Mean, median & mode for every kind of series

Mean, median and mode in one line each

The mean is the fair share: add up everything and divide by the number of items. The median is the middle item once the data is in order. The mode is the value that appears most often. Each one answers a different question, so the calculator gives all three.

Three ways to find the mean, and why they agree

The three methods give the same answer. They only differ in how much arithmetic you do, which is why your teacher may ask for one of them.

Direct method

Multiply each value (or class midpoint) by its frequency, add those products to get Σfx (or Σfm), and divide by N. For the classes 0–10, 10–20, 20–30, 30–40 and 40–50 with frequencies 5, 8, 12, 7 and 3, the midpoints are 5, 15, 25, 35 and 45. Σfm = 825 and N = 35, so the mean is 825 ÷ 35 = 23.57.

Assumed mean method

Pick a convenient value A near the middle. Take deviations d = m − A, multiply by f, and the mean is A + Σfd ÷ N. With A = 25 the deviations are −20, −10, 0, 10 and 20, Σfd = −50, and the mean is 25 − 50 ÷ 35 = 23.57.

Step deviation method

When the classes are the same width h, divide the deviations by h to get smaller numbers: d′ = (m − A) ÷ h. Here h = 10, so d′ is −2, −1, 0, 1 and 2, Σfd′ = −5, and the mean is 25 + (−5 ÷ 35) × 10 = 23.57. Step deviation only works when the class widths are equal.

The median of a continuous series

Find N ÷ 2, which is 17.5 for N = 35. Read down the cumulative frequencies (cf): 5, 13, 25, 32, 35. The first cf that reaches 17.5 is 25, so the median class is 20–30. Then median = L + (N ÷ 2 − cf) ÷ f × h, where L is the lower limit of the median class, cf is the cumulative frequency of the class before it, f is its frequency and h its width. That gives 20 + (17.5 − 13) ÷ 12 × 10 = 23.75.

The mode of a continuous series

The modal class is the one with the highest frequency, here 20–30 with 12. Then mode = L + (f1 − f0) ÷ (2f1 − f0 − f2) × h, where f0 and f2 are the frequencies of the classes just before and after the modal class. That gives 20 + (12 − 8) ÷ (24 − 8 − 7) × 10 = 24.44.

Inclusive classes

Some tables write classes as 10–19 and 20–29, so 19 and 20 look like separate numbers. Real values can sit between 19 and 20, so the classes must be stretched to 9.5–19.5 and 19.5–29.5. The gap is 20 − 19 = 1, and half of it, 0.5, is added to each upper limit and taken from each lower limit. Do this first, then find the median and mode.

Reading the histogram

Each bar is a class and its height is the frequency. (When the classes are not all the same width, the height is the frequency per unit of width, so that each bar’s area matches its frequency.) With equal classes the tallest bar is the modal class, which is where the mode sits. The median line is the point that splits the total area in half, so about half the items lie on each side of it.