Question

The following distribution shows the marks of 230 students in a particular subject. If the median marks are 46, then find the values of \(x\) and \(y\).

Marks	Number of Students
10 -- 20	12
20 -- 30	30
30 -- 40	\(x\)
40 -- 50	65
50 -- 60	\(y\)
60 -- 70	25
70 -- 80	18

Hint:

Use the cumulative frequency just before the median class, and substitute all known values into the median formula systematically.

Answer 1

Total number of students = 230
Median class = 40 -- 50 (since cumulative frequency just before it must be ≤ 115)
Let’s denote the frequencies and calculate the cumulative frequency (CF) column: \begin{center} \begin{tabular}{|c|c|c|} \hline Class Interval & Frequency (f) & Cumulative Frequency (CF)
\hline 10 -- 20 & 12 & 12
20 -- 30 & 30 & 42
30 -- 40 & $x$ & $42 + x$
40 -- 50 & 65 & $42 + x + 65 = 107 + x$
50 -- 60 & $y$ & $107 + x + y$
60 -- 70 & 25 & $132 + x + y$
70 -- 80 & 18 & $150 + x + y$
\hline \end{tabular} \end{center} We are told: \[ \text{Total frequency } = 230 \Rightarrow 150 + x + y = 230 \Rightarrow x + y = 80 \quad \text{(Equation 1)} \] Step 1: Identify median class details:
Median class = 40 -- 50, so:
\[ l = 40,\quad f = 65,\quad h = 10,\quad N = 230,\quad \frac{N}{2} = 115,\quad CF = 42 + x \] Step 2: Apply median formula: \[ \text{Median} = l + \frac{\frac{N}{2} - CF}{f} \cdot h\\ 46 = 40 + \frac{115 - (42 + x)}{65} \cdot 10\\ 46 = 40 + \frac{73 - x}{65} \cdot 10 \] \[ 6 = \frac{(73 - x) \cdot 10}{65} \Rightarrow \frac{730 - 10x}{65} = 6\\ 730 - 10x = 390 \Rightarrow 10x = 340 \Rightarrow x = 34 \] Step 3: Use Equation 1 to find $y$
\[ x + y = 80 \Rightarrow 34 + y = 80 \Rightarrow y = 46 \] Answer: $x = 34$, $y = 46$