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Home  >>  CBSE XI  >>  Math  >>  Statistics
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From the prices of shares A and Y below. Find out which is more stable in value.

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Toolbox:
  • 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} }$$ \times 100$
Step 1:
Let Assumed mean $A=52$ for share X
Step 2:
Mean =$ A+ \large\frac{\sum d_i}{n} $
$\qquad= 52 +\large\frac{(-10)}{10}$
$\qquad= 52-1$
$\qquad= 51$
Step 3:
Standard deviation $\sigma= \sqrt {\large\frac{\sum d_i^2}{n} - \bigg( \large\frac{\sum d_i^2}{n} \bigg)^2 }$
$\qquad= \sqrt { \large\frac{360}{10} - \bigg( \large\frac{-10}{10}\bigg)^2}$
$\qquad= \sqrt {36-1}$
$\qquad= \sqrt {35}$
$\qquad= 5.92$
Step 4:
Coefficient variation $=\large\frac{\sigma}{\bar {x} }$$ \times 100$
$\qquad= \large\frac{5.92}{51}$$ \times 100$
$\qquad= \large\frac{592}{51}$$=11.60$
For share y, Let Assumed mean $A=105$
Step 1:
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.
answered Jul 1, 2014 by meena.p
 

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