Question

From the following data, calculate standard deviation of the two groups A and B. Which group is more consistent?

Class Interval	Group A	Group B
5−10	2	9
10−15	9	11
15−20	29	18
20−25	54	32
25−30	11	27
30−35	5	13

Answer 1

For group A

Class Interval	Frequency (f)	Mid-Values (m)	fm	m2	fm2
5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35	2 9 29 54 11 5	7.5 12.5 17.5 22.5 27.5 32.5	15 112.5 507.5 1215 302.5 162.5	56.25 156.25 306.25 506.25 756.25 1056.25	112.5 1406.25 8881.25 27337.5 8318.75 5281.25
	Σf = 110		Σfm = 2315		Σfm2 = 51337.5

$$\begin{gathered} {\overline{X}}_A=\frac{\Sigma fm}{\Sigma f}=\frac{2315}{110}=21.05 \\ {\mathrm{\sigma }}_{\mathrm{A}}=\sqrt{\frac{\Sigma f{m}^2}{\Sigma f}-{\left(\overline{X}\right)}^2} \\ =\sqrt{\frac{51337.5}{110}-{\left(21.05\right)}^2} \\ =\sqrt{466.70-443.10} \\ =\sqrt{23.6} \\ =4.86 \\ \mathrm{C}.\mathrm{V}.ofgroupA=\frac{{\mathrm{\sigma }}_{\mathrm{A}}}{{\overline{\mathrm{X}}}_A}\times 100=\frac{4.86}{21.05}\times 100=23.08\% \end{gathered}$$

For group B

Class Interval	Frequency (f)	Mid-Values (m)	fm	m2	fm2
5 − 10 10 − 15 15 − 20 20 − 25 25 − 30 30 − 35	9 11 18 32 27 13	7.5 12.5 17.5 22.5 27.5 32.5	67.5 137.5 315 720 742.5 422.5	56.25 156.25 306.25 506.25 756.25 1056.25	506.25 1718.75 5512.5 16200 20418.75 13731.25
	Σf = 110		Σfm = 2405		Σfm2 = 58087.5

$$\begin{gathered} {\overline{X}}_B=\frac{\Sigma fm}{\Sigma f}=\frac{2405}{110}=21.86 \\ {\mathrm{\sigma }}_{\mathrm{B}}=\sqrt{\frac{\Sigma f{m}^2}{\Sigma f}-{\left(\overline{X}\right)}^2} \\ =\sqrt{\frac{58087.5}{110}-{\left(21.86\right)}^2} \\ =\sqrt{528.06-477.85} \\ =\sqrt{50.21} \\ =7.08 \\ \mathrm{C}.\mathrm{V}.ofgroupB=\frac{{\sigma }_B}{{\overline{X}}_B}\times 100=\frac{7.08}{21.86}\times 100=32.4\% \end{gathered}$$

Group A is more consistent as C.V. of group A is less than C.V. of group B.