1. **Problem Statement:** Find the median of grouped data. 2. **Formula for Median of Grouped Data:** $$\text{Median} = L + \left(\frac{\frac{n}{2} - F}{f}\right) \times h$$ where: - $L$ = lower boundary of the median class - $n$ = total number of observations - $F$ = cumulative frequency before the median class - $f$ = frequency of the median class - $h$ = class width 3. **Explanation:** - First, find the total number of observations $n$ by adding all frequencies. - Then, calculate $\frac{n}{2}$ to locate the median position. - Identify the median class as the class where the cumulative frequency just exceeds $\frac{n}{2}$. - Use the values of $L$, $F$, $f$, and $h$ from the median class to apply the formula. 4. **Example:** Suppose the grouped data is: | Class Interval | Frequency | |---------------|-----------| | 10 - 20 | 5 | | 20 - 30 | 8 | | 30 - 40 | 12 | | 40 - 50 | 5 | - Total frequency $n = 5 + 8 + 12 + 5 = 30$ - $\frac{n}{2} = 15$ - Cumulative frequencies: 5, 13, 25, 30 - Median class is 30 - 40 (since cumulative frequency 25 just exceeds 15) - $L = 30$, $F = 13$, $f = 12$, $h = 10$ 5. **Calculate Median:** $$\text{Median} = 30 + \left(\frac{15 - 13}{12}\right) \times 10 = 30 + \left(\frac{2}{12}\right) \times 10 = 30 + \frac{20}{12} = 30 + 1.67 = 31.67$$ 6. **Answer:** The median of the grouped data is approximately $31.67$.