From the data given below state which group is more variable, A or B?

Where A = 45,

and y_i = \( \dfrac{ (x_i – A)}{h}\)

Here h = class size = 20 – 10

h = 10

So, x̅ = 45 + ((\( \dfrac{-6}{150}\)) × 10)

= 45 – 0.4

= 44.6

\( Variance(σ^2)= \)

\( \dfrac{h^2}{N^2}[N\displaystyle\sum{f_i}y_i^2-(\displaystyle\sum{f_i}y_i)^2] \)

σ² = \( (\dfrac{10^2}{150^2}\) [150(342) – (-6)²]

= (\( \dfrac{100}{22500}\)) [51,300 – 36]

= (\( \dfrac{100}{22500}\)) × 51264

= 227.84

Hence, standard deviation = σ 

= \( \sqrt{227.84}\)

= 15.09

C.V for group A = \( \dfrac{σ}{x̅}\) × 100

= (\( \dfrac{15.09}{44.6} \)) × 100

= 33.83

Now, for group B.

Where A = 45,

h = 10

So, x̅ = 45 + ((\( \dfrac{-6}{150}\)) × 10)

= 45 – 0.4

= 44.6

\( Variance(σ^2)=\)

\( \dfrac{h^2}{N^2}[N\displaystyle\sum{f_i}y_i^2-(\displaystyle\sum{f_i}y_i)^2] \)

σ² = (\( (\dfrac{10^2}{150^2}\)) [150(366) – (-6)²]

= (\( \dfrac{100}{22500}\)) [54,900 – 36]

= (\( \dfrac{100}{22500}\)) × 54,864

= 243.84

Hence, standard deviation = σ 

= \(\sqrt{ 243.84}\) 

= 15.61

C.V for group B = (\( \dfrac{σ}{x̅}\)) × 100

= (\( \dfrac{15.61}{44.6} \)) × 100

= 35

By comparing C.V. of group A and group B.

C.V of Group B > C.V. of Group A

So, Group B is more variable.