Question

Calculate standard deviation of the following data, using step-deviation method.

Age	20−30	30−40	40−50	50−60	60−70	70−80	80−90
Frequency	3	61	132	153	140	51	2

Answer 1

Age	Mid value m	Frequency (f)	Deviation assumed from average (dx = m − A)	${\mathit{d}}_{\mathbf{x}}\mathbf{\text{'}}\mathbf{=}\frac{{\mathit{d}}_{\mathbf{x}}}{\mathit{c}}\mathbf{=}\frac{{\mathbf{d}}_{\mathbf{x}}}{\mathbf{10}}$	$\mathit{f}{\mathit{d}}_{\mathit{x}}\mathbf{\text{'}}\mathbf{=}\mathit{f}\mathbf{\times }{\mathit{d}}_{\mathit{x}}\mathbf{\text{'}}$	$\mathit{f}{\mathit{d}}_{\mathit{x}}{\mathbf{\text{'}}}^{\mathbf{2}}\mathbf{=}\mathit{f}\mathbf{\times }{\mathit{d}}_{\mathit{x}}{\mathbf{\text{'}}}^{\mathbf{2}}$
20−30 30 −40 40 −50 50 −60 60 −70 70 −80 80 −90	25 35 45 $\boxed{55=A}$ 65 75 85	3 61 132 153 140 51 2	−30 −20 −10 0 10 20 30	−3 −2 −1 0 1 2 3	−9 −122 −132 0 140 102 6	27 244 132 0 140 204 18
		N = 542			Σfdx' = -15	Σfdx'2 = 765

$$\begin{gathered} \mathrm{Standard}\mathrm{deviation}\left(\mathrm{\sigma }\right)=\sqrt{\frac{\mathrm{\Sigma }f{d}_x{\text{'}}^2}{N}-{\left(\frac{\mathrm{\Sigma }f{d}_x\text{'}}{N}\right)}^2}\times c \\ \mathrm{or},\mathrm{\sigma }=\sqrt{\frac{765}{542}-{\left(\frac{-15}{542}\right)}^2}\times 10 \\ \mathrm{or},\mathrm{\sigma }=\sqrt{1.41-{\left(-0.03\right)}^2}\times 10 \\ \mathrm{or},\mathrm{\sigma }=\sqrt{1.41-0.0009}\times 10 \\ \mathrm{or},\mathrm{\sigma }=1.19\times 10 \\ \Rightarrow \sigma =11.9 \end{gathered}$$

Hence, standard deviation of the above data is 11.9