Question

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

Direct 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: Use the formula of mean.

Arithmetic Mean, $\overline{x}=\frac{1}{N}\underset{i=1}{\overset{n}{\sum f_i}}{x}_i$

Where,

$n=$Number of class intervals.

$N=$Sum of the frequencies.

$x_i=$Mid value of the class intervals.

$f_i=$ Values of the frequency.

Step 2 : Make required frequency distribution table.

At first, we have to prepare the table values for $x_i$ and $f_i$ for finding the values for ${\displaystyle \sum _{}}{f}_i{x}_i$

Class interval	$x_i$	$f_i$	$f_i{x}_i$
$110-120$ $120-130$ $130-140$ $140-150$ $150-160$ $160-170$ $170-180$ $180-190$ $190-200$ $200-210$ $210-220$	$115$ $125$ $135$ $145$ $155$ $165$ $175$ $185$ $195$ $205$ $215$	$2$ $5$ $11$ $21$ $26$ $34$ $36$ $28$ $16$ $7$ $4$	$230$ $625$ $1,485$ $3,045$ $4,030$ $5,610$ $6,300$ $5,180$ $3,120$ $1,435$ $860$
		$\sum _{i=1}^{11}{f}_i=N=190$	$\sum _{i=1}^{11}{f}_i{x}_i=31,920$

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

$\sum _{i=1}^{11}{f}_i{x}_i=31,920$.

By Substituting the obtained values in the Arithmetic mean formula,

We will get, Arithmetic Mean, $\overline{x}=\frac{1}{N}\underset{i=1}{\overset{n}{\sum f_i}}{x}_i$

$=\frac{1}{190}\times 31,920=168$

Final Answer:

Hence, the mean value of the given frequency distribution is $168$.