The ages of students in a small primary school were recorded in the table below....

Explanation

\(\frac {N + 1}{2} = \frac {107 + 1}{2}\) 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