Class IntervalFrequencyClass boundariesClass Mid-point1.5−1.921.45−1.951.72.0−2.4211.95−2.452.22.5−2.942.45−2.952.73.0−2.9152.95−3.453.23.5−3.9103.45−3.953....

$\begin{array}{cccc}\text{Class Interval}& Frequency& \text{Class boundaries}& \text{Class Mid-point}\\ 1.5-1.9& 2& 1.45-1.95& 1.7\\ 2.0-2.4& 21& 1.95-2.45& 2.2\\ 2.5-2.9& 4& 2.45-2.95& 2.7\\ 3.0-2.9& 15& 2.95-3.45& 3.2\\ 3.5-3.9& 10& 3.45-3.95& 3.7\\ 4.0-4.4& 5& 3.95-4.45& 4.2\\ 4.5-4.9& 3& 4.45-4.95& 4.7\end{array}$

The median of the distribution above is

Answer Details

Median = $\frac{a+b}{fm}$ ($\frac{1}{2}\sum f-C{F}_b$)
= 2.95 + $\frac{0.5}{15}$(2.-7)
= 2.95 + $\frac{0.5}{15}$ x 13
= 2.95 + 0. 43
= 3.38
= 3.4