Measures of Central Tendency

One value to stand for a whole set of data: the arithmetic mean (direct, assumed mean, step deviation, weighted), the median, quartiles, percentiles and the mode.

Key ideas

The mean uses every value but is pulled by extreme values; the median is the middle value and resists them; the mode is the most frequent value.
Use the mean for average marks, the median for a typical income where a few people are very rich, and the mode for the most popular size.
The assumed mean and step deviation methods give the same mean with smaller numbers.
A weighted mean multiplies each value by its importance before averaging.
The sum of deviations from the mean is always zero.
Quartiles split ordered data into four equal parts and percentiles into a hundred; Q2 and the 50th percentile are the median.
Data may have one mode, two (bimodal), or none.

Remember

Mean	ΣX ÷ N; for a series Σfx ÷ Σf (class marks for classes)
Assumed mean	A + Σfd ÷ Σf, where d = X − A
Step deviation	A + (Σfd′ ÷ Σf) × c, where d′ = (X − A) ÷ c
Weighted mean	ΣWX ÷ ΣW
Missing value	N × mean − sum of the others
Median position	(N + 1) ÷ 2 th item; Q1 at (N + 1) ÷ 4, Q3 at 3(N + 1) ÷ 4
Median (classes)	L + ((N ÷ 2 − c.f.) ÷ f) × h
Mode (classes)	L + D₁ ÷ (D₁ + D₂) × h; D₁, D₂ = modal frequency minus the classes before and after

Watch out

Finding the median without putting the data in order first.
For an even number of items, taking one middle value instead of the mean of the two.
In the median formula, using the cumulative frequency of the median class itself instead of the class before it.
Using the mean for income data with one huge value.

Worked example

Find the mode of this distribution.
Class: 40–60, 60–80, 80–100, 100–120, 120–140
Frequency: 19, 27, 22, 18, 24

Working: modal class 60–80: L = 60, D1 = 8, D2 = 5, h = 20; L + D1 ÷ (D1 + D2) × h

Answer: 72.31