The ages of students in a small primary school were recorded in the table below.
| Age | 5-6 | 7-8 | 9-10 |
| Frequency | 29 | 40 | 38 |
Estimate the median.
7.725
6.225
7.5
6.5
Correct answer is A
\(\frac {N + 1}{2} = \frac {107 + 1}{2}\) 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