Let X, Y represent the marks in subject A and B respectively,
\(\bar{X}=39.5, \quad \bar{Y}=47.5 \)
\(\sigma_x=10.8 \quad \sigma_y=16.8 \quad r=0.42 . \)
\(b_{y x}=r \frac{\sigma_y}{\sigma_x}=0.42\left(\frac{16.8}{10.8}\right)=0.653\)
Regression equation of Y on X
\(Y-\bar{Y}=b_{y x}(X-\bar{X})\)
Y - 47.5 = 0.653(X - 39.5)
Y = 0.653 X - 25.79 + 47.5
Y = 0.653 X + 21.7z
At X = 52,
Y = 33.956 + 21.71 = 55.67
Hence if the candidate secures 52 marks in subject A, he will secure 55.67 marks in subject B.

2X = 8 - 3Y
Regression equation of X on Y is
\(X=4-\frac{3}{2} Y \)
\(b_{x y}=-\frac{3}{2}<1\)
Regression equation of Y on X be
2Y = 5 - X
\(Y=\frac{5}{2}-\frac{X}{2} \)
\(b_{y x}=-\frac{1}{2}<1\)
\(\because\) both are negative, r is - ve.
\(r =-\sqrt{b_{x y} \cdot b_{y x}} \)
\(=-\sqrt{\left(\frac{1}{2}\right)\left(\frac{3}{2}\right)}=\frac{-\sqrt{3}}{2} \)
\(=-\frac{1.732}{2}=-0.866\)

\(\bar{X} =\frac{45}{9}=5 \quad \bar{Y}=\frac{90}{9}=10 \)
\(b_{y x} =\frac{N \Sigma X Y-\Sigma X \Sigma Y}{N \Sigma X^2-(\Sigma X)^2} \)
\(=\frac{9(530)-4050}{9(285)-2025}=\frac{720}{540}=1.33\)
Regression equation of Y on X
\(Y-\overset{-}{Y}=b_{yx}(X-\overset{-}{X})\)
Y - 10 = 1.33(X - 5)
Y = 1.33X - 6.65 + 10
Y = 1.33X + 3.35

\(\bar{X} =\frac{\Sigma X}{N}=\frac{652}{10}=65.2 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{1169}{10}=116.9 \)
\(b_{y x} =\frac{N \Sigma d x d y-(\Sigma d x)(\Sigma d y)}{N \Sigma d x^2-(\Sigma d x)^2} \)
\(=\frac{1840-18}{780-4} \)
\(=\frac{1822}{776}=2.348\)
Regression equation of Y on X
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 116.9 = 2.348(X - 65.2)
Y = 2.348 X - 153.09 + 116.9
Y = 2.3 X - 36.19
If  X = 69
Y = 2.348(69) - 36.19
= 162.019 - 36.19
= 125.82 lb .
Weight of the student is 125.82 lb

\(\bar{X} =\frac{\Sigma X}{n} \)
\(=\frac{320}{10}=32 \)
\(\bar{Y} =\frac{\Sigma Y}{n} \)
\(=\frac{380}{10}=38 \)
\(b_{x y} =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{N \Sigma Y^2-(\Sigma Y)^2} \)
\(=\frac{10(12067)-(320)(380)}{10(14838)-(380)^2} \)
\(=\frac{120670-121600}{148380-144400} \)
\(=-\frac{930}{3980} \)
= -0.234
\(b_{y x} =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{N \Sigma X^2-(\Sigma X)^2} \)
\(=-\frac{930}{10(10380)-(320)^2} \)
\(=-\frac{930}{10(10380)-(320)^2} \)
\(=-\frac{930}{1400}=-0.6642\)
(a) Regression equation of X on Y is
\(X-\bar{X} =b_{x y}(Y-\bar{Y}) \)
\(X-32 =-0.234(Y-32) \)
\(X-32 =-0.234 Y+8.892 \)
\(X =32-0.23 Y+8.892 \)
\(X =-0.234 Y+40.892\)
Regression equation of Y on X
\(Y-38 =-0.664(X-32) \)
\(Y =-0.664 X+21.248+38 \)
\(Y =-0.664 X+59.12\)
\((b)\ r=\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \cdot \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=-\frac{930}{\sqrt{3980 \times 1400}}=-\frac{930}{2360.508}\)
r = -0.3939
(c) If X = 30
Y = -0.664(30) + 59.248
Y = -19.92 + 59.25
= 39.33