To find the median of the given frequency distribution, follow these steps:

1. Determine the total number of observations (N): $N = 6 + 10 + 14 + 12 + 8 + 4 = 54$

2. Find the median class: The median is the value that separates the higher half from the lower half. Since $N = 54$, the median will be at the $\frac{N+1}{2}$-th position, which is the 27.5-th position. We need to find the cumulative frequency that first exceeds 27.5.

From the cumulative frequency table:

• The cumulative frequency of the first three classes is:
  • $6$ (for 42-49)
  • $16$ (for 50-57)
  • $30$ (for 58-65)

The 27.5-th position lies in the 58-65 class.

3. Identify the values for the median formula:

   • Lower class boundary (L) of the median class: 57.5
   • Cumulative frequency (F) of the class before the median class: 16
   • Frequency (f) of the median class: 14
   • Class width (C) of the median class interval: 8

4. Apply the median formula: $\text{Median} = L + \left(\frac{\frac{N}{2} - F}{f}\right) \times C$ $\text{Median} = 57.5 + \left(\frac{27 - 16}{14}\right) \times 8$ $\text{Median} = 57.5 + \left(\frac{11}{14}\right) \times 8$ $\text{Median} = 57.5 + 0.7857 \times 8$ $\text{Median} = 57.5 + 6.2856$ $\text{Median} \approx 63.79$

There seems to be a calculation error in the problem options provided. The calculated median of approximately 63.79 does not match any of the options listed.

Would you like further details or have any questions?

Here are some related questions to deepen your understanding:

1. What is the mean of the data set?
2. How do you calculate the mode of a grouped frequency distribution?
3. What is the range of the data set?
4. How do you interpret the cumulative frequency graph?
5. How would you find the standard deviation of this data?

Tip: When dealing with grouped data, always double-check the class intervals and ensure that the class boundaries are correctly identified.