1. **State the problem:** Find the median of the grouped data using the formula for median in grouped data. 2. **Given data:** - Class intervals and frequencies: - Below 140: 4 - 140 – 145: 7 - 145 – 150: 18 - 150 – 155: 11 - 155 – 160: 6 - 160 – 165: 5 - Cumulative frequencies: - Below 140: 4 - 140 – 145: 11 - 145 – 150: 29 - 150 – 155: 40 - 155 – 160: 46 - 160 – 165: 51 - Total number of observations $n = 51$ 3. **Find the median class:** - Median position is at $\frac{n}{2} = \frac{51}{2} = 25.5$ - The cumulative frequency just greater than or equal to 25.5 is 29, corresponding to class interval 145 – 150. - So, median class is 145 – 150. 4. **Identify values for the formula:** - Lower limit of median class $l = 145$ - Frequency of median class $f = 18$ - Cumulative frequency of class preceding median class $cf = 11$ - Class size $h = 150 - 145 = 5$ 5. **Apply the median formula:** $$\text{Median} = l + \left(\frac{n}{2} - cf\right) \times \frac{h}{f}$$ 6. **Substitute values:** $$\text{Median} = 145 + \left(25.5 - 11\right) \times \frac{5}{18}$$ 7. **Calculate inside the parentheses:** $$25.5 - 11 = 14.5$$ 8. **Calculate the fraction:** $$\frac{5}{18} \approx 0.2778$$ 9. **Multiply:** $$14.5 \times 0.2778 \approx 4.0278$$ 10. **Add to lower limit:** $$145 + 4.0278 = 149.0278$$ **Final answer:** $$\boxed{149.03}$$ (rounded to two decimal places)