Question

Consider the following frequency distribution :
$\begin{array}{cccccc}\text{Class:}& 10-20& 20-30& 30-40& 40-50& 50-60\\ \text{Frequency:}& \alpha & 110& 54& 30& \beta \end{array}$

If the sum of all frequencies is $584$ and median is $45,$ then $|\alpha –\beta |$ is equal to

Answer 1

$\begin{array}{ccccc}\text{C.I.}& x_i& f_i& x_i\cdot f_i& \text{C.F.}\\ 10-20& 15& \alpha & 15\alpha & \alpha \\ 20-30& 25& 110& 2750& 110+\alpha \\ 30-40& 35& 54& 1890& 164+\alpha \\ 40-50& 45& 30& 1350& 194+\alpha \text{(median class)}\\ 50-60& 55& \beta & 55\beta & 194+\alpha +\beta \\ & & 194+\alpha +\beta & 5990+15\alpha +55\beta & \end{array}$

$\because \alpha +\beta =584-194$
$\Rightarrow \alpha +\beta =390\dots \left(1\right)$
and median
$40+\left(\frac{\frac{194+\alpha +\beta }{2}-\beta }{30}\right)\times 10=45$
From equation $\left(1\right)$
$\Rightarrow 40+\left(\frac{194+\alpha -292}{30}\right)10=45$
$\Rightarrow \alpha =113\dots \left(2\right)$
So $\beta =277$
$|\alpha -\beta |=164$