To calculate the median from the given frequency distribution table, follow these steps:

Data:

Steps:

1. Find the total frequency ($N$): Add all the frequencies: $N = 7 + 13 + 15 + 20 + 30 + 33 + 28 + 14 = 160$

2. Locate the median class: The median corresponds to the class that contains the $\frac{N}{2}$-th observation. Here: $\frac{N}{2} = \frac{160}{2} = 80$ Now, we calculate the cumulative frequency (CF) to find the class that contains the 80th observation.

The cumulative frequency just greater than 80 is 85, which corresponds to the class 35–40. Thus, the median class is 35–40.

3. Apply the Median Formula: The formula for the median is: $\text{Median} = L + \left( \frac{\frac{N}{2} - CF}{f_m} \right) \times h$ Where:

• $L = 35$ (lower boundary of the median class),
• $N = 160$ (total frequency),
• $CF = 55$ (cumulative frequency before the median class),
• $f_m = 30$ (frequency of the median class),
• $h = 5$ (class width).

Substitute the values: $\text{Median} = 35 + \left( \frac{80 - 55}{30} \right) \times 5$ Simplify: $\text{Median} = 35 + \left( \frac{25}{30} \right) \times 5$ $\text{Median} = 35 + \left( 0.833 \right) \times 5$ $\text{Median} = 35 + 4.17$ $\text{Median} = 39.17$

Final Answer:

The median is approximately 39.17.

Do you need clarification on any step or additional details? Let me know!

Related Questions:

1. What is the significance of the median in statistics?
2. How is the class width determined in a frequency distribution table?
3. What is the difference between mean, median, and mode?
4. Can you explain cumulative frequency and its importance?
5. How would you calculate the mean from the same data?

Tip:

The median class is identified by the cumulative frequency just greater than $\frac{N}{2}$.