Question

If the arithmetic mean of the following frequency distribution is 57.6 and the sum of the observations is 50, then the missing frequencies x and y are

Classes	0 - 20	20 - 40	40 - 60	60 - 80	80 - 100	100 - 120
Frequency	7	x	12	7	y	5

• A. $x=10,y=8$
• B. $x=8,y=10$
• C. $x=9,y=10$
• D. $x=10,y=9$

Answer 1

The correct option is C $x=9,y=10$

Classes	Frequency $f_i$	Mid-points $x_i$	$u_i$	$f_i{u}_i$
0 - 20	7	10	-2	-14
20 - 40	x	30	-1	-x
40 - 60	12	50	0	0
60 - 80	7	70	1	7
80 - 100	y	90	2	2y
100 - 120	5	110	3	15
Total	31 + x + y			8 - x + 2y

Let the assumed mean, A = 50
No. of all observations = 50
⇒ 31 + x + y = 50
∴ x + y = 19 …(1)

MEAN $=57.6\Rightarrow A+\frac{{\sum }_{i=1}^6{f}_i{u}_i}{{\sum }_{i=1}^6{f}_i}\times h=57.6$

$\Rightarrow 50+\frac{(8-x+2y)\times 20}{50}=57.6$

$\Rightarrow \frac{(8-x+2y)\times 2}{5}=7.6$

$\Rightarrow 8-x+2y=19$

$\Rightarrow -x+2y=11$ ...(2)

Adding (1) and (2), we get:

$x+y=19$
$-x+2y=11$
$3y=30\Rightarrow y=10$

Substituting y = 10 in (1), we get:

$x+y=19$

$\Rightarrow x+10=19$

$\therefore x=9$