To calculate the median for grouped data from the provided table, we need to follow these steps:

1. Identify the total frequency (N): The total frequency is given as $N = 40$.

2. Find $\frac{N}{2}$: To find the median class, we compute $\frac{N}{2} = \frac{40}{2} = 20$.

3. Locate the median class: The median class is the class interval where the cumulative frequency first exceeds 20. From the cumulative frequency column:

   • The cumulative frequency for the class 36-38 is 15.
   • The cumulative frequency for the class 39-41 is 22.

   Since 22 exceeds 20, the median class is $39-41$.

4. Apply the median formula: The formula for the median in grouped data is:

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

   • $L$ = Lower boundary of the median class = 38.5
   • $F$ = Cumulative frequency before the median class = 15
   • $f$ = Frequency of the median class = 7
   • $h$ = Class width = 3 (since the class intervals are 3 units wide)

5. Substitute the values:

   $\text{Median} = 38.5 + \left( \frac{20 - 15}{7} \right) \times 3$

   $\text{Median} = 38.5 + \left( \frac{5}{7} \right) \times 3$

   $\text{Median} = 38.5 + \left( 0.714 \right) \times 3$

   $\text{Median} = 38.5 + 2.142$

   $\text{Median} = 40.642$

So, the median score is approximately 40.64.

Would you like further details on this, or do you have any questions?

Here are some follow-up questions based on the concept of median for grouped dataTo calculate the median for grouped data from the provided table, we need to follow these steps:

1. Identify the total frequency (N): The total frequency is given as $N = 40$.

2. Find $\frac{N}{2}$: To find the median class, we compute $\frac{N}{2} = \frac{40}{2} = 20$.

3. Locate the median class: The median class is the class interval where the cumulative frequency first exceeds 20. From the cumulative frequency column:

   • The cumulative frequency for the class 36-38 is 15.
   • The cumulative frequency for the class 39-41 is 22.

   Since 22 exceeds 20, the median class is $39-41$.

4. Apply the median formula: The formula for the median in grouped data is:

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

   • $L$ = Lower boundary of the median class = 38.5
   • $F$ = Cumulative frequency before the median class = 15
   • $f$ = Frequency of the median class = 7
   • $h$ = Class width = 3 (since the class intervals are 3 units wide)

5. Substitute the values:

   $\text{Median} = 38.5 + \left( \frac{20 - 15}{7} \right) \times 3$

   $\text{Median} = 38.5 + \left( \frac{5}{7} \right) \times 3$

   $\text{Median} = 38.5 + \left( 0.714 \right) \times 3$

   $\text{Median} = 38.5 + 2.142$

   $\text{Median} = 40.642$

So, the median score is approximately 40.64.

Would you like further details on this, or do you have any questions?

Here are some follow-up questions based on the concept of median for grouped

1. How is the class width determined in a grouped frequency table?
2. Why do we use the lower boundary of the median class in the formula?
3. Can the median be calculated for ungrouped data differently?
4. How would the calculation change if the class widths were uneven?
5. What is the significance of cumulative frequency in finding the median?

Tip: Always ensure that the cumulative frequency is calculated correctly, as it is crucial for identifying the correct median class.