Median of Grouped Data and Missing Frequencies for CBSE Class 10

When evaluating income levels across a nation, the arithmetic mean can be heavily distorted by a handful of multi-billionaires, painting an artificially wealthy picture of the population. To eliminate the distorting influence of extreme outliers, economists and statisticians rely on the Median—the exact physical middle observation that partitions an ordered population into two halves.

In CBSE Class 10 Mathematics, Chapter 13 (Statistics), calculating the Median of Grouped Data and solving for missing frequencies ($x$ and $y$) are two of the most heavily tested 4-mark and 5-mark questions in the board examination.

What You Will Learn

• Definition of the Median as a measure of central tendency
• How to construct a Cumulative Frequency ($cf$) column
• Locating the Median Class using $\frac{N}{2}$
• The Median Formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$
• The critical distinction: $cf$ of the preceding class vs. $f$ of the median class
• Complete, step-by-step solution to the classic two missing frequencies ($x$ and $y$) problem
• Sanity checks and common student pitfalls

1. What is the Median?

Definition

The Median is the measure of central tendency that gives the value of the middle-most observation in a dataset when the observations are arranged in ascending or descending order of magnitude.

In grouped data, individual values are grouped inside class intervals. To find the median, we first construct a running total column called the cumulative frequency ($cf$), locate the Median Class, and interpolate using the median formula.

2. What is Cumulative Frequency ($cf$)?

The cumulative frequency of a class interval is the running total obtained by adding the frequency of that class to the sum of frequencies of all preceding classes:

    Class Interval      Frequency (fi)      Cumulative Frequency (cf)
    0 - 10                    5             5
    10 - 20                   8             5 + 8 = 13
    20 - 30                  12             13 + 12 = 25
    30 - 40                   7             25 + 7 = 32  <-- Last entry = Total N!

• The last entry in the $cf$ column must always equal the total frequency ($N = \sum f_i$).

3. How to Identify the Median Class

1. Calculate the total frequency: $N = \sum f_i$.
2. Compute $\frac{N}{2}$.
3. Inspect the cumulative frequency ($cf$) column from top to bottom.
4. Locate the first class whose cumulative frequency is greater than (or equal to) $\frac{N}{2}$.
5. That interval is the Median Class!

4. The Median Formula for Grouped Data

The Median Formula

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

    Parameters Breakdown:
    l  = Lower limit of the MEDIAN class
    N  = Total frequency (Σ fi)
    cf = Cumulative frequency of the class PRECEDING the median class (f_before!)
    f  = Simple frequency of the MEDIAN class itself
    h  = Class size (Upper limit - Lower limit)

The Single Most Common Error in Statistics: <u>In the median formula, students frequently substitute the cumulative frequency of the median class itself. THIS IS FATAL! $cf$ represents the cumulative frequency of the class PRECEDING the median class! Only $f$ belongs to the median class itself.</u>

5. Solved CBSE Board Examination Problems

Solved Example 1: Standard Median Calculation

Problem: Find the median of the following frequency distribution of 68 consumers:

Solution:

1. Construct the Cumulative Frequency Table:

2. Locate the Median Class: $\frac{N}{2} = \frac{68}{2} = \mathbf{34}$ The cumulative frequency just greater than $34$ is $42$, which belongs to the class $125 - 145$. Therefore, the Median Class is $125 - 145$.
3. List Parameters:
   • Lower limit: $l = \mathbf{125}$
   • Class size: $h = 145 - 125 = \mathbf{20}$
   • Frequency of median class: $f = \mathbf{20}$
   • Cumulative frequency of preceding class: $cf = \mathbf{22}$
4. Apply the Median Formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$ $\text{Median} = 125 + \left( \frac{34 - 22}{20} \right) \times 20$ Notice that $20$ in numerator and denominator cancels completely: $\text{Median} = 125 + (34 - 22) = 125 + 12 = \mathbf{137\text{ units}}$
5. Therefore, <u>the median monthly consumption is $137\text{ units}$</u>.

Solved Example 2: The Two Missing Frequencies ($x$ and $y$) (NCERT 5-Mark Classic)

Problem: The median of the following data is $28.5$. Find the values of $x$ and $y$, if the total frequency is $60$:

Solution:

1. Construct the Cumulative Frequency Table:

2. Formulate Equation 1 from Total Frequency: $\sum f_i = 45 + x + y = 60$ $x + y = 60 - 45 \implies \mathbf{x + y = 15} \quad \text{--- (1)}$

3. Locate the Median Class from the Given Median:

   • We are given that $\text{Median} = 28.5$.
   • The value $28.5$ lies in the interval $20 - 30$.
   • Therefore, the Median Class is strictly $20 - 30$!

4. List Parameters:

   • $l = 20$
   • $h = 10$
   • $f = 20$
   • $\frac{N}{2} = \frac{60}{2} = 30$
   • $cf = \mathbf{5 + x}$ (cumulative frequency of class preceding $20 - 30$).

5. Apply the Median Formula: $\text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h$ $28.5 = 20 + \left( \frac{30 - (5 + x)}{20} \right) \times 10$ $28.5 - 20 = \left( \frac{30 - 5 - x}{2} \right)$ $8.5 = \frac{25 - x}{2}$ $8.5 \times 2 = 25 - x \implies 17 = 25 - x$ $x = 25 - 17 = \mathbf{8}$

6. Substitute $x = 8$ into Equation (1) to Find $y$: $x + y = 15 \implies 8 + y = 15 \implies y = 15 - 8 = \mathbf{7}$

7. Therefore, <u>the missing frequencies are $x = 8$ and $y = 7$</u>.

6. Summary and Examination Tips

Exam Tip: In missing frequency problems with given median (e.g., $\text{Median} = 28.5$), DO NOT look at $N/2$ to find the median class! The median class is determined DIRECTLY by seeing which interval contains the number $28.5$ (here, $20 - 30$)!

Common Mistake: Forgetting parentheses when subtracting $cf$. In $30 - (5 + x)$, the minus sign applies to both terms: $30 - 5 - x = 25 - x$. Writing $30 - 5 + x$ is a catastrophic algebraic blunder!