Question

Calculate the median for the following data:

$\begin{array}{ccccccc}\text{Age(in years)}& \text{19-25}& 26-32& 33-39& 40-46& 47-53& 54-60\\ \text{Frequency}& 35& 96& 68& 102& 35& 4\end{array}$

Answer 1

The given series is of inclusive form. Converting it into exclusive form and preparing the cumulative frequency table, we get

$\begin{array}{ccc}\text{Class}& \text{Frequency}\left(f_i\right)& C.F\\ 18.5-25.5& 35& 35\\ 25.5-32.5& 96& 131\left[cf\right]\\ \left[L\right]32.5-39.5& 68\left[F\right]& 199\\ 39.5-46.5& 102& 301\\ 46.5-53.5& 35& 306\\ 53.5-60.5& 4& 340\\ & N=\sum f_i=340& \end{array}$

From the table, $N=340$

$\frac{N}{2}=\frac{340}{2}=170$

Now,

The cumulative frequency just greater than $170$ is $199$ and the corresponding class is $32.5-39.5$.

Thus, the median class is $32.5-39.5$.

$L=32.5$

$h=7$

$F=68$

$c.f$. = cf preceding class is $131$

Median $=L+\left(\frac{\frac{N}{2}-F}{cf}\right)\times h$

$=32.5+\left(\frac{170-68}{131}\right)\times 7$

$=32.5+\left(\frac{102}{131}\right)\times 7$

$=32.5+\frac{714}{131}$

$=32.5+4$

$=36.5$

Hence median is $36.5$ years.