Question

If the mean of the following frequency distribution is $188,$ find the missing frequencies x and y, if the sum of all frequencies is $100.$

Class	$0–80$	$80–160$	$160–240$	$240–320$	$320–400$
Frequency	$20$	$25$	$x$	$y$	$10$

Answer 1

Here, class size, $h=80$

Let the assumed mean, $A$ be $200.$

Class	Mid-Values(xi)	Frequency(fi)	$d_i=x_i-A=x_i-200$	$u_i=\frac{x_i-200}{80}$	$f_i{u}_i$
$0-80$	$40$	$20$	$-160$	$-2$	$-40$
$80-160$	$120$	$25$	$-80$	$-1$	$-25$
$160-240$	$200$	$x$	$0$	$0$	$0$
$240-320$	$280$	$y$	$80$	$1$	$y$
$320-400$	$360$	$10$	$160$	$2$	$20$
		$\sum f_i=x+y+55$			$\sum f_i{u}_i=-45+y$

$\sum f_i=x+y+55=100$ (Given)

$\Rightarrow x+y=45\dots \dots \left(i\right)$

We know

Mean $=A+h\times \frac{\sum f_i{u}_i}{\sum f_i}$

$\therefore 200+80\times \frac{-45+y}{100}=188$ (Given)

$\Rightarrow \frac{-45+y}{100}=\frac{188-200}{80}=-0.15$

$\Rightarrow -45+y=-15$

$\Rightarrow y=45-15=30$

Substituting the value of $y$in $eq.\left(i\right),$ we get

$x+30=45$

$\Rightarrow x=45-30=15$

Thus, the missing frequencies $x$ and $y$ are $15$ and $30$ respectively.