If the median of the following data is 32.5 , then  the value of x and y:

Given a frequency distribution table with:

• Class intervals: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70
• Frequencies: f₁, 5, 9, 12, f₂, 3, 2
• Total frequency: 40
• Median: 32.5

We need to find the values of x (f₁) and y (f₂).

Step-by-Step Solution

Step 1: Set Up the Frequency Equation

Since the total frequency is 40:

f₁ + 5 + 9 + 12 + f₂ + 3 + 2 = 40

Simplifying: f₁ + f₂ + 31 = 40

f₁ + f₂ = 9 ... (Equation 1)

Step 2: Create Cumulative Frequency Table

Step 3: Identify the Median Class

For grouped data:

• N (total frequency) = 40
• N/2 = 20

The median class is where cumulative frequency just exceeds N/2.

Since the median is 32.5 (which falls in 30-40), the median class is 30-40.

This means:

• Cumulative frequency before median class: f₁ + 14
• This should be less than 20: f₁ + 14 < 20
• Therefore: f₁ < 6

Step 4: Apply the Median Formula

The median formula for grouped data:

Median = L + [(N/2 - cf)/f] × h

Where:

• L = Lower boundary of median class=30
• N/2 = 20
• cf = Cumulative frequency before median class=f₁ + 14
• f = Frequency of median class=12
• h = Class width = 10

Substituting values:

32.5 = 30 + [(20 - (f₁ + 14))/12] × 10

32.5 - 30 = [(20 - f₁ - 14)/12] × 10

2.5 = [(6 - f₁)/12] × 10

2.5 = (6 - f₁) × 10/12

2.5 × 12 = (6 - f₁) × 10

30 = 60 - 10f₁

10f₁ = 30

f₁ = 3

Step 5: Find f₂

From Equation 1: f₁ + f₂ = 9

3 + f₂ = 9

f₂ = 6

Final Answer

x = 3 and y = 6 (Option a: 3,6)

Verification

Let's verify:

• Total frequency: 3 + 5 + 9 + 12 + 6 + 3 + 2 = 40 ✓
• Cumulative frequency before median class: 3 + 5 + 9 = 17
• Using median formula: 30 + [(20-17)/12] × 10 = 30 + 2.5 = 32.5 ✓

Key Concepts for Median in Grouped Data

1. Median is the middle value that divides the distribution into two equal parts
2. Median class is identified where cumulative frequency ≥ N/2
3. The median formula helps find the exact median value within the median class
4. Understanding cumulative frequency distribution is crucial for solving such problems

This problem demonstrates the practical application of median calculation in statistics, commonly tested in competitive exams like JEE, NEET, and other standardized tests.