Question

Calculate mean and standard deviation of the following data by short-cut method:

Class Interval	10−20	20−30	30−40	40−50	50−60	60−70	70−80
Frequency	5	10	15	20	25	18	7

Answer 1

C.I.	Mid Value (m)	Frequency (f)	Deviation from Assumed mean dx = m − A (A = 45)	fdx	$f\times {d_x}^2=f{d_x}^2$
10 − 20 20 − 30 30 − 40 40 − 50 50 − 60 60 − 70 70 − 80	15 25 35 $\boxed{45=\mathrm{A}}$ 55 65 75	5 10 15 20 25 18 7	−30 −20 −10 0 10 20 30	−150 −200 −150 0 250 360 210	4500 4000 1500 0 2500 7200 6300
		Σf = N = 100		Σfdx = 320	Σfdx2 = 26000

$$\begin{gathered} \mathrm{Calculating}\mathrm{Mean}\mathrm{and}\mathrm{standard}\mathrm{deviation}\mathrm{through}\mathrm{Short}-\mathrm{cut}\mathrm{method}: \\ \mathrm{Mean}\left(\overline{\mathrm{X}}\right)=\mathrm{A}+\frac{\mathrm{\Sigma }f{d}_x}{\mathrm{\Sigma }f} \\ \mathrm{or},\overline{\mathrm{X}}=45+\frac{320}{100} \\ \mathrm{or},\overline{\mathrm{X}}=45+3.2 \\ \Rightarrow \overline{\mathrm{X}}=48.2 \\ S\tan darddeviation\left(\mathrm{\sigma }\right)=\sqrt{\frac{\mathrm{\Sigma }f{d_x}^2}{N}-{\left(\frac{\mathrm{\Sigma }f{d}_x}{N}\right)}^2} \\ \mathrm{or},\mathrm{\sigma }=\sqrt{\frac{26000}{100}-{\left(\frac{320}{100}\right)}^2} \\ \mathrm{or},\mathrm{\sigma }=\sqrt{260-{\left(3.2\right)}^2} \\ \mathrm{or},\mathrm{\sigma }=\sqrt{260-10.24} \\ \mathrm{or},\mathrm{\sigma }=\sqrt{249.76} \\ \Rightarrow \mathrm{\sigma }=15.80 \end{gathered}$$