Find mean deviation about median for grouped frequency data

Question

Find the mean deviation about the median for the following grouped data:

Class	0-10	10-20	20-30	30-40	40-50
Frequency	5	8	12	6	4

(NCERT Class 11, Chapter 15 — Statistics)

Solution — Step by Step

Total frequency: $N = 5 + 8 + 12 + 6 + 4 = 35$ , so $N/2 = 17.5$ .

Cumulative frequencies: 5, 13, 25, 31, 35.

The median class is the one where cumulative frequency first exceeds $N/2 = 17.5$ . That’s the 20-30 class (CF = 25).

\text{Median} = l + \frac{N/2 - CF}{f} \times h

where $l = 20$ (lower limit), $CF = 13$ (cumulative frequency before median class), $f = 12$ (frequency of median class), $h = 10$ (class width).

\text{Median} = 20 + \frac{17.5 - 13}{12} \times 10 = 20 + \frac{4.5}{12} \times 10 = 20 + 3.75 = 23.75

Using class mid-points ( $x_i$ ): 5, 15, 25, 35, 45

| $x_i$ | $f_i$ | $|x_i - 23.75|$ | $f_i|x_i - 23.75|$ | |--------|--------|-----------------|-------------------| | 5 | 5 | 18.75 | 93.75 | | 15 | 8 | 8.75 | 70.00 | | 25 | 12 | 1.25 | 15.00 | | 35 | 6 | 11.25 | 67.50 | | 45 | 4 | 21.25 | 85.00 |

\text{MD(Median)} = \frac{\sum f_i|x_i - \text{Median}|}{N} = \frac{93.75 + 70 + 15 + 67.5 + 85}{35}

= \frac{331.25}{35}

\boxed{\text{MD(Median)} \approx 9.46}

Why This Works

Mean deviation measures the average absolute distance of data points from a central value (here, the median). Unlike variance (which uses squared deviations), mean deviation uses absolute values, making it more intuitive — it tells you, on average, how far each observation is from the median.

We use mid-points of class intervals as representative values because we don’t know the exact distribution within each class. The median is calculated using the standard interpolation formula for grouped data.

Alternative Method — Mean deviation about the mean

If the question asks for MD about the mean instead, use the same process but replace the median with the mean:

Mean $= \frac{\sum f_i x_i}{N} = \frac{5(5) + 8(15) + 12(25) + 6(35) + 4(45)}{35} = \frac{735}{35} = 21$

Then compute $\sum f_i |x_i - 21|$ and divide by $N$ .

For CBSE, always present the solution in a table format — it’s cleaner and earns full step marks. Include columns for $x_i$ , $f_i$ , $|x_i - M|$ , and $f_i|x_i - M|$ . The examiner can follow your work easily, reducing the chance of losing marks for a calculation slip.

Common Mistake

The most common error: confusing mean deviation about the mean with mean deviation about the median. The question specifies which central value to use — read it carefully. Also, students sometimes forget to take the absolute value of $(x_i - \text{Median})$ , getting positive and negative deviations that partially cancel out. The absolute value is essential — without it, you’d get a much smaller (and incorrect) answer.

Question

Solution — Step by Step

Why This Works

Alternative Method — Mean deviation about the mean

Common Mistake

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