Question

Find the Arithmetic mean for each of the following frequency distribution by:

(iii) Step deviation method:

Class interval	Frequency
$110-120$ $120-130$ $130-140$ $140-150$ $150-160$ $160-170$ $170-180$ $180-190$ $190-200$ $200-210$ $210-220$	$2$ $5$ $11$ $21$ $26$ $34$ $36$ $28$ $16$ $7$ $4$

Answer 1

Step 1: Make the required frequency distribution table:

At, first, we have to prepare the table values for $x_i$ and $f_i$ for finding the value of $x_i=\sum f_i{d}_i$.

Class interval	$x_i$	$f_i$	$d_i=x_i-165$	$f_i{d}_i$
$110-120$ $120-130$ $130-140$ $140-150$ $150-160$ $160-170$ $170-180$ $180-190$ $190-200$ $200-210$ $210-220$	$115$ $125$ $135$ $145$ $155$ $165$ $175$ $185$ $195$ $205$ $215$	$2$ $5$ $11$ $21$ $26$ $34$ $36$ $28$ $16$ $7$ $4$	$-50$ $-40$ $-30$ $-20$ $-10$ $0$ $10$ $20$ $30$ $40$ $50$	$-100$ $-200$ $-330$ $-420$ $-260$ $0$ $360$ $560$ $480$ $280$ $200$
		$\sum _{i=1}^{11}{f}_i=N=190$		$\sum _{i=1}^{11}{f}_i{d}_i=570$

Step 2: Calculate the mean:

From the above table, we get,
$\sum _{i=1}^{11}{f}_i=N=190$

$\sum _{i=1}^{11}{f}_i{d}_i=570$

Arithmetic mean, $\overline{x}=A+\frac{1}{N}\sum _{i=1}^n{f}_i{d}_i$

$\overline{x}=165+\frac{1}{190}\times 570$

$=168$

Hence, the mean value of the frequency distribution by the step deviation method is$168$.