In both the series items are in small number. Therefore correlation coefficient can also be calculated without taking deviations from actual means or assumed mean.

r = \(\frac { N\Sigma XY-(\Sigma X)(\Sigma Y) }{ \sqrt { N\Sigma X^{ 2 }-(\Sigma X)^{ 2 }\times \sqrt { N\Sigma { Y }^{ 2 }-(\Sigma Y)^{ 2 } } } } \)
= \(\frac { 7(676)-(70)(63) }{ \sqrt { 7(728)-(70)^{ 2 }\times \sqrt { 7(651)-(63)^{ 2 } } } } \)
= \(\frac { 322 }{ 339.48 } \)
r = +0.95

\(r =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{9(7393)-(180)(360)}{\sqrt{9(3720)-(180)^2} \sqrt{9(14746)-(360)^2}} \)
\(=\frac{66537-64800}{\sqrt{33480-32400} \times \sqrt{132714-129600}} \)
\(=\frac{1737}{\sqrt{1080 \times 3114}}=\frac{1737}{1833.88}=0.9472\)

Let X is considered for Subject 1 and Y is considered for Subject 2

\(\rho =1-\frac { 6\Sigma d^{ 2 } }{ N(N^{ 2 }-1) } \)
\(\rho =1-\frac { 6(142) }{ 11(11^{ 2 }-1) } \)
\(\rho =1-\frac { 852 }{ 1320 } \)
= 0 354

Let us consider X for price P and Y for stock S. Then the mean and SD for P is considered as \(\bar { X } \) = 100 and σ_x = 8 respectively and the mean and SD of S is considered as \(\bar { Y } \) = 103 and σ_y = 4. The correlation coefficient between the series is r(X, Y) = 0.4
Let the regression line X on Y be
\(X-\bar { X } =r\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } (Y-\bar { Y } )\)
X-100 = (0.4)\(\frac{8}{4}\)(Y-103)
X–100 = 0.8(Y–103 )
X–0.8Y–17.6 = 0 (or)  X = 0.8Y+17.6
The regression line Y on X be \(Y-\bar { Y } =r\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } (X-\bar { X } )\)
Y-103 = (0.4)\(\frac{4}{8}\)(X-100)
Y–103 = 0.2 (X–100 )
Y–103 = 0.2 X–20
Y = 0.2 X + 83 (or) 0.2 X–Y + 83 = 0

\(\bar{X}=36, \bar{Y}=85 \)
\(\sigma_x=11, \sigma_y=8, r=0.66 \)
\(b_{y x}=r \frac{\sigma_y}{\sigma_x}=0.66\left(\frac{8}{11}\right)=0.48 \)
\(b_{x y}=r\left(\frac{\sigma_x}{\sigma_y}\right)=0.66\left(\frac{11}{8}\right)=0.91\)
(i) Regression equation of Y on X is
\(Y-\bar{Y} =b_{y x}(X-\bar{X}) \)
Y - 85 = 0.48 X - 17.28
Y = 0.48 X + 67.72 .
Regression equation of X on Y is
\(X-\bar{X} =b_{x y}(Y-\bar{Y}) \)
X - 36 = 0.91Y - 77.35
X = 0.91 Y - 41.35
(ii) If X = 10
Y = 4.8 + 67.72 = 72.52