Mean deviation for grouped data

When values repeat, the data comes as a frequency table. Each distance now counts as many times as its value appears, so multiply by the frequency.

Discrete data (separate values), about the mean.

• N = Σf = 20 and Σfx = 3 + 12 + 30 + 28 + 27 = 100, so the mean is 100/20 = 5.
• |x – 5| = 4, 2, 0, 2, 4; times f: 12, 8, 0, 8, 12, which add to 40.
• M.D. = Σf|x – x̄| / N = 40/20 = 2.

About the median (discrete). Make the cumulative frequency (c.f.) column; the median is the value whose c.f. first reaches N/2.

• N/2 = 12; the first c.f. that reaches 12 is 16, so the median is 6.
• |x – 6| × f: 12, 10, 0, 12, 8, which add to 42. M.D. = 42/24 = 1.75.

Continuous data (classes). Use each class's mid-point as its x.

• Mean: Σfx = 25 + 120 + 350 + 280 + 225 = 1000, N = 40, mean = 25.
• f|x – 25|: 100, 80, 0, 80, 100, total 360. M.D.(mean) = 360/40 = 9.

Median of a continuous table: find the class where the c.f. first reaches N/2, then median = l + (N/2 – C)/f × h, where l is that class's lower limit, C the c.f. before it, f its frequency and h its width. Here N/2 = 20, the class is 20-30 (c.f. before it 13): 20 + (7/14) × 10 = 25.

Step-deviation saves work. Take a middle mid-point a (here 25) and the width h (10): d = (x – a)/h gives –2, –1, 0, 1, 2. Then x̄ = a + (Σfd/N) × h. Here Σfd = –10 – 8 + 0 + 8 + 10 = 0, so x̄ = 25, the same answer with small numbers.

Tidy the classes first.

• Classes with gaps, like 11-15, 16-20, must be made continuous: move each boundary half the gap, giving 10.5-15.5, 15.5-20.5.
• 'Less than' or 'up to' totals (up to 15: 12 people, up to 30: 30 people) must be turned into ordinary class frequencies by subtracting: 0-15 has 12, 15-30 has 30 – 12 = 18.