\(\frac {N + 1}{2} = \frac {107 + 1}{2}\) th = 54th value i.e the median age is in the interval "7 - 8" (29 + 25 = 54) Median = \(1_m + (\frac {^{\sum f} /_2 - f_b}{f_m})\)c \(l_m\)=lower class boundary of the median age = 6.5 \(\sum f\) = 107 \(f_b\) = sum of all frequencies before the median age = 29 \(f_m\)=frequency of the median age = 40 c = class width = 8.5 - 6.5 = 2 Median = 6.5 + \((\frac {^{107}/_2 - 29}{40})\) x 2 = 6.5 + \((\frac {53.5 - 29}{40})\) x 2 = 6.5 + \((\frac {24.5}{40})\) x 2 = 6.5 + 1.225 \(\therefore\) median age = 7.725