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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*}
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