b_yx=\(\frac { n\sum { XY-(\sum { X)(\sum { Y) } } } }{ n{ \sum { X } }^{ 2 }-\left( \sum { { X }^{ 2 } } \right) } =\frac { 5(44)-(15)(15) }{ 5(49)-({ 15) }^{ 2 } } =\frac { -1 }{ 5 } \)
b_xy=\(\frac { n\sum { XY-(\sum { X)(\sum { Y) } } } }{ n{ \sum { Y } }^{ 2 }-\left( \sum { { Y) }^{ 2 } } \right) } =\frac { 5(44)-(15)(15) }{ 5(49)-({ 15) }^{ 2 } } =\frac { -1 }{ 5 } \)
The co-efficient of correlation is
r= \(\sqrt { { b }_{ xy }.{ b }_{ yx } } =\sqrt { \left( -\frac { 1 }{ 5 } \right) \left( -\frac { 1 }{ 5 } \right) } \)=0.2
As b_xy and b_yx are both negative, r=-0.2

The regression co-efficient of X on Y is
b_xy= \(\frac { N\sum { XY-(\sum { X)(\sum { Y) } } } }{ N.\sum { { Y }^{ 2 }-(\sum { { Y) }^{ 2 } } } } =\frac { 5(300)-(20)(40) }{ 5(345)-({ 40) }^{ 2 } } \)=5.6
Also, \(\bar {X}\) =\(\frac { \sum { X } }{ N } =\frac { 20 }{ 5 } \)=4
and \(\bar {Y}\)=\(\frac { \sum { Y } }{ N } =\frac { 40 }{ 5 } \)=8
\(\therefore\)The regression equation of X on Y is
X- \(\bar {X}\)=b_xy(Y-\(\bar {Y}\) )
X-4=5.6(Y-8)
X-4=5.6Y-44.8
X=5.6Y-40.8
When Y=6, X=5.6(6)-40.8=33.6-40.8
\(\therefore\)X=-7.2

Let 2 and 4 be the assumed means for the X and Y series respectively.

Correlation Co-efficient
r(X,Y) =\(​​​​\frac { N\sum { dXdY-(\sum { dX)(\sum { Y) } d } } }{ \sqrt { n.\sum { d{ x }^{ 2 }-{ (\sum { dx } })^{ 2 } } } \sqrt { n\sum { d{ y }^{ 2 }-(\sum { d{ y) }^{ 2 } } } } } \)
=\(\frac { 7(0)-(-14)(0) }{ \sqrt { 56(7)-({ -14) }^{ 2 } } \sqrt { 84(7)-0 } } \)=0
 ∴ r(X, Y)=0

Correlation co-efficient
r(X, Y) =\(\frac { N\sum { XY-(\sum { X)(\sum { Y) } } } }{ \sqrt { n\sum { { X }^{ 2 }-{ (\sum { X) } }^{ 2 } } } \sqrt { n\sum { { Y }^{ 2 }-({ \sum { Y) } }^{ 2 } } } } \)
=\(\frac { 25(50)-(125)(100) }{ \sqrt { (1650)(25)-{ (125) }^{ 2 } } \sqrt { (1460)(25)-{ (100) }^{ 2 } } } \)
=\(\frac { -450 }{ \sqrt { 1125\sqrt { 1060 } } } \)
r =-0.41

Given \(\bar{X}\)=50, \(\bar{Y}\)=50, \({ \sigma }_{ x }\)=3.5, \({ \sigma }_{ y }\)=4.5 and r=0.5
b_xy=\(r.\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } =\frac { (0.5)(3.5) }{ 4.5 } \)=0.38
b_yx=\(r.\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } =\frac { (0.5)(4.5) }{ 3.5 } \)=0.64
The regression equation of X on Y is
X-\(\bar{X}\) =b_xy(Y-\(\bar{Y}\))
X-50=0.38(Y-50)
X=0.38Y+31
\(\therefore\)The regression equation Y on X is
Y-\(\bar{Y}\)=b_yx(X-\(\bar{X}\))
Y-50=0.64(X-50)
Y=0.64X+18