Question

Find the standard deviation for the following distribution:

x :	4.5	14.5	24.5	34.5	44.5	54.5	64.5
f :	1	5	12	22	17	9	4

Answer 1

x:	4.5	14.5	24.5	34.5	44.5	54.5	64.5
f:	1	5	12	22	17	9	4

Median value of $x$ is 34.5.

$x_i$	$f_i$	$d_i=x_i-34.5$	$u_i=\frac{x_i-34.5}{10}$	$f_i{u}_{i_{}}$	${u_i}^2$	$f_i{u_i}^2$
4.5	1	$-$30	$-$3	$-$3	9	9
14.5	5	$-$20	$-$2	$-$10	4	20
24.5	12	$-$10	$-$1	$-$12	1	12
34.5	22	0	0	0	0	0
44.5	17	10	1	17	1	17
54.5	9	20	2	18	4	36
64.5	4	30	3	12	9	36
	$N=\sum f_i=70$			$\sum f_i{u}_i=22$		$\sum f_i{u_i}^2=130$

$\mathrm{Var}\left(X\right)=h^2\left[\frac{1}{N}\sum _{i=1}^n{f}_i{u_i}^2-{\left(\frac{1}{N}\sum _{i=1}^n{f}_i{u}_i\right)}^2\right]$

We have
$N=70,\sum _{i=1}^n{f}_i{u}_i=22,\sum _{i=1}^n{f}_i{u_i}^2=130,h=10$

Plugging all the values in the formula of variance:
$$\begin{gathered} \mathrm{Var}\left(X\right)={10}^2\left[\frac{1}{70}\times 130-{\left(\frac{1}{70}\times 22\right)}^2\right] \\ =100\left[\frac{130}{70}-{\left(\frac{22}{70}\right)}^2\right] \\ =100\left[\frac{13}{7}-\frac{121}{1225}\right] \\ =100\left[1.857-0.0987\right] \\ =100\left[1.7583\right] \\ =175.83 \end{gathered}$$

Standard deviation, $\mathrm{SD}=\sqrt{\mathrm{Var}\left(X\right)}$
$$\begin{gathered} \mathrm{SD}=\sqrt{\mathrm{Var}\left(X\right)} \\ =\sqrt{175.83} \\ =13.26 \end{gathered}$$