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# If $b_{yx}=1.6$ and $b_{xy}=0.4$ and the angle between two regression lines is $\theta$ then $\tan \theta$ is

$\begin {array} {1 1} (A)\;0.18 & \quad (B)\;0.16 \\ (C)\;0.3 & \quad (D)\;0.24 \end {array}$

$r = \sqrt{1.6 \times 0.4}$
$= 0.8$
$b_{yx} = r\large\frac{\sigma_y}{\sigma_x} \Rightarrow \large\frac{\sigma_y}{\sigma_x}$
$= \large\frac{1.6}{0.8}$ $= 2.0$
$m_1 = \large\frac{1}{r}$ $\large\frac{\sigma_y}{\sigma_x}$
$= \large\frac{1}{0.8}$ $\times 2.0 = \large\frac{5}{2}$ $= 2.5$
$m_2 = r\large\frac{\sigma_y}{\sigma_x}$ $= 0.8 \times 2$
$= 1.6$
$\therefore \tan \theta = \large\frac{m_1-m_2}{1+m_1m_2}$
$= \large\frac{0.9}{5}$
$= 0.18$
Ans : (A)