Question

If the median of the following distribution is 32.5, then find the values of x and y.
Class: 0-10, 10-20, 20-30, 30-40, 40-50, 50-60, 60-70
Frequency: x, 5, 9, 12, y, 3, 2 [Total = 40]

• A. x = 3, y = 6
• B. x = 5, y = 4
• C. x = 4, y = 5
• D. x = 3, y = 5

Hint:

For any median equation with decimals like \(2.5 = (6 - x) \times \frac{10}{12}\):
Simplify the fraction to \(\frac{5}{6}\) first:
\[ 2.5 = (6-x) \times \frac{5}{6} \implies 2.5 \times \frac{6}{5} = 6 - x \implies 3 = 6 - x \implies x = 3 \]
This algebraic cleaning reduces calculation time!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given a grouped frequency distribution table with a total frequency of 40. Two of the frequencies are unknown: \(x\) in interval 0--10 and \(y\) in interval 40--50. The median of the distribution is given as 32.5. We need to find the values of \(x\) and \(y\).

Step 2: Key Formula or Approach:
1. The total frequency is \(\sum f_i = 40\). This gives our first linear equation:
\[ x + 5 + 9 + 12 + y + 3 + 2 = 40 \]
2. Find the cumulative frequency (\(cf\)) for each class interval.
3. Use the median formula:
\[ \text{Median} = l + \left( \frac{\frac{N}{2} - cf}{f} \right) \times h \]
where \(l\) is the lower limit of the median class, \(N = 40\), \(cf\) is the cumulative frequency of the preceding class, \(f\) is the frequency of the median class, and \(h\) is the class width.

Step 3: Detailed Explanation:
1. Formulate the first equation using total frequency:
\[ x + 5 + 9 + 12 + y + 3 + 2 = 40 \]
\[ x + y + 31 = 40 \]
\[ x + y = 9 \quad \text{--- (Equation 1)} \]
2. Construct the cumulative frequency (\(cf\)) table:
- 0--10: \(cf = x\)
- 10--20: \(cf = x + 5\)
- 20--30: \(cf = x + 14\)
- 30--40: \(cf = x + 26\)
- 40--50: \(cf = x + y + 26\)
- 50--60: \(cf = x + y + 29\)
- 60--70: \(cf = x + y + 31\)
3. Determine the median class:
Since the median is given as 32.5, it lies in the class interval 30--40.
Therefore, the median class is 30--40.
- Identify the parameters:
Lower limit of median class, \(l = 30\)
Frequency of median class, \(f = 12\)
Cumulative frequency of preceding class, \(cf = x + 14\)
Class width, \(h = 10\)
Total frequency, \(N = 40 \implies \frac{N}{2} = 20\)
4. Substitute these parameters into the median formula:
\[ 32.5 = 30 + \left( \frac{20 - (x + 14)}{12} \right) \times 10 \]
Subtract 30 from both sides:
\[ 2.5 = \left( \frac{6 - x}{12} \right) \times 10 \]
Multiply both sides by 12:
\[ 30 = 10(6 - x) \]
Divide by 10:
\[ 3 = 6 - x \implies x = 3 \]
5. Substitute \(x = 3\) into Equation 1 to find \(y\):
\[ 3 + y = 9 \implies y = 6 \]
So, the values are \(x = 3\) and \(y = 6\).

Step 4: Final Answer:
The values of the unknown frequencies are \(x = 3\) and \(y = 6\), which corresponds to option (A).