The mean as a balancing point

From two numbers to many

Take any two numbers and find their mean: for 3 and 7 it is (3 + 7) ÷ 2 = 5, and for 8 and 9 it is 8.5. On a dot plot the mean always lands exactly halfway between the two. With three numbers it is still the ‘centre’, but not always the midpoint of the extremes. The book’s dot plots show 6, 7, 8 with mean 7; 3, 6, 9 with mean 6; 2, 4, 9 with mean 5; and 4, 11, 15 with mean 10. With four values, 11, 13, 17, 19 have mean 15 and 5, 6, 15, 16 have mean 10.5.

Balancing the distances

What makes the mean the centre? Measure how far each value is from the mean. For 2, 4 and 9, with mean 5, the values below are 3 and 1 away, a total of 4, and the value above is 4 away. For 10, 10, 11 and 17, with mean 12, the distances below are 2 + 2 + 1 = 5 and above, 5. The total distance on the left (LHS) always equals the total on the right (RHS): the mean is the balancing point.

Could another point balance too? Try 13 for the data 10, 10, 11, 17: every distance on the left grows and the distance on the right shrinks, so the totals no longer match. Any point below 12 fails the other way. So the balancing point is unique.

Including and removing values

If a new value above the mean is included, the right side gains distance, so the balancing point has to move up: the mean increases. A value below the mean pulls it down, and a value equal to the mean adds no distance, so nothing changes. Removing values works in reverse: removing a value above the mean lowers it. In the fair-share picture, a newcomer who brings more than the fair share raises everyone’s share.

The mean can stay fixed when several values are included, as long as their distances balance. With mean 9, including 7 and 11 (2 below and 2 above) keeps it at 9, and so does including 7, 7 and 13 (2 + 2 below, 4 above). Two values above and one below also work, such as 10 and 11 with 6: 1 + 2 above balance 3 below.

Changing every value

What if every value goes up by the same amount? The mean of 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5 is 79 ÷ 11 ≈ 7.18. Add 10 to every value and the whole dot plot slides 10 to the right, so the new mean is about 17.18; subtract 1 and it is about 6.18. Algebra shows why: if n values add up to n × a, adding 3 to each adds 3n to the sum, and dividing by n gives a + 3. If every value is doubled, the mean doubles too: multiplying each value by 5 multiplies the sum, and so the mean, by 5.

Using the mean to find unknowns

Because sum = mean × number of values, the mean can reveal a hidden number. Coach Balwan’s ten wrestlers average 39.2 kg, so they weigh 392 kg together; the nine readable weights add up to 349 kg, so the smudged one is 43 kg. Venkayya’s 15 coconut trees average 25.6, a total of 384; one count was 3 too high, so the true total is 381 and the correct mean is 25.4.

Read on: line graphs, mean and median from a frequency table, the chapter notes and the NCERT solutions.