Question

What is standard deviation of the following series

Measurement	Frequency
$0-10$	$1$
$10-20$	$3$
$20-30$	$4$
$30-40$	$2$

• A. $81$
• B. $76$
• C. $9$
• D. $2.26$

Answer 1

The correct option is C

$9$

Explanation for the correct option

Step 1: Make a tabular for the given problem

Let the mean of the series be $25$.

The class interval formula is $c=\mathrm{Upper}\mathrm{class}-\mathrm{lower}\mathrm{class}$.

The class interval for the given series is $c=10-0=10$

Measurement	Frequency$\left(f_i\right)$	Midpoint $\left(x_i\right)$	$u_i=\frac{\left(x_i-a\right)}{c}$	$f_i{u}_i$	$f_i{u^2}_i$
$0-10$	$1$	$5$	$\frac{5-25}{10}=-2$	$-2$	$4$
$10-20$	$3$	$15$	$\frac{15-25}{10}=-1$	$-3$	$3$
$20-30$	$4$	$25$	$\frac{25-25}{10}=0$	$0$	$0$
$30-40$	$2$	$35$	$\frac{35-25}{10}=1$	$2$	$2$
Total	$10$			$-3$	$9$

Step 2: Calculate the standard deviation

The formula of standard deviation is $\sigma =\sqrt{c^2\left[\frac{\sum f_i{u_i}^2}{\sum f_i}-{\left(\frac{\sum f_i{u}_i}{\sum f_i}\right)}^2\right]}$

The standard deviation is $\sigma =\sqrt{{10}^2\left[\frac{9}{10}-{\left(\frac{-3}{10}\right)}^2\right]}$

$=\sqrt{100\left[0.9-0.09\right]}$

$$\begin{gathered} =\sqrt{100\times 0.81} \\ =\sqrt{81} \\ =9 \end{gathered}$$

Hence option (C) is the correct option.