Question

Find mean for the following data by using: (i) Direct Method: (ii) Short-cut Method; (iii) Step Deviation Method.

X	100−200	200−300	300−400	400−500	500−600
f	10	18	12	20	40

Answer 1

Class Interval	Mid values (m)	(f)	fm	d=m−A	fd	$\mathit{d}\mathbf{\text{'}}\mathbf{=}\frac{\mathit{d}}{\mathit{i}}\mathbf{=}\frac{\mathit{d}}{\mathbf{100}}$	fd'
100−200 200 −300 300 −400 400 −500 500 −600	150 250 $\boxed{350=A}$ 450 550	10 18 12 20 40	1500 4500 4200 9000 22000	−200 −100 0 100 200	−2000 −1800 0 2000 8000	−2 −1 0 1 2	−20 −18 0 20 80
		$\mathit{\Sigma }\mathit{f}$=100	$\mathit{\Sigma }\mathit{f}\mathit{m}$=41200		$\mathit{\Sigma }\mathit{f}\mathit{d}$=6200		$\mathit{\Sigma }\mathit{f}\mathit{d}\mathbf{\text{'}}$=62

(i) Calculating mean using direct method:

$$\begin{gathered} \overline{X}=\frac{\Sigma fm}{\mathrm{\Sigma }f} \\ here, \\ m=\frac{l_1+l_2}{2} \\ \overline{X}=\frac{41200}{100} \\ \Rightarrow \overline{X}=412 \end{gathered}$$

(ii) Calculating mean using short cut method:

$\overline{X}=A+\frac{\Sigma fd}{\Sigma f}$

Here,

A represents assumed mean.
d represents the deviation of the values from the assumed mean.

Here, we take 350 as the assumed mean. So, we take deviations of each item in the series from 350.

$$\begin{gathered} \overline{\mathit{X}}=350+\frac{6200}{100} \\ or,\overline{X}=350+62 \\ \Rightarrow \overline{X}=412 \end{gathered}$$

(iii) Calculating mean height step-deviation method:

$\overline{X}=A+\frac{\Sigma fd\text{'}}{\Sigma f}\times i$

$$\begin{gathered} \overline{\mathit{X}}=\mathrm{A}+\frac{\mathrm{\Sigma }fd\text{'}}{\mathrm{\Sigma }f}\times i \\ or,\overline{\mathit{X}}=350+\frac{62}{100}\times 100 \\ \Rightarrow \overline{X}=412 \end{gathered}$$

Hence, the mean of the above series is 412.