Question

If sum of ten observations is 210,sum of their squares is 4612.5. find out C.V.​

Answer 1

$\large\underline{\sf{Solution-}}$

$\sf \: Let \: x_1, \: x_2, \: x_3, \: - - - -,x_{10} \: be \: 10 \: observations.$

• According to statement,

$\sf \: sum \: of \: 10 \: observations, \: \displaystyle\sum_{i=1}^n{x _i} \: = \: 210$

So,

• Mean of 10 observations is

$\rm :\longmapsto\:\bf \:{\overline x } = \bf \:\dfrac{\: \displaystyle\sum_{i=1}^n({x _i} )}{n}$

$\rm :\longmapsto\:{\overline x } = \dfrac{210}{10}$

$\bf\implies \:{\overline x } = 21 - - - (1)$

Again,

According to statement,

• Sum of the squares of 10 observations = 4612.5

$\rm :\implies\:\: \displaystyle\sum_{i=1}^n{x _i}^{2} = 4612.5$

We know,

• Variance of given observations is evaluated by

$\rm :\longmapsto\:Variance \: = \: \dfrac{\displaystyle\sum_{i=1}^n{x _i}^{2}}{n} \: - \: {({\overline x })}^{2}$

$\rm :\longmapsto\:Variance = \dfrac{4612.5}{10} - {(21)}^{2}$

$\rm :\longmapsto\:Variance = 461.25 - 441$

$\bf\implies \:Variance = 20.25$

Now,

• Standard deviation of given data is evaluated by

$\rm :\longmapsto\:Standard \: deviation \: ( \sigma \: ) = \sqrt{Variance}$

$\rm :\longmapsto\:Standard \: deviation \: ( \sigma \: ) = \sqrt{20.25}$

$\bf :\longmapsto\:Standard \: deviation \: ( \sigma \: ) = 4.5$

Now,

We know that,

• Coefficient of variation, C.V. is evaluated as

$\rm :\longmapsto\:C.V. = \dfrac{\sigma \:}{{\overline x }} \times 100\%$

$\rm :\longmapsto\:C.V. = \dfrac{4.5}{21} \times 100\%$

$\bf\implies \:C.V. = 21.43\% \: approx.$