Question

An incomplete frequency distribution is given as below:

Variable	10-20	20-30	30-40	40-50	50-60	60-70	70-80
Frequency	12	30	?	65	?	25	18

Given that the total frequency is 229 and median is 46, find the missing frequencies:

• A. 34, 45
• B. 33, 34
• C. 32, 33
• D. 31, 32

Hint:

To save time on equations:
Look at the options and find the pair whose sum equals 79 (since \(229 - 150 = 79\)):
- Option 1: \(34 + 45 = 79\) (Correct).
- Option 2: \(33 + 34 = 67\).
- Option 3: \(32 + 33 = 65\).
- Option 4: \(31 + 32 = 63\).
This simple check immediately isolates Option 1 without doing any median calculations!

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a classic statistical problem involving a grouped frequency distribution with missing frequencies.
We use the total frequency (\(N\)) and the median formula for grouped data to set up a system of linear equations to solve for the two unknown frequencies.
Note: The class interval "70-80" with frequency "18" is included in standard reference problems to yield a total sum of 229.
Key Formula or Approach:
The formula for the median of a grouped frequency distribution is:
\[ \text{Median} = L + \left( \frac{\frac{N}{2} - CF}{f} \right) \times h \]
Where:
- \(L\) is the lower limit of the median class.
- \(N\) is the total frequency.
- \(CF\) is the cumulative frequency of the class preceding the median class.
- \(f\) is the frequency of the median class.
- \(h\) is the width of the class interval.

Step 2: Detailed Explanation:
Let the missing frequencies for the classes 30-40 and 50-60 be \(f_1\) and \(f_2\) respectively.
We are given:
- Total frequency, \(N = 229\)
- Median = 46
Summing all the frequencies:
\[ 12 + 30 + f_1 + 65 + f_2 + 25 + 18 = 229 \]
\[ 150 + f_1 + f_2 = 229 \implies f_1 + f_2 = 79 \quad \text{--- (Equation 1)} \]
Since the median is 46, the median class is 40-50.
For this class:
- Lower limit, \(L = 40\)
- Class width, \(h = 10\)
- Class frequency, \(f = 65\)
- Cumulative frequency of preceding class, \(CF = 12 + 30 + f_1 = 42 + f_1\)
- \(\frac{N}{2} = \frac{229}{2} = 114.5\)
Now, substitute these values into the median formula:
\[ 46 = 40 + \left( \frac{114.5 - (42 + f_1)}{65} \right) \times 10 \]
\[ 6 = \left( \frac{72.5 - f_1}{65} \right) \times 10 \]
Multiply both sides by 65:
\[ 390 = (72.5 - f_1) \times 10 \]
Divide by 10:
\[ 39 = 72.5 - f_1 \implies f_1 = 72.5 - 39 = 33.5 \approx 34 \]
Substitute \(f_1 = 34\) into Equation 1:
\[ 34 + f_2 = 79 \implies f_2 = 79 - 34 = 45 \]
Therefore, the missing frequencies are 34 and 45.

Step 3: Final Answer:
The correct option is (A).