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

Given Data:
Total number of students = 230
Median marks = 46

 

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

Step 1: Find total frequency equation
\[ 12 + 30 + x + 65 + y + 25 + 18 = 230 \] \[ 150 + x + y = 230 \implies x + y = 80 \quad ...(1) \]
Step 2: Median class and class size
Median = 46 lies in class interval 40 – 50 (since cumulative frequency reaches median here)
Class size, \( h = 50 - 40 = 10 \)

Step 3: Calculate cumulative frequencies (cf) up to each class
- cf up to 10–20 = 12
- cf up to 20–30 = 12 + 30 = 42
- cf up to 30–40 = 42 + x = 42 + x
- cf up to 40–50 = 42 + x + 65 = 107 + x

Total \( n = 230 \)
Median position \( = \frac{n}{2} = \frac{230}{2} = 115 \)

Step 4: Use median formula
\[ \text{Median} = l + \left( \frac{\frac{n}{2} - F}{f_m} \right) \times h \] Where:
- \( l = 40 \) (lower limit of median class)
- \( F = \) cumulative frequency before median class = \(42 + x\)
- \( f_m = \) frequency of median class = 65
- \( h = 10 \)
- Median = 46

Substitute values:
\[ 46 = 40 + \left( \frac{115 - (42 + x)}{65} \right) \times 10 \] \[ 46 - 40 = \left( \frac{115 - 42 - x}{65} \right) \times 10 \] \[ 6 = \left( \frac{73 - x}{65} \right) \times 10 \] \[ 6 = \frac{10 (73 - x)}{65} \] Multiply both sides by 65:
\[ 390 = 10 (73 - x) \] \[ 390 = 730 - 10x \] \[ 10x = 730 - 390 = 340 \] \[ x = 34 \]
Step 5: Find \(y\) using equation (1)
\[ x + y = 80 \implies 34 + y = 80 \implies y = 46 \]
Final Answer:
\[ x = 34, \quad y = 46 \]