The mean balances the data
Put every number on a seesaw. The mean is the pivot where it stays level.
Where should the pivot go?
Four equal weights sit on a beam at the places 10, 10, 11 and 17. You can put the pivot (the support) anywhere under the beam.
Where should the pivot go to keep the beam level?
What this lesson covers
The idea
The mean is the unique centre of the data at which the total distance to the values below it equals the total distance to the values above it; it is not always the midpoint of the extremes.
Where should the pivot go?
Four equal weights sit on a beam at the places 10, 10, 11 and 17. You can put the pivot (the support) anywhere under the beam.
Where should the pivot go to keep the beam level?
- At 10, where most of the weights are
- At 12
- At 13.5, halfway between 10 and 17
Level the beam
Each number is a weight on the beam. Slide the pivot (or drag it) until the beam is level. Watch the distances to the weights on each side. Do all three sets.
The mean is the balancing point
The mean is the unique centre of the data at which the total distance to the values below it equals the total distance to the values above it; it is not always the midpoint of the extremes.
For 10, 10, 11, 17 the mean is (10 + 10 + 11 + 17) ÷ 4 = 12 The values below 12 are 2, 2 and 1 away: total 5. The value above is 5 away: total 5. Both sides are equal.
The midpoint of the extremes is (10 + 17) ÷ 2 = 13.5. At 13.5 the left side adds to 9.5 and the right side to only 3.5, so the beam tips. For two numbers the mean is halfway between them (3 and 7 give 5), but for more numbers it need not be.
There is only one such centre. Move the pivot to the right of 12 and all the left distances grow while the right distances shrink. The same happens the other way.
Notes
The mean is the unique centre of the data at which the total distance to the values below it equals the total distance to the values above it; it is not always the midpoint of the extremes.
Check yourself
A beam has weights at 2, 4 and 12. Where is the balancing point, the mean?
Answer: 6
(2 + 4 + 12) ÷ 3 = 18 ÷ 3 = 6. Check: the left distances are 4 and 2 (total 6) and the right distance is 6.
Which statement is true for the data 10, 10, 11, 17?
The pivot under the beam with 10, 10, 11 and 17 is moved from 12 to 13. What happens?
The mean of four numbers is 20. Three of them are 12, 15 and 17. What is the fourth number?
Answer: 36
The distances below 20 are 8, 5 and 3: total 16. So the fourth number is 16 above 20: 36. Check: (12 + 15 + 17 + 36) ÷ 4 = 80 ÷ 4 = 20.
- The mean is halfway between 10 and 17.. Halfway is 13.5, but the mean is 12. The midpoint of the extremes is not always the mean.
- Any pivot between 10 and 17 balances the beam.. Only one pivot balances it. Move it away from 12 and one side gets heavier.
- The mean is 11, the middle number.. The mean is (10 + 10 + 11 + 17) ÷ 4 = 12. The middle number is the median idea, not the balancing point.
- The mean 12 is where the distances on both sides are equal. — correct. Yes! The distances on the left total 5 and on the right 5.
- The distances on both sides get bigger.. Moving right shortens every distance to the weights on the right.
- The left distances get bigger and the right distances get smaller, so the left side is heavier. — correct. Yes! The left total becomes 8 and the right total only 4, so the beam tips left.
- Nothing changes: the beam is still level.. It tips. The left total is now 8 and the right total is 4.
- The right distances get bigger.. The pivot moved towards the right weights, so their distances got smaller.