Explanation
Given: $$\displaystyle\sum\limits_{i = 1}^{18} {\left( {x_i - 8} \right) = 9} $$
$$\displaystyle\sum\limits_{i = 1}^{18} {{{\left( {x_i - 8} \right)}^2} = 45} $$
Let $$X$$ be a random variable taking values $${x_1},........{x_{18.}}$$
Then $$X-8$$ has the values $${x_1} - 8,....,{x_{18}} - 8.$$
Now, $$E\left( {X - 8} \right) =\displaystyle {{\displaystyle\sum\limits_{i = 1}^{18} {\left( {x_i - 8} \right)} } \over {18}}$$
$$ = \displaystyle{9 \over {18}} = {1 \over 2}$$
And $$E\left[ {{{\left( {X - 8} \right)}^2}} \right] = \displaystyle{{\sum\limits_{i = 1}^{18} {{{\left( {x_i - 8} \right)}^2}} } \over {18}}$$
$$ =\displaystyle {{45} \over {18}} = {5 \over 2}$$
Thus, $$Var\left( {X - 8} \right) = E\left[ {{{\left( {X - 8} \right)}^2}} \right] - \left[ {E{{\left( {X - 8} \right)}}} \right]^2$$
$$ =\displaystyle {5 \over 2} - {\left( {{1 \over 2}} \right)^2}$$
$$ = \displaystyle{5 \over 2} - {1 \over 4}$$
$$ = \displaystyle{9 \over 4}$$
We know $$Var\left( {1.X - 8} \right) = {1^2}Var\left( X \right)$$, thus,
$$Var\left( X \right) = \displaystyle{9 \over 4}$$
Standard deviation of $$X = \sqrt {Var\left( X \right)} $$
$$ = \displaystyle\sqrt {{9 \over 4}} = {3 \over 2}$$
If the sum of squares of deviations of $$15$$ observations from their mean $$20$$ is $$240$$, then what is the value of coefficient of variation (CV)?
Please disable the adBlock and continue. Thank you.
