Understanding the Problem: Median in a Cumulative Frequency Distribution

The question provides a cumulative frequency distribution table showing the number of students who scored less than certain marks. We are given that the median value for this distribution is 24. Our goal is to find the value of the unknown '\(k\)'.

Key Concepts: Median and Cumulative Frequency

• Median: The median is the middle value in a dataset that separates the higher half from the lower half. For grouped data, it's calculated using a specific formula.
• Cumulative Frequency Distribution (Less than type): This type of distribution shows the total number of observations up to a certain value. The table given is in this format.
• Median Value: The median value (24) tells us that half of the students scored less than 24 and the other half scored more than 24. Since 24 falls between 20 and 30, the "median class" is the interval 20-30.

Setting up the Data

Let's organize the data and calculate the frequencies for each class interval. The total number of students (\(N\)) is the final cumulative frequency.

From the table:

• The median value is 24.
• This means the median class is 20-30.
• The lower boundary of the median class (\(L\)) is 20.
• The cumulative frequency of the class preceding the median class (\(CF\)) is 30.
• The frequency of the median class (\(f\)) is \(k\).
• The width of the median class (\(w\)) is \(30 - 20 = 10\).
• The total number of students (\(N\)) is \(55+k\).
• The position of the median is \(N/2 = (55+k)/2\).

Applying the Median Formula

The formula for calculating the median for grouped data is:

\(\text{Median} = L + \left(\frac{N/2 - CF}{f}\right) \times w\)

Step-by-Step Calculation for k

1. Substitute the known values into the median formula: \(24 = 20 + \left(\frac{(55+k)/2 - 30}{k}\right) \times 10\)
2. Isolate the fraction term: Subtract 20 from both sides. \(24 - 20 = \left(\frac{(55+k)/2 - 30}{k}\right) \times 10\) \(4 = \left(\frac{(55+k)/2 - 30}{k}\right) \times 10\)
3. Simplify the equation: Divide both sides by 10. \(\frac{4}{10} = \frac{(55+k)/2 - 30}{k}\) \(0.4 = \frac{(55+k)/2 - 30}{k}\)
4. Multiply both sides by \(k\): \(0.4k = \frac{55+k}{2} - 30\)
5. Simplify the right side by finding a common denominator: \(0.4k = \frac{55+k - (30 \times 2)}{2}\) \(0.4k = \frac{55+k - 60}{2}\) \(0.4k = \frac{k - 5}{2}\)
6. Multiply both sides by 2: \(2 \times 0.4k = k - 5\) \(0.8k = k - 5\)
7. Solve for \(k\): Rearrange the terms to group \(k\) together. \(5 = k - 0.8k\) \(5 = 0.2k\) \(k = \frac{5}{0.2}\) \(k = \frac{5}{1/5}\) \(k = 5 \times 5\) \(k = 25\)

Verification

Let's check if \(k=25\) yields a median of 24.

• If \(k=25\), the total number of students (\(N\)) is \(55+25 = 80\).
• The median position is \(N/2 = 80/2 = 40\).
• The median class is 20-30.
• \(L=20\), \(CF=30\), \(f=k=25\), \(w=10\).
• Median = \(20 + \left(\frac{40 - 30}{25}\right) \times 10\)
• Median = \(20 + \left(\frac{10}{25}\right) \times 10\)
• Median = \(20 + (0.4) \times 10\)
• Median = \(20 + 4 = 24\).

The calculated median matches the given median value. Therefore, the value of \(k\) is 25.