\(\rho =1-\frac{6 \sum d^2}{n\left(n^2-1\right)}=1-\frac{6(8)}{8(63)} \)
\(=1-\frac{6}{63}=1-0.095=0.905\)

\(\bar{X} =\frac{\Sigma X}{n}=\frac{15}{5}=3, \bar{Y}=\frac{\Sigma Y}{n}=\frac{25}{5}=5 \)
\(b_{x y} =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{N \Sigma Y^2-(\Sigma Y)^2}=\frac{5(83)-(15)(25)}{5(135)-(25)^2} \)
\(=\frac{415-375}{675-625}=\frac{40}{50}=0.8 \)
\(b_{y x} =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{N \Sigma X^2-(\Sigma X)^2}=\frac{40}{5(55)-225} \)
\(=\frac{40}{50}=0.8\)
Regression equation of X on Y is
\(X-\overset { - }{ X } =b_{ xy }(Y-\overset { - }{ Y } )\)
X - 3 = 0.8(Y - 5)
X - 3 = 0.8Y - 4 + 3
X = 0.8Y - 1
If Y = 8, X = 6.4 - 1 = 5.4
Regression equation of Y on X is
\(Y-\overset { - }{ Y } =b_{ yx }(X-\overset { - }{ X } )\)
Y - 5 = 0.8(X - 3)
Y = 0.8X - 2.4 + 5
Y = 0.8X + 2.6
At X = 12, Y= 9.6 + 2.6 = 12.2

\(\bar{X} =\frac{\Sigma X}{N}=\frac{338}{7}=48.29 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{361}{7}=51.57 \)
\(b_{y x} =\frac{N \Sigma d x d y-\Sigma d x \Sigma d y}{N \Sigma d x^2-(\Sigma d x)^2} \)
\(=\frac{7(728)+6}{7(774)-4} \)
\(=\frac{5102}{5414}=0.942\)
Regression equation of Y on X is
\(Y-\overset{-}{Y}=b_{yx}(X-\overset{-}{X})\)
Y - 51.57 = 0.942(X - 48.29)
Y = 0.942X - 45.49 + 51.57
= 0.942X + 6.08
Y = 0.942(30) + 6.08
Y = 34.34 (In Crores of rupees)

\(\bar{X} =\frac{\Sigma X}{N}=\frac{1686}{10}=168.6 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{1690}{10}=169 \)
\(b_{x y} =\frac{N \Sigma d x d y-(\Sigma d x)(\Sigma d x)}{N \Sigma d y^2-(\Sigma d y)^2} \)
\(=\frac{10(248)-0}{10(446)-0}=\frac{248}{446}=0.556 \)
\(b_{y x} =\frac{N \Sigma d x d y-(\Sigma d x)(\Sigma d y)}{N \Sigma d x^2-(\Sigma d x)^2} \)
\(=\frac{2480}{4100-36}=\frac{2480}{4064}=0.6102\)
Regression equation of X on Y
\(X-\bar{X}=b_{x y}(Y-\bar{Y}) \)
X - 168.6 = 0.556(Y - 169)
X = 0.556 Y + 168.6-93.964
X = 0.556 Y + 74.64
Regression equation of Y on X
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 169 = 0.6102(X - 168.6)
Y = 0.6102 X - 102.8 + 169
Y = 0.6102X + 66.12
If X = 164
Y= 100.07 + 66.12
= 166.19
Height of son is 166.19

(i) First convert the given equations Y on X and X on Y in standard form and find their regression coefficients respectively.
Given regression lines are
3X–2Y = 5      ... (1)
X–4Y = 7        ... (2)
Let the line of regression of X on Y is
3X–2Y = 5
3X = 2Y+5
X = \(\frac{1}{3}\)(2Y+5)
X = \(\frac{1}{3}\)(2Y+5)
X = \(\frac { 2 }{ 3 } Y+\frac { 5 }{ 3 } \)
∴ Regression coefficient of X on Y is
b_xy =\(\frac{2}{3}\)(<1)
Let the line of regression of Y on X is
X–4Y = 7
–4Y = –X+7
4Y = X–7
Y = \(\frac{1}{4}\)(X-7)
Y = \(\frac{1}{4}\)X-\(\frac{7}{4}\)
∴ Regression coefficient of Y on X is
b_yx = \(\frac{1}{4}\)(<1)
(ii) Coefficient of correlation
Since the two regression coefficients are positive then the correlation coefficient is also positive and it is given by
r = \(\sqrt { { b }_{ yx }.{ b }_{ xy } } \)
= \(\sqrt { \frac { 2 }{ 3 } .\frac { 1 }{ 4 } } \)
= \(\sqrt { \frac { 1 }{ 6 } } \)
= 0.4082
∴ r = 0.4082