Question

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

Assumed mean 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 using assumed mean method.

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

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.

$A=$Assumed mean (Total Mid value of a grouped data of class intervals)

$d_i=$Deviation (Mid value of the class intervals $-$Assumed mean).

Step 2: Make the required frequency distribution table.

First, we have to prepare the table values to $x_i$ find the values for $f_i{d}_i$

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

From the above table we get the values of,

$\sum _{i=1}^{11}{f}_i=N=190$ ,

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

$A=165$.

Substitute the obtained values in the Arithmetic mean formula,

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

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

$=165+3$

$=168$

Final answer:

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