Question

From the prices of shares X and Y given below : find out which is more stable in value :

$\begin{array}{ccccccccccc}X:& 35& 54& 52& 53& 56& 58& 52& 50& 51& 49\\ Y:& 108& 107& 105& 105& 106& 107& 104& 103& 104& 101\end{array}$

Answer 1

$\begin{array}{ccc}{x}_i& |d|=(x_i-Mean)& d^2\\ 35& -13& 169\\ 24& -24& 576\\ 52& 4& 16\\ 53& 5& 25\\ 56& 8& 64\\ 58& 10& 100\\ 52& 4& 16\\ 50& 2& 16\\ 51& 3& 9\\ 49& 1& 1\\ 480& & 980\end{array}$

$\overline{X}=\frac{1}{n}\sum x_i=\frac{1}{10}\left[480\right]=48$

var (x) = $\frac{1}{n}\{\sum (x_i-\overline{x}{)}^2\}=\frac{1}{10}\left\{980\right\}=98$

S. D (x) = $\sqrt{var\left(x\right)}=\sqrt{98}=9.9$

Coefficient of variation = $\frac{S.D}{\overline{X_1}}\times 100=\frac{9.9}{48}\times 100=20.6$

$\begin{array}{ccc}x& d=(x-Mean)& d^2\\ 35& -13& 169\\ 24& -24& 576\\ 52& 4& 16\\ 53& 5& 25\\ 56& 8& 64\\ 58& 10& 100\\ 52& 4& 16\\ 50& 2& 4\\ 51& 3& 9\\ 49& 1& 1\\ 480& & 980\end{array}$

$\overline{(}x)=\frac{1}{n}\sum x_i=\frac{1}{10}[1050]=105$

var(x) = $\frac{1}{n}\{\sum (x_i-\overline{x}{)}^2\}=\frac{1}{40}\left\{40\right\}=4$

$\therefore$ S.D .(x) = $\sqrt{var\left(x\right)}=\sqrt{4}=2$

Coefficient of variation for shares Y$=\frac{S.D}{\overline{X}}\times 100=\frac{2}{105}\times 100=1.90$

Since the coefficient of variation for shares Y is smaller than the coefficient of variation for shares X, they are more stable.