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- The formula required to solve this problem are : Mean $ A+ \large\frac{\sum f_i d_i}{\sum f_i} $$ \times h$
- Standard deviation $\sigma= \sqrt {\large\frac{\sum d_i^2}{n} - \bigg( \large\frac{\sum d_i^2}{n} \bigg)^2 }$$ \times h$
- Coefficient variation $=\large\frac{\sigma}{\bar {x} }$$ \times 100$

Step 2:

Mean =$ A+ \large\frac{\sum d_i}{n} $

$\qquad= 52 +\large\frac{(-10)}{10}$

$\qquad= 52-1$

$\qquad= 51$

Step 3:

Standard deviation $\sigma= \sqrt {\large\frac{\sum d_i^2}{n} - \bigg( \large\frac{\sum d_i^2}{n} \bigg)^2 }$

$\qquad= \sqrt { \large\frac{360}{10} - \bigg( \large\frac{-10}{10}\bigg)^2}$

$\qquad= \sqrt {36-1}$

$\qquad= \sqrt {35}$

$\qquad= 5.92$

Step 4:

Coefficient variation $=\large\frac{\sigma}{\bar {x} }$$ \times 100$

$\qquad= \large\frac{5.92}{51}$$ \times 100$

$\qquad= \large\frac{592}{51}$$=11.60$

Step 2:

Mean =$ A+ \large\frac{\sum d_2}{n} $

$\qquad= 105 +\large\frac{0}{10}$

$\qquad=105$

Step 3:

Standard deviation $\sigma= \sqrt {\large\frac{\sum d_i^2}{n} - \bigg( \large\frac{\sum d_i^2}{n} \bigg)^2 }$

$\qquad= \sqrt {\large\frac{40}{10 } -\bigg(\frac{0}{10}\bigg)^2}$

$\qquad= \sqrt {4}$

$\qquad=2$

Step 4:

Coefficient variation $=\large\frac{\sigma}{\bar {x} }$$ \times 100$

$\qquad=\large\frac{2}{105} $$ \times 100 $

$\qquad= \large\frac{200}{105}$

$\qquad= 1.90$

COMPARISON:

Shares, whose coefficient variation is lesser is considered to be more stable.

$\therefore $ CV of share Y is lesser as compared to CV of share X.

$\therefore $ Share Y is more stable.

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