Median of Grouped Data

Median and Cumulative Frequency

The median is the middle value — half the data lies below it, half above. For grouped data we cannot point to a single middle observation, so we build up a cumulative frequency (cf) column: for each class, the cf is the running total of frequencies up to and including that class.

The last cf equals the total frequency $n = \sum f_i$.

Finding the Median Class

Compute $\dfrac{n}{2}$. The median class is the first class whose cumulative frequency is greater than (and nearest to) $\dfrac{n}{2}$.

For example, if $n = 30$, then $\dfrac{n}{2} = 15$; the median class is the one where the running cf first reaches or passes 15.

Key Point: Use $\dfrac{n}{2}$ (not $\dfrac{n+1}{2}$) for grouped data, and find the class where the cf first crosses it.

The Median Formula

$$\boxed{\text{Median} = l + \left(\dfrac{\frac{n}{2} - cf}{f}\right)\times h}$$

where

• $l$ = lower limit of the median class,
• $n$ = total frequency $\sum f_i$,
• $cf$ = cumulative frequency of the class just before the median class,
• $f$ = frequency of the median class,
• $h$ = class size.

Key Point: $cf$ is the cumulative frequency of the class before the median class — not of the median class itself.

A Worked Median

For the marks data (frequencies $2,3,7,6,6,6$), the cumulative frequencies are $2, 5, 12, 18, 24, 30$. Here $n=30$, so $\dfrac{n}{2}=15$.

The cf first passes 15 in the class $55\text{-}70$ (cf $=18$), so that is the median class: $l=55$, $cf=12$, $f=6$, $h=15$. $$\text{Median} = 55 + \dfrac{15-12}{6}\times 15 = 55 + \dfrac{3}{6}\times 15 = 55 + 7.5 = 62.5.$$

[Board Important] So for this data: mean $=62$, median $=62.5$, mode $=52$.

Solved Examples

Example 1: Median of the marks data

Find the median: frequencies $2,3,7,6,6,6$ for classes $10\text{-}25,\dots,85\text{-}100$.

Solution:

1. cf: $2,5,12,18,24,30$; $n=30$, $\dfrac{n}{2}=15$.
2. Median class $55\text{-}70$ (cf 18): $l=55$, $cf=12$, $f=6$, $h=15$.
3. Median $= 55 + \dfrac{15-12}{6}\times 15 = 62.5$.

Final Answer: Median $= 62.5$.

Takeaway: Build cf, find $\tfrac n2$, then substitute.

Example 2: Median height

Heights (cumulative, 'less than'): below 140 → 4, 140-145 → 7, 145-150 → 18, 150-155 → 11, 155-160 → 6, 160-165 → 5 (total 51).

Solution:

1. cf: $4, 11, 29, 40, 46, 51$; $n=51$, $\dfrac{n}{2}=25.5$.
2. cf first passes 25.5 in $145\text{-}150$ (cf 29): $l=145$, $cf=11$, $f=18$, $h=5$.
3. Median $= 145 + \dfrac{25.5-11}{18}\times 5 = 145 + \dfrac{72.5}{18} = 149.03$ cm.

Final Answer: Median height $\approx 149.03$ cm.

Takeaway: For 'less than' data, the differences of the cumulative counts give the class frequencies.

Example 3: A missing frequency from the median

The median of a distribution is 28.5 with $n = 60$. The classes are $0\text{-}10,10\text{-}20,20\text{-}30,30\text{-}40,40\text{-}50,50\text{-}60$ with frequencies $5, x, 20, 15, y, 5$. Given the median class is $20\text{-}30$ and $\sum f = 60$, and the median formula holds, we illustrate the setup.

Solution:

1. $5 + x + 20 + 15 + y + 5 = 60 \Rightarrow x + y = 15$.
2. $\dfrac{n}{2}=30$; median class $20\text{-}30$: $l=20$, $cf=5+x$, $f=20$, $h=10$.
3. $28.5 = 20 + \dfrac{30-(5+x)}{20}\times 10 \Rightarrow 8.5 = \dfrac{25-x}{2} \Rightarrow 17 = 25 - x \Rightarrow x = 8$, so $y = 7$.

Final Answer: $x = 8$, $y = 7$.

Takeaway: Combine the total-frequency equation with the median equation to find two unknowns.

Example 4: Locating the median class

For frequencies $5, 8, 12, 10, 5$ (classes $0\text{-}5,5\text{-}10,10\text{-}15,15\text{-}20,20\text{-}25$), find the median class.

Solution:

1. cf: $5, 13, 25, 35, 40$; $n=40$, $\dfrac{n}{2}=20$.
2. cf first passes 20 in $10\text{-}15$ (cf 25).

Final Answer: Median class $= 10\text{-}15$.

Takeaway: The median class is where the running cf first reaches or crosses $\tfrac n2$.