Median of grouped data

The median is the middle value of a data set , the value that splits the data into two halves of equal size. For an ungrouped data of $n$ values arranged in increasing order, the median is the $((n+1)/2)$th value (if $n$ is odd) or the average of the $(n/2)$th and $(n/2 + 1)$th values (if $n$ is even).

For grouped data, the calculation is more involved. We can't pinpoint a single middle value because each class contains many observations. Instead, we find the median class , the class that contains the $(n/2)$th observation , and use a formula to estimate where within that class the median lies.

Step 1: Cumulative frequency

Compute the cumulative frequency table. This tells you how many observations fall up to and including each class.

For example:

Here $n = 28$ and $n/2 = 14$.

Step 2: Find the median class

The median class is the class whose cumulative frequency is the first to be $\geq n/2$.

For the example, $n/2 = 14$. The cumulative frequencies are $4, 10, 20, \ldots$ , the first to reach or exceed $14$ is $20$, which is in the class $20-30$. So the median class is $20-30$.

Step 3: Apply the formula

$\text{Median} = L + \frac{(n/2) - F}{f} \times h,$

where:

• $L$ = lower limit of the median class.
• $n$ = total frequency.
• $F$ = cumulative frequency before the median class.
• $f$ = frequency of the median class.
• $h$ = class size.

For the example: $L = 20, n/2 = 14, F = 10$ (cf of class $10-20$), $f = 10, h = 10$.

$\text{Median} = 20 + \frac{14 - 10}{10} \times 10 = 20 + 4 = 24$.

Intuition behind the formula

The median is the value at which exactly $n/2$ observations lie below. The cumulative frequency before the median class accounts for $F$ observations; we need $n/2 - F$ more to reach the median position. Within the median class, these $n/2 - F$ observations are spread across the class of size $h$, evenly. So we move $((n/2 - F)/f)$ of the way through the class, starting from its lower limit.

Worked examples

Example 1. Find the median: classes $0-10, 10-20, 20-30, 30-40, 40-50$ with frequencies $7, 10, 16, 8, 9$.

$cf$: $7, 17, 33, 41, 50$. $n = 50, n/2 = 25$. First $cf \geq 25$ is $33$, in class $20-30$.

$L = 20, F = 17, f = 16, h = 10$.

$\text{Median} = 20 + \frac{25 - 17}{16} \times 10 = 20 + 5 = 25$.

Example 2. Marks of $40$ students:

Find the median.

$cf$: $5, 13, 27, 34, 40$. $n/2 = 20$. First $cf \geq 20$ is $27$, in class $20-30$. $L = 20, F = 13, f = 14, h = 10$.

$\text{Median} = 20 + \frac{20 - 13}{14} \times 10 = 20 + 5 = 25$.

Example 3. Class size $5$, classes start at $0$. Frequencies: $3, 6, 9, 7, 5$. Find the median.

Classes: $0-5, 5-10, 10-15, 15-20, 20-25$. $cf$: $3, 9, 18, 25, 30$. $n/2 = 15$. First $cf \geq 15$: $18$. Median class: $10-15$. $L = 10, F = 9, f = 9, h = 5$.

$\text{Median} = 10 + \frac{15 - 9}{9} \times 5 = 10 + 10/3 \approx 13.33$.

Example 4. The median of a data with classes $0-10, 10-20, 20-30, 30-40$ and frequencies $f_1, 15, 25, 10$ is $24$. Find $f_1$ given $n = 60$.

$cf$: $f_1, f_1 + 15, f_1 + 40, f_1 + 50 = 60 \Rightarrow f_1 = 10$.

$n/2 = 30$. $cf$: $10, 25, 50, 60$. First $\geq 30$: $50$ (in class $20-30$). $L = 20, F = 25, f = 25, h = 10$. Median $= 20 + (30-25)/25 \times 10 = 20 + 2 = 22$.

But the given median is $24$, not $22$. So $f_1 = 10$ doesn't match. Let me re-verify (the question's setup may be designed to yield $f_1$ matching $24$): $24 = 20 + (30 - (f_1 + 15))/25 \times 10 \Rightarrow 4 = (15 - f_1)/2.5 \Rightarrow 15 - f_1 = 10 \Rightarrow f_1 = 5$.

So $f_1 = 5$, with total $= 5 + 15 + 25 + 10 = 55 \neq 60$. The problem is over-constrained. (This is the kind of detail to be careful with on board exams.)

Example 5. A more-than table gives cumulative frequency: more than $0$: $50$; more than $10$: $46$; more than $20$: $35$; more than $30$: $20$; more than $40$: $5$; more than $50$: $0$. Find the median.

Class frequencies: $50 - 46 = 4$ (class $0-10$), $46 - 35 = 11$ ($10-20$), $35 - 20 = 15$ ($20-30$), $20 - 5 = 15$ ($30-40$), $5$ ($40-50$). Total $50$.

$cf$ (less-than form): $4, 15, 30, 45, 50$. $n/2 = 25$. First $\geq 25$: $30$, in class $20-30$. $L = 20, F = 15, f = 15, h = 10$.

$\text{Median} = 20 + \frac{25 - 15}{15} \times 10 = 20 + 20/3 = 26.67$.

Try it yourself

1. Find the median: classes $0-10, 10-20, 20-30, 30-40$ with frequencies $4, 9, 7, 5$.
2. Find the median: classes $5-15, 15-25, 25-35, 35-45, 45-55$ with frequencies $5, 7, 11, 6, 4$.
3. Find the median height (cm): classes $140-150, 150-160, 160-170, 170-180$ and frequencies $10, 25, 15, 5$.
4. The mean of a data is $25$ and the mode is $20$. Find the median (use empirical relation).
5. A data set has total frequency $40$ and the median falls in the class $20-30$ with $L = 20, F = 15, f = 12, h = 10$. Find the median.
6. A frequency table has class size $5$ starting at $5$. Frequencies $3, 6, 8, 7, 4, 2$. Find the median.
7. A less-than table: $<10: 5, <20: 15, <30: 30, <40: 45, <50: 50$. Find the median.
8. The median of a data set with classes $0-10, 10-20, 20-30, 30-40$ is $25$. Total frequency is $50$ and the median class has frequency $20$. The cf before the median class is $f_0$. Find $f_0$.
9. Find the median: data given as cumulative frequencies $<5: 2, <10: 6, <15: 12, <20: 18, <25: 22$.
10. Compare mean, median, and mode for the data: $10, 12, 15, 17, 20, 22, 25, 28, 30, 32$.
11. The mean is $50$ and median is $48$. Find the mode (empirical).
12. A frequency distribution has classes $0-10, 10-20, 20-30, 30-40$ with cumulative frequencies $5, 15, 30, 40$. Find the median.

Pitfalls / Insight

(1) Always use the less-than cumulative frequency. If the problem gives a more-than table, first convert it.

(2) The median class is the first class whose cumulative frequency $\geq n/2$. Not the last class with $cf \leq n/2$ , that would be the previous class.

(3) The class size $h$ must be uniform for the formula to be straightforward. If not, you need to convert to uniform classes.

(4) The median is robust to outliers; the mean is not. If the data has a long tail (skew), the median is often the more "typical" value.

(5) For perfectly symmetric data, mean = median = mode. For right-skewed, mean > median > mode. For left-skewed, mean < median < mode.