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# If $r_{xy}$ between two variates $x$ and $y$ is 0.6 $cov (x,y)=4.8\: \sigma_x^2=9$ then $\sigma_y$ is

$\begin {array} {1 1} (A)\;\large\frac{8}{9} & \quad (B)\;\large\frac{5}{8} \\ (C)\;\large\frac{8}{3} & \quad (D)\;None \: of \: these \end {array}$

Using $r = \large\frac{cov (x,y)}{\sigma_x \sigma_y}$
Given $r = 0.6$
$cov (x,y)=4.8$
$\sigma_x^2=9$
$\sigma_x= 3$
We have
$0.6 = \large\frac{4.8}{3 \sigma_y}$
$0.6 \times 3 \sigma_y = 4.8$
$3 \sigma_y=\large\frac{4.8}{0.6}$
$\sigma_y=\large\frac{ \not{4.8}^8}{3 \times \not{0.6}}$
$\sigma_y=\large\frac{8}{3}$
Hence Ans : (C)