Question

Find the missing frequency in the following distribution if N is 60 and the median is 40.

$\begin{array}{cccccc}\text{Marks)}& 0-10& 10-30& 30-60& 60-80& 80-90\\ \text{Frequency}& 5& F1& F2& 8& 2\end{array}$

Answer 1

$\begin{array}{ccc}\text{Marks}& \text{Frequency}& \text{Cumulative frequency}\\ 0-10& 5& 5\\ 10-30& F_1& 5+F_1\\ 30-60& F_2& 5+F_1+F_2\\ 60-80& 8& 13+F_1+F_2\\ 80-90& 2& 15+F_1+F_2\\ & N=60& \end{array}$

$\because N=60$

$\because 15+F_1+F_2=60$

$F_1+F_2=\mathrm{45........}\left(i\right)$

$\text{Median Size}=\frac{N}{2}=\frac{60}{2}=30\text{th item.}$

$\because M=40\Rightarrow \text{Median Class is}30-60.$

$h=$ Class size $=30$

$l=$ lower limit of median class $=30$

$C.F=$ cumulative frequency of precceeding median class $=5+F_1$

$f=$ frequency of median class $=F_2$

Median, $M=l+\frac{\frac{N}{2}-C.f}{f}\times h$

$\Rightarrow 40=30+\frac{\frac{60}{2}-(5+F_1)}{F_2}\times 30$

$\Rightarrow 10=\frac{(25-F_1)}{F_2}\times 30$

$\Rightarrow 10F_2=750-30F_1$

$\Rightarrow 30F_1+10F_2=750$

$\Rightarrow 3F_1+F_2=\mathrm{75.......}\left(ii\right)$

$\text{Solving eq (i) and eq (ii)}$, we get

$\Rightarrow 2F_1=30\Rightarrow F_1=15\therefore F_2=30$