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

Class 11 | Exercise 15.3 | Maths | NCERT | Statistics

Solution:-

Calculate the coefficient of variance for each series for comparing the variability or dispersion of two series. The series having greater C.V. is said to be more variable than the other. The series having lesser C.V. is said to be more consistent than the other.

We know that Co-efficient of variation (C.V.) =

$\left( \sigma /\text{ }\bar{X} \right)\text{ }\times \text{ }100$

Where,

$\sigma$

= standard deviation,

$\bar{X}$

= mean

For Group A

Mean,

$\bar{X}=A+\frac{\sum\limits_{i=1}^{a}{{{f}_{i}}{{y}_{i}}}}{N}\times h$

Where A =

$45$

and

${{y}_{i}}~=\text{ }({{x}_{i}}~\text{ }A)/h$

Here h = class size =

$20-10$

h =

$10$

So,

$\bar{X}\text{ }=\text{ }45\text{ }+\text{ }\left( \left( -6/150 \right)\text{ }\times \text{ }10 \right)$

$45\text{ }\text{ }0.4$

$44.6$

We know that variance,

${{\sigma }^{2}}=\frac{{{h}^{2}}}{{{N}^{2}}}\left[ N\sum{{{f}_{i}}y_{i}^{2}-(\sum{{{f}_{i}}{{y}_{i}}{{)}^{2}}}} \right]$

${{\sigma }^{2}}~=\text{ }({{10}^{2}}/{{150}^{2}})\text{ }[150\left( 342 \right)\text{ }\text{ }{{\left( -6 \right)}^{2}}]$

$\left( 100/22500 \right)\text{ }\left[ 51,300\text{ }\text{ }36 \right]$

$\left( 100/22500 \right)\text{ }\times \text{ }51264$

$227.84$

Therefore, standard deviation =

$\sigma =\sqrt{227.84}$

$15.09$

∴ C.V for group A =

$\left( \sigma /\text{ }\bar{X} \right)\text{ }\times \text{ }100$

$\left( 15.09/44.6 \right)\text{ }\times \text{ }100$

$33.83$

Now, for group B.

Mean,

$\bar{X}=A+\frac{\sum\limits_{i=1}^{a}{{{f}_{i}}{{y}_{i}}}}{N}\times h$

Where A =

$45$

h =

$10$

So,

$\bar{X}\text{ }=\text{ }45\text{ }+\text{ }\left( \left( -6/150 \right)\text{ }\times \text{ }10 \right)$

$45\text{ }\text{ }0.4$

$44.6$

We know that variance,

${{\sigma }^{2}}=\frac{{{h}^{2}}}{{{N}^{2}}}\left[ N\sum{{{f}_{i}}y_{i}^{2}-(\sum{{{f}_{i}}{{y}_{i}}{{)}^{2}}}} \right]$

${{\sigma }^{2}}~=\text{ }({{10}^{2}}/{{150}^{2}})\text{ }[150\left( 366 \right)\text{ }\text{ }{{\left( -6 \right)}^{2}}]$

$\left( 100/22500 \right)\text{ }\left[ 54,900\text{ }\text{ }36 \right]$

$\left( 100/22500 \right)\text{ }\times \text{ }54,864$

$243.84$

Hence, standard deviation =

$\sigma \text{ }=\text{ }\sqrt{243.84}$

$15.61$

∴ C.V for group B =

$\left( \sigma /\text{ }\bar{X} \right)\text{ }\times \text{ }100$

$\left( 15.61/44.6 \right)\text{ }\times \text{ }100$

$35$

Compare C.V. of group A and group B.

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

Therefore, Group B is more variable.

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