Shifting and stretching the data

Including one value moves the mean. Can you include two values and leave the mean where it was? Yes: put one above and one below, at equal distances from the mean. The balance does not change. (Removing two such values works the same way: take away 7 and 11 from data whose mean is 9, and the mean is still 9.)

Three values? Also possible, if the pulls cancel. For example, two values below the mean and one value above, far enough away to balance both. Try it and see.

Now a different question. What if every value changes by the same amount?

Take the data 8, 3, 10, 13, 4, 6, 7, 7, 8, 8, 5. Add 10 to each value: 18, 13, 20, 23, 14, 16, 17, 17, 18, 18, 15. What happens to the mean?

It goes up by exactly 10, and nothing else changes. The dot plot is the same picture slid 10 places along. The mean slides with it, and its position among the dots stays the same. A small check: 3, 7, 5 has mean 5; add 2 to each (5, 9, 7) and the mean is 7.

Here is why, using algebra. Let the values be x₁, x₂, …, xₙ with mean a, so (x₁ + x₂ + … + xₙ)/n = a. Add 3 to every value:

((x₁ + 3) + (x₂ + 3) + … + (xₙ + 3))/n = (x₁ + x₂ + … + xₙ + 3n)/n.

Split the fraction: (x₁ + … + xₙ)/n + 3n/n = a + 3. The 3n/n is just 3, because there are exactly n threes, one for each value.

Multiplying works the same way. Multiply every value by 5 and the mean becomes (5x₁ + 5x₂ + … + 5xₙ)/n = 5(x₁ + … + xₙ)/n = 5a (by the distributive property). Double every value and the mean doubles.

In short: shift the data and the mean shifts by the same amount; stretch the data and the mean stretches by the same factor.