Question

Mean life of following distribution in the following distribution is $57.6$ hours. If sum of frequency is $50$ then find the missing frequency $x$ and $y$

$0-20$	$7$
$20-40$	$x$
$40-60$	$12$
$60-80$	$y$
$80-100$	$8$
$100-120$	$5$

• A. $x=7,y=11$
• B. $x=11,y=7$
• C. $x=13,y=5$
• D. $x=8,y=10$

Answer 1

Answer: (D)

$x=8,y=10$

$C.I$	$Frequency$ $f_i$	$Mid-value$ $x_i$	$f_i{x}_i$
$0-20$	$7$	$10$	$70$
$20-40$	$x$	$30$	$30x$
$40-60$	$12$	$50$	$600$
$60-80$	$y$	$70$	$70y$
$80-100$	$8$	$90$	$720$
$100-120$	$5$	$110$	$550$
	$\sum f_i=32+x+y$		$\sum f_i{x}_i=1940+30x+70y$

$\Rightarrow$ $\sum f_i=32+x+y$

$\Rightarrow$ $50=32+x+y$

$\therefore$ $x+y=18$ ------- ( 1 )

$\Rightarrow$ $\overline{x}=\frac{\sum f_i{x}_i}{\sum f_i}$

$\Rightarrow$ $57.6=\frac{1940+30x+70y}{50}$

$\Rightarrow$ $2880=1940+30x+70y$

$\Rightarrow$ $30x+70y=940$

$\therefore$ $3x+7y=94$ ----- ( 2 )

Multiplying equation ( 1 ) by $3$ we get,

$\Rightarrow$ $3x+3y=54$ ------ ( 3 )

Subtracting equaton ( 3 ) from equation ( 2 ) we get,

$4y=40$

$\therefore$ $y=10$

Substituting $y=10$ in equation ( 1 ),

$x+10=18$

$\therefore$ $x=8$

$\therefore$ $x=8$ and $y=10$