To find the median from the given frequency distribution, we can follow these steps:

Step 1: Calculate Total Frequency (N)

The total frequency $N$ is the sum of all the frequencies:

$N = 5 + 7 + 9 + 8 + 6 = 35$

Step 2: Find the Median Class

The median class is the class where the cumulative frequency reaches or exceeds $\frac{N}{2}$.

$\frac{N}{2} = \frac{35}{2} = 17.5$

We now locate where this 17.5 falls in the cumulative frequency column:

• 14 (cumulative frequency for class 31–35)
• 23 (cumulative frequency for class 36–40)

Thus, the median class is 36–40, as 17.5 falls between 14 and 23.

Step 3: Apply the Median Formula

The formula to calculate the median is:

$\text{Median} = L + \left( \frac{\frac{N}{2} - F}{f_m} \right) \times h$

Where:

• $L$ = lower boundary of the median class = 35.5
• $N/2$ = 17.5
• $F$ = cumulative frequency of the class before the median class = 14
• $f_m$ = frequency of the median class = 9
• $h$ = class width = $40.5 - 35.5 = 5$

Step 4: Calculate the Median

$\text{Median} = 35.5 + \left( \frac{17.5 - 14}{9} \right) \times 5$ $\text{Median} = 35.5 + \left( \frac{3.5}{9} \right) \times 5$ $\text{Median} = 35.5 + 0.3889 \times 5$ $\text{Median} = 35.5 + 1.9445 \approx 37.44$

Final Answer:

The median score is approximately 37.44.

Would you like more details or have any questions about this solution? Here are some related questions you might consider:

1. How do you determine the class width in a grouped frequency table?
2. What is the significance of the cumulative frequency in finding the median?
3. How would the calculation change if the total frequency $N$ was an even number?
4. What are some alternative methods to find the median for different types of data sets?
5. Can the median be calculated directly from the "less than" cumulative frequency column?

Tip: When solving for the median in grouped data, it's essential to identify the correct median class, as it determines the rest of your calculation!