Question

Find the mean of the following data.

Class-interval	$0-10$	$10-20$	$20-30$	$30-40$	$40-50$
Frequency	$12$	$11$	$8$	$10$	$9$

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 $a=25$.

Here class size is $h=10$.

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

The table can be completed as below,

Class-interval	Mid value $x_i$	Number of students $f_i$	$$\begin{gathered} u_i=\frac{x_i-a}{h} \\ {u}_i=\frac{x_i-25}{10} \end{gathered}$$	$f_i{u}_i$
$0-10$	$5$	$12$	$-2$	$-24$
$10-20$	$15$	$11$	$-1$	$-11$
$20-30$	$25$	$8$	$0$	$0$
$30-40$	$35$	$10$	$1$	$10$
$40-50$	$45$	$9$	$2$	$18$
		$\sum _{}{f}_i=50$		$\sum _{}{f}_i{u}_i=-7$

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 value in the mean formula we get,

$$\begin{gathered} \Rightarrow \overline{x}=25+10\left(-\frac{7}{50}\right) \\ \Rightarrow \overline{x}=25-1.4 \\ \therefore \overline{x}=23.6 \end{gathered}$$

Hence, the mean of the given data is $23.6$.