Table 1.6.5. Interval Frequency Distribution \([a_k,b_k]\) \(f_k\) [0,5) 5 [5,10) 7 [10,20) 4 [20,23) 3 [23,30) 6 The total cummulative frequency is 25 and so \(m = \frac{25}{2} = 12.5\) which lies in the k = 3 interval [10,20) and \(F_2 = 12\text{.}\) Therefore \begin{equation*} \text{median} = (20-10) \frac{12.5-12}{4} + 10 = 11.25 \end{equation*}