Mumbai University > COMPS > Sem 4 > Applied Mathematics 4

Marks : 04

Year : DEC 2015

1) Mean $\overline x$ and $\overline y$ are obtained by solving the two given equations

$y = 1/2,$ & $x = 3$

2) If the line $x + 6y = 6$ is the line of regression of y on x, then of regression of y on x, then

by $=-x + 6 $ i.e $y = -1/6x + 1 \space\space byx = -1$

If the line $3x + 2y = 10$ is the line of regression of x on y then

$$3x = -2y + 10 i.e x = -2y/3 + 10/3$$

$Bxy = -2/3\\ r=\sqrt{byx\times bxy}=\sqrt{\dfrac {-1}6\times \dfrac {-2}3}=\sqrt{\dfrac 19}=\dfrac 13 $

Since byx and bxy are negative.

X is greater.

$$\therefore x=\dfrac {-1}3$$