Question

Calculate the mean deviation about the median for the following data.

Class	16- 20	21- 25	26- 30	31- 35	36- 40	41- 45	46- 50	51- 55
Frequency	5	6	12	14	26	12	16	9

• A. $9.35$
• B. $7.35$
• C. $26$
• D. $14.35$

Answer 1

Answer: (B)

Converting the given series into an exclusive series, we prepare the table given below.

Class	Frequency $f_i$	$C$	Midvalue $(x_i)$
15.5-20.5	5	5	18
20.5-25.5	6	11	23
25.5-30.5	12	23	28
30.5-35.5	14	37	33
35.5-40.5	26	63	38
40.5-45.5	12	75	43
45.5-50.5	16	91	48
50.5-55.5	9	100	53
	N = 100

Thus $N=100$ and therefore, $\frac{N}{2}=50$
$\Rightarrow$ median class is $35.5-40.5$
$\Rightarrow l=35.5$, $f=26$, $h=5$ and $C=37$
$\therefore$ Median $=l+\frac{\left(\frac{N}{2}-C\right)}{f}\times h$
$=\left\{35.5+\frac{\left(50-37\right)}{26}\times 5\right\}$
$\left(35.5+2.5\right)=38$.
Thus, $M=38$.
Now, we prepare the table given below.

$f_i$	$x_i$	$|x_i-M|$	$f_i|x_i-M|$
5	18	20	100
6	23	15	90
12	28	10	120
14	33	5	70
26	38	0	0
12	43	5	60
16	48	10	160
9	53	15	135
N=100			735

Thus, $\sum f_i\left|x_i-M\right|=735$ and $N=100$
$\therefore M.D.\left(M\right)=\frac{\sum f_i|x_i-M|}{N}$
$=\frac{735}{100}=7.35$.
Hence, the mean deviation about the median is $7.35$.