Given \(\bar{x}\)  =18, \(\bar{y}\) = 100, \({ \sigma }_{ x }\)=14,  \({ \sigma }_{ y }\)=20 and r=0.8
(i) The regression line of y on x is
y-\(\bar{y}\) =r.\(\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } (x-\bar { x } )\)
y-100=\(\left( \frac { 8 }{ 10 } \right) \left( \frac { 20 }{ 14 } \right) (x-18)\)
y-100=\(\frac{8}{7}\)(x-18)
y=\(\frac{8}{7}\)x+100-\(\frac{144}{7}\)
y=\(\frac{8}{7}\)x+\(\frac{556}{7}\)
y=1.14x+79.41 -----(1)
The regression line of x on y is
x-\(\bar{x}\) =\(\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } (y-\bar { y } )\)
x-18=\(\left( \frac { 8 }{ 10 } \right) \left( \frac { 14 }{ 20 } \right) \)(y-100)
x-18=\(\frac{56}{100}\)(y-100)
x-18=0.56(y-100)
x=0.56y-56+18
x=0.56y-38 ---(2)
(ii) Putting x=70 in (1) we get,
y=1.14(70)+79.41=79.80+79.41
y=159.21
(iii) Putting y=90 in (2) we get,
x=0.56(90)-38=50.40-38
x=12.4

Here n=5,
b_yx=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ n\sum { { x }^{ 2 }-{ (\sum { x } ) }^{ 2 } } } \)
=\(\frac { 5(44)-(15)(15) }{ 5(49)-{ (15) }^{ 2 } } \)
=\(\frac { 220-225 }{ 245-225 } =\frac { -5 }{ 20 } =\frac { -1 }{ 4 } \)
b_xy=\(\frac { n\sum { xy } -(\sum { x } )(\sum { y } ) }{ n\sum { { y }^{ 2 }-{ (\sum { y } ) }^{ 2 } } } \)
=\(\frac { 5(44)-15(15) }{ 5(49)-{ (15) }^{ 2 } } \)
\(\frac { 220-225 }{ 245-225 } =\frac { -5 }{ 20 } =\frac { -1 }{ 4 } \)
b_xy and b_yx are negative, r is also negative.
\(\therefore\)Correlation co-efficient
r(x,y)=-\(\sqrt { { b }_{ xy }.{ b }_{ yx } } \)
=-\(\sqrt { \left( \frac { -1 }{ 4 } \right) \left( \frac { -1 }{ 4 } \right) } =\frac { -1 }{ 4 } \)
r=-0.25

The regression lines are
2x+3y=7 ...(1)
and 5x+4y=9  ...(2)
Let (1) represents the regression line of x on y.
\(\therefore\) (2) represents the regression line of y on x.
From (1),  2x=-3y+y
x=-\(\frac{3}{2}y+\frac{7}{2}\)
\(\therefore { b }_{ yx }\)=-\(\frac{3}{2}\)
From (2) 4y=-5x+9
y=-\(\frac {5}{4}+\frac{9}{4}\)
\(\therefore { b }_{ yx }\)=-\(\frac {5}{4}\)
\(\therefore\)r²=b_xy.b_yx
=\(\left( \frac { -3 }{ 2 } \right) \left( \frac { -5 }{ 4 } \right) \)
=\(\frac{15}{8}\) Which is greater than one. Which is not possible.
\(\therefore\)Our choice of regression lines is not correct. Hence, regression line of x on y 5x+4y=9.

Here n=10, d₁²=200, d₂²=214 and d₃²=60
\(\therefore \rho (x,y)\)=\(1-\frac { 6\sum { { d }_{ 1` }^{ 2 } } }{ n({ n }^{ 2 }-1) } \)
=\(1-\frac { 6\times 200 }{ 10(100-1) } =1-\frac { 40 }{ 33 } =-\frac { 35 }{ 165 } \)
\(\therefore \rho (y,z)=1-\frac { 6\sum { { d }_{ 2 }^{ 3 } } }{ n({ n }^{ 2 }-1) } \)
=\(1-\frac { 6\times 214 }{ 10(100-1) } =-\frac { 49 }{ 165 } \)
\(\therefore \rho (z,x)=1-\frac { 6\sum { { d }_{ 3 }^{ 2 } } }{ n({ n }^{ 2 }-1) } =1-\frac { 6\times 60 }{ 10(100-1) } \)
=\(1-\frac { 4 }{ 11 } =\frac { 7 }{ 11 } =\frac { 105 }{ 165 } \)
Since \(\rho (z,x)\) is maximun, the pair of judges x and y has the nearest approaches to ciommon likings in music.

Let the assumed mean of x and y be 16 and 12 resp[ectively.

n=7
\(\overline { X } =\frac { \sum { x } }{ n } =\frac { 136 }{ 7 } \)=19.43
\(\overline { Y } =\frac { \sum { y } }{ n } =\frac { 82 }{ 7 } \)=11.71
Regression co-efficients
b_xy=\(\frac { n\sum { dxdy-(\sum { dx } )(\sum { dy } ) } }{ n\sum { d{ y }^{ 2 }-{ (\sum { dy } ) }^{ 2 } } } =\frac { 7(157)-24(-2) }{ 7(94)-({ -2) }^{ 2 } } \)=1.75
b_yx=\(\frac { n\sum { dxdy-(\sum { dx } )(\sum { dy } ) } }{ n\sum { d{ y }^{ 2 }-{ (\sum { dy } ) }^{ 2 } } } =\frac { 7(157)-24(-2) }{ 7(374)-({ 24) }^{ 2 } } \)=0.56
The regression equation of x on y is
x-\(\bar{x}\) =b_xy(y-\(\bar{y}\))
x-19.43=1.75(y-11.71)
x=1.75y-1.06   ....(1)
(ii) The regression equation of y on x is
y-\(\bar{y}\)=b_yx(x-\(\bar{x}\))
y-11.71=0.56(x-19.43)
y=0.56x+0.82 ....(2)
(iii) The co-efficient of correlation
r=\(\sqrt { { b }_{ xy }.{ b }_{ yx } } =\sqrt { (0.56)(1.75) } \)=0.989
(iv) when y=5,  x=(1.75)(5)-1.06 [put y=5 in(1)]
x=7.69
(v) when x=6, y=0.56(6)+09.82 [put x=6 in (2)]
y=4.18