Let \(\bar { X } =\frac { \sum { X } }{ N } =\frac { 310 }{ 10 } \)=31
Here N=10 \(\bar { Y } =\frac { \sum { Y } }{ N } =\frac { 310 }{ 10 } \)=48

Co-efficient correlation 
r(x,y) =\(\frac { \sum { xy } }{ \sqrt { { \sum { x } }^{ 2 } } \sqrt { { \sum { y } }^{ 2 } } } =\frac { -492 }{ \sqrt { 330 } .\sqrt { 784 } } \)
\(\Rightarrow\)\(\frac { -492 }{ 18.1659\times 28 } =\frac { -492 }{ 508.6452 } \)
\(\Rightarrow\) r=-0.9673

Here N=5

\(\overline { X } =\frac { \sum { X } }{ N } =\frac { 30 }{ 5 } \)=6
\(\overline { Y } =\frac { \sum { Y } }{ N } =\frac { 40 }{ 5 } \)=8
\({ b }_{ xy }=\frac { N\sum { XY-(\sum { X } )(Y) } }{ N\sum { { X }^{ 2 }-{ (\sum { Y } ) }^{ 2 } } } =\frac { 5(214)-(30)(40) }{ 5(340)-{ (40) }^{ 2 } } \)
=\(\frac { 1070-1200 }{ 1700-1600 } =-\frac { 130 }{ 100 } \)=-1.3
b_yx=\({ b }_{ xy }=\frac { N\sum { XY-(\sum { X } )(Y) } }{ N\sum { { X }^{ 2 }-{ (\sum { X } ) }^{ 2 } } } =\frac { 5(214)-(30)(40) }{ 5(2200)-{ (30) }^{ 2 } } \)
\(=\frac { 1070-1200 }{ 1700-900 } =\frac { -130 }{ 200 } \)=-0.65
Regression equation of X on Y is
\(\Rightarrow\)X-\(\bar{X}\)=b_xy(Y-\(\bar{Y}\))
X-6=-1.3(Y-8)
X=-1.3Y+10.4+6
X=-1.3Y+16.40
Regression equation of Y on X is
Y-\(\bar{Y}\) =b_yx(X-\(\bar{X}\))
Y-8=-0.65(X-6)
Y=-0.065X+3.9+8
Y=-0.65X+11.90

Here N=5

\(\bar { X } =\frac { \sum { X } }{ N } =\frac { 70 }{ 5 } \)=14
\(\bar { Y } =\frac { \sum { Y } }{ N } =\frac { 40 }{ 5 } \)=8
b_xy=\(\frac { N\sum { XY } }{ N{ \sum { Y } }^{ 2 }-{ (\sum { Y } ) }^{ 2 } } =\frac { 5(600)-70(40) }{ 5(360)-{ (40) }^{ 2 } } \)
=\(\frac { 3000-2800 }{ 1800-1600 } =\frac { 200 }{ 200 } \)=1
Regression equation of X on Y is
X-\(\bar { X }\) =b_xy(Y-\(\bar { Y }\))
\(\Rightarrow\)X-14=1(Y-8)
\(\Rightarrow\)X=Y-8+14
\(\Rightarrow\)X=Y+6
When Y=20, X=20+6
\(\Rightarrow\)X=26
Hence, when Y=20, X=26

We will find the correct values of \(\sum\)x, \(\sum\)x², \(\sum\)y², and \(\sum\)xy by delecting the old values and adding new ones.
\(\therefore\)\(\sum\)x=125-(6+8)+(6+8)=125
\(\sum\)y=100-(14+6)+(12+8)=100
x²=650-(6²+8²)+(8²+6²)=650
y²=460-(14²+6²)+(12²+8²)=436
and xy =508-(14x6+8x6)+(12x8+6x8)=520
\(\therefore\) Correlation Co-efficient
r(x,y)=\(\frac { N\sum { xy } -(\sum { x } )(\sum { y } ) }{ \sqrt { N{ \sum { x } }^{ 2 }-{ (\sum { x } ) }^{ 2 } } \sqrt { N{ \sum { y } }^{ 2 }-{ (\sum { y } ) }^{ 2 } } } \)
\(\Rightarrow\)\(\frac { 25(520)-125(100) }{ \sqrt { 25(650)-{ (125) }^{ 2 } } \sqrt { 25(436)-{ (100) }^{ 2 } } } \)
\(\Rightarrow\) r(x,y)=0.66

We know that the lines of regression intersect at the mean values of x and y.
Given lines are 4x+3y =-7  ....(i)
and      3x+4y =-8  ....(ii)
\(\Rightarrow\)(1)x3    12x+9y =-21
                  (-)   (+)    (-)
\(\Rightarrow\)(2)x4    12x+16y=-32
             _______________
Subtracting,  -7y =+11   \(\Rightarrow\) y =-\(\frac{11}{7}\)
Subtarcting y=-\(\frac{11}{7}\) in (1) we get,
4x+3(\(\frac{11}{7}\)) =-7
4x-\(\frac{33}{7}\) =-7
4x=-7+\(\frac{33}{7}\) =\(\frac{-49+33}{7}\)
4x=\(\frac{-16}{7}\)
x=\(\frac{-16}{4\times7}=\frac{-4}{7}\)
\(\bar{X}\)=\(\frac{-4}{7}\) and \(\bar{Y}\)=\(\frac{11}{7}\)
(ii) Let us take the equation 4x+3y+7=0 as the line of regression of X on Y and 3x+4y+8=0 as the line of regression of Y on X.
Then, \(\Rightarrow\)4x+3y+7 =-3y-7
x=\(\frac{-3}{4}\) y -\(\frac{7}{4}\)
\(\therefore\)b_xy=\(\frac{-3}{4}\)
and 3x+4y+8=0  \(\Rightarrow\)4y=-3x-8
\(\Rightarrow\)y=\(\frac{-3}{4}x-\frac{8}{4}\)
\(\Rightarrow\)b_yx=\(\frac{-3}{4}\)
(iii) The correlation co-efficient is
r=\(\sqrt { { b }_{ xy }-{ b }_{ yx } } =\sqrt { \left( \frac { -3 }{ 4 } \right) \left( \frac { -3 }{ 4 } \right) } =\frac { 3 }{ 4 } \)=0.75
As b_xy and b_yx are both negative,
r=-0.75