Question

If the two regression lines of a bivariate distribution are 4x - 5y + 33 = 0 and 20x - 9y - 107 = 0

(i) calculate \(\bar{x}\) and \(\bar{y}\), the arithmetic means of x and y respectively.

(ii) estimate the value of x when y = 7. 

(iii) find the variance of y when σ_x = 3.

Answer 1

The given regression lines are,

4x - 5y + 33 = 0

20x - 9y - 107 = 0

The regression lines intersect at the point \((\bar{x},\bar{y})\)

(i) Equation are

4x - 5y + 33 = 0

20x - 9y - 107 = 0

Multiply equation (1) by 5 and subtracting equation (2) from it, we get

Putting the value of y in equation (1), we get

4x - 5y = -33

4x - 5 x 17 = -33

4x - 85 = -33

4x = -33 + 85

4x = -33 + 85

4x = 52

x = 13

\(\bar{x}=13,\bar{y}=17\)

(ii) The line of regression of x and y gives the value of x.

Let equation (1) is the regression line of y on x then, line (2) is the regression line of x on y.

Rewriting equation (1) and (2), We get

\(y=\frac{4}{5}x +\frac{33}{5}\,and\,x=\frac{9}{20}y +\frac{107}{20}\)

⇒ \(b_{yx}=\frac{4}{5}\,and\,b_{xy}=\frac{9}{20}\)

Now, 

\(b_{yx}.b_{xy}=\frac{4}{5}\times\frac{9}{20}=\frac{9}{25}<1\)

which means line (2) is the regression line of x on y.

putting y = 7 in the equation (2), we get

20x - 9y - 107 = 0

20x - 63 - 107 = 0

20x - 170 = 0

20x = 170

x = 170/20

x = 8.5

(iii) Given, σ_x = 3 then variance (σ_x²) = 9

Now, 

\(r=\pm\sqrt{b_{yx}.b_{xy}}\)

\(=\pm\sqrt{\frac{4}{5}\times\frac{9}{20}}\)

\(=\pm\sqrt{\frac{9}{25}}=\pm\frac{3}{5}\)

\(\therefore \,r=\frac{3}{5}\) (∵ r is positive as b_xy is positive)

\(r\frac{\sigma_y}{\sigma _x}=b_{yx}\)

\(\frac{3}{5}\times \frac{\sigma _y}{3}=\frac{4}{5}\)

\(\sigma _y=4\)

∴ Variance of y (σ_y²) = 16