Question

For the following pairs of observations $(1,10),(2,9),(3,9),(4,8),(5,6),(6,12),(7,4),(8,3),(9,13),(10,1),$ then $Cov(x,y)=$

• A. $1.05$
• B. $2.45$
• C. $–4.45$
• D. $–4.9$

Answer 1

The correct option is C

$–4.45$

Finding covariance of (x, y):

Given data $(1,10),(2,9),(3,9),(4,8),(5,6),(6,12),(7,4),(8,3),(9,13),(10,1),$

Here, $$\begin{gathered} x=1,2,3,4,5,6,7,8,9,10\& \\ y=10,9,8,7,6,5,4,3,2,1 \end{gathered}$$

$\begin{array}{rcl}\mathbf{M}\mathbf{e}\mathbf{a}\mathbf{n}\mathbf{}\mathbf{\left(}\overline{\mathbf{x}}\mathbf{\right)}\mathbf{}& \mathbf{=}& \mathbf{}\frac{\mathbf{s}\mathbf{u}\mathbf{m}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{b}\mathbf{s}\mathbf{e}\mathbf{r}\mathbf{v}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{}}{\mathbf{n}\mathbf{u}\mathbf{m}\mathbf{b}\mathbf{e}\mathbf{r}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{o}\mathbf{b}\mathbf{s}\mathbf{e}\mathbf{r}\mathbf{v}\mathbf{a}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}}\\ & & x=\frac{1+2+3+4+5+6+7+8+9+10}{10}=5.5\\ & & similarly,\\ & & \overline{y}=5.5\end{array}$

$\begin{array}{|lll|}\hline x-\overline{x}& y-\overline{y}& (x-\overline{x})(y-\overline{y}\\ -4.5& 2.5& -11.25\\ -3.5& 1.5& -5.25\\ -2.5& 1.5& -3.75\\ -1.5& 0.5& -0.75\\ -0.5& -1.5& 0.75\\ 0.5& 4.5& 2.25\\ 1.5& -3.5& -5.25\\ 2.5& -4.5& -11.25\\ 3.5& 5.5& 19.25\\ 4.5& -6.5& -29.25\\ \hline\end{array}$

$\begin{array}{rcl}\mathbf{C}\mathbf{o}\mathbf{v}\mathbf{a}\mathbf{r}\mathbf{i}\mathbf{a}\mathbf{n}\mathbf{c}\mathbf{e}\mathbf{(}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{)}& \mathbf{=}& \frac{\mathbf{\sum }\mathbf{(}\mathbf{}\mathbf{x}\mathbf{–}\overline{\mathbf{x}}\mathbf{}\mathbf{)}\mathbf{(}\mathbf{}\mathbf{y}\mathbf{–}\overline{\mathbf{y}}\mathbf{}\mathbf{)}}{\mathbf{n}}\\ Cov(x,y)& =& \frac{-44.5}{10}\\ Cov(x,y)& =& -4.45\end{array}$

Hence, option $\mathbf{\left(}\mathit{C}\mathbf{\right)}$ is correct answer.