Question

Find the mean of the following frequency distribution.

Class-interval	$15-25$	$25-35$	$35-45$	$45-55$	$55-65$
Frequency	$60$	$35$	$22$	$18$	$15$

Answer 1

Step 1: Finding the value of ${\mathit{u}}_{\mathbf{i}}$

For determining the mean, we will use the step deviation method.

Let the assumed mean be $a=40$.

Here class size is $h=10$.

$u_{\mathrm{i}}$ can be found by substituting the corresponding values in the formula $u_i=\frac{x_i-a}{h}$.

Marks	Mid value $x_i$	Number of students $f_i$	$$\begin{gathered} u_i=\frac{x_i-a}{h} \\ {u}_i=\frac{x_i-40}{10} \end{gathered}$$	$f_i{u}_i$
$15-25$	$20$	$60$	$-2$	$-120$
$25-35$	$30$	$35$	$-1$	$-35$
$35-45$	$40$	$22$	$0$	$0$
$45-55$	$50$	$18$	$1$	$18$
$55-65$	$60$	$15$	$2$	$30$
		$\sum _{}{f}_i=150$		$\sum _{}{f}_i{u}_i=-107$

Step 2: Finding the mean

The formula used to determine the mean is $\overline{x}=a+h\left(\frac{\sum _{}{f}_i{u}_i}{\sum _{}{f}_i}\right)$.

where

$a$ is the assumed mean,

$h$ is the class size,

$\sum _{}{f}_i$ is the total frequency, and

$u_i=\frac{x_i-a}{h}$

Substituting the values in the mean formula we get,

$$\begin{gathered} \Rightarrow \overline{x}=40+10\left(-\frac{107}{150}\right) \\ \Rightarrow \overline{x}=40-7.13 \\ \therefore \overline{x}=32.87 \end{gathered}$$

Hence the mean of the frequency distribution is $32.87$.