1 Answer
Solution :-
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Variance for y =
\( \dfrac{1}{n^2}[ N\displaystyle\sum y_i^2-(\displaystyle\sum y_i)^2] \)
= \( \dfrac{1}{10}\times 40=4\)
Standard deviation (σ2) = \( \sqrt{Variance}\)
= \( \sqrt{4}=2\)
CV (shares X) =
\( \dfrac{σ_2}{y}\times 100\) = \( \dfrac{2}{105}\times 100\)
= 1.9 = 11.58
C.V. of prices of shares X is greater than the C.V. of prices of shares Y.
Thus, the prices of shares Y are more stable than the prices of shares X.
Answered by Pragya Singh | 1 year agoRelated Questions
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