Question

Two lines of regression are given by 5x+7y-22=0 and 6x+2y-22=0if the variance of y is 15 find the standard deviation of x

Answer 1

Answer:

6x +2y-22=0 is the line of xon y

if the variance of y is 15 then variance of x is given as follows

(2/6)^2 *15 = 4/36*15 =1.66666

and sd is 1.29099

Answer 2

Answer:

The standard deviation of $x$ is $4.5$ .

Step-by-step explanation:

Consider the equations of lines of regression as follows:

$5x+7y-22=0$ . . . . . (1)

$6x+2y-22=0$ . . . . . (2)

Rewrite equation (1) as follows:

⇒ $7y=22-5x$

⇒ $y=\frac{22}{7} -\frac{5x}{7}$

⇒ $y=3.14-0.17x$

⇒ $y=-0.17x+3.14$ . . . . . (3) (Regression equation of $y$ on $x$)

Then, the regression equation of $y$ on $x$ is

$b_{yx}=r\frac{\sigma_y}{\sigma_x}$,

where $r$ = correlation coefficient and $\sigma_y ,\ \sigma_x$ = variance of $y$ and $x$ respectively.

⇒ $-0.17=r\frac{\sigma_y}{\sigma_x}$ . . . . . (4)

Rewrite equation (2) as follows:

⇒ $2x=22-2y$

⇒ $x=\frac{22}{6} -\frac{2y}{6}$

⇒ $x=3.66-0.33y$

⇒ $x=-0.33y+3.66$ . . . . . (5) (Regression equation of $x$ on $y$)

Then, the regression equation of $x$ on $y$ is

$b_{xy}=r\frac{\sigma_y}{\sigma_x}$,

⇒ $-0.33=r\frac{\sigma_x}{\sigma_y}$ . . . . . (6)

Multiply (4) and (6), we get

⇒ $(-0.17)(-0.33)=r^2$

⇒ $0.0561=r^2$

⇒ $r=-0.23$ (As both $b_{xy}, b_{yx}$ are negative so $r$ will also be negative)

From (4),

⇒ $-0.17=-0.23\frac{15}{\sigma_x}$ (Since $r=-0.23$ and $\sigma_y=15$)

⇒ $\sigma_x=(-0.23)(15)/(-0.17)$

⇒ $\sigma_x = 20.29$

As we know,

The Standard deviation of $x$ = $\sqrt{\sigma_x}$

S.D = $\sqrt{20.29}$

      = 4.5

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