28 votes)
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    • duskpin seed style avatar for user ashna
      the mean is a measure of central data so it will always be in the middle of the data plot. this is why your data plot should be balanced on both sides. but don't do the literal middle of the data plot because that won't always be correct. you have to consider all of the data you have.
      (20 votes)
  • hopper happy style avatar for user Herobrine
    can someone please explain the last question to me
    (15 votes)
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    • blobby green style avatar for user Tony.YCHsieh
      You can take "mean" as a balance point between left and right, making the both side has exactly the same strength.

      In this case, we can see that the energy above the mean number is (7-5)*2=4, so you need 4 energy below the mean to balance the strength. Then, below the mean, we already have (5-4) = 1 energy, therefore we still need 3 energy to make it balanced.
      As a result, if you plot a dot at "2", you will have (5-2) = 3 energy.
      (7 votes)
  • blobby green style avatar for user shaithcock001
    I dont really understand this
    (9 votes)
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    • winston default style avatar for user Kaia
      Probably too late but here's an example:
      1 2. 3. 4. 5 6. 7. 8. 9. 10. 11.
      the numbers with dots are points.
      5 is the mean
      the distance from 5 to 4 is one. the distance from 5 to 3 is two. the distance from 5 to 2 is three. 1+2+3=6 so the distance to the left of the mean is 6. the same on the other side.
      Does that make sense?
      (4 votes)
  • piceratops sapling style avatar for user Desiree Gibbs
    i get it but the bar is messing me up
    (6 votes)
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  • starky sapling style avatar for user joseph hwangbo
    WHY SO MANY GRAPH it's not like i don't know how to do it just asking
    (5 votes)
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  • blobby green style avatar for user 646643
    This was a little hard for me so I don't get it
    (7 votes)
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  • mr pink orange style avatar for user isabellephillips
    i still do not understand the balancing point of mean. how can you break it down a little easier for me?
    (5 votes)
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  • aqualine ultimate style avatar for user Pachniak, Addison
    How to find the mean of the data on the plots.
    (6 votes)
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  • blobby green style avatar for user Alexander
    Hi! How can I prove that in all cases the total distance below the mean is equal to the total distance above the mean?
    (4 votes)
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    • primosaur seed style avatar for user Ian Pulizzotto
      The statement that the total distance below the mean equals the total distance above the mean is equivalent to the statement that the sum of the directed distances from the mean is 0.

      Let n be the number of data values, and let x_1, x_2, ... , x_n be the data values. Let m be the mean of these values.

      We need to prove that sum j from 1 to n of (x_j - m) = 0.

      By definition of mean, m = (sum j from 1 to n of x_j)/n.

      So mn = sum j from 1 to n of x_j.

      Therefore,
      sum j from 1 to n of (x_j - m)
      = (sum j from 1 to n of x_j) - (sum j from 1 to n of m)
      = (sum j from 1 to n of x_j) - mn
      = (sum j from 1 to n of x_j) - (sum j from 1 to n of x_j)
      = 0.

      Have a blessed, wonderful day!
      (5 votes)
  • male robot johnny style avatar for user surya.bond.prakash
    distance below the mean is equal to distance above the mean.....please can you guys tell an example in a real life.
    (5 votes)
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