To find the mode from the given data, we use the formula for the mode in grouped data:

$\text{Mode} = L + \left( \frac{f_m - f_1}{2f_m - f_1 - f_2} \right) \times h$

Where:

• $L$ = Lower boundary of the modal class.
• $f_m$ = Frequency of the modal class.
• $f_1$ = Frequency of the class preceding the modal class.
• $f_2$ = Frequency of the class succeeding the modal class.
• $h$ = Class width.

Step-by-step:

1. Identify the modal class: This is the class interval with the highest frequency. Here, the highest frequency is $13$, so the modal class is $31-40$.

2. Find $L$: The lower boundary of the modal class $31-40$ is $31$.

3. Find $f_m$, $f_1$, and $f_2$:

   • $f_m = 13$ (frequency of the modal class).
   • $f_1 = 7$ (frequency of the class preceding the modal class, $21-30$).
   • $f_2 = 8$ (frequency of the class succeeding the modal class, $41-50$).

4. Find $h$: The class width is $10$ (difference between consecutive class limits).

Substitute into the formula:

$\text{Mode} = 31 + \left( \frac{13 - 7}{2(13) - 7 - 8} \right) \times 10$

$\text{Mode} = 31 + \left( \frac{6}{26 - 15} \right) \times 10$

$\text{Mode} = 31 + \left( \frac{6}{11} \right) \times 10$

$\text{Mode} = 31 + 5.45$

$\text{Mode} \approx 36.45$

Final Answer:

The mode of the data is approximately 36.45.

Let me know if you want more clarification or have any questions. Here are some related questions to explore:

1. How is the mode different from the mean in grouped data?
2. What is the impact of overlapping class intervals on mode calculation?
3. How would you calculate the mean and median for the same dataset?
4. What changes in the formula if the modal class is the first or last interval?
5. Why is it necessary to use interpolation for grouped data?

Tip: For grouped data, visualizing the frequency distribution as a histogram can help identify the modal class intuitively.