Let the sales be X and advertisement expenditure be Y
Given \(\bar { X } \) = 40, \(\bar { Y } \) = 6, σ_x = 10, σ_y = 1.5 and r = 0.9
Equation of line of regression x on y is
X -\(\bar { X } \) = r\(\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } (Y-\bar { Y } )\)
X - 40 = (0.9)\(\frac{10}{1.5}\)(Y - 6)
X - 40 = 6Y - 36
X = 6Y + 4
When advertisement expenditure is 10 crores i.e., Y = 10 then sales X = 6(10) + 4 = 64 which implies sales is 64.

Given Σxy = 120, Σx² = 90, Σy² = 640
Then r = \(\frac { \Sigma xy }{ \sqrt { \Sigma { x }^{ 2 }\Sigma { y }^{ 2 } } } =\frac { 120 }{ \sqrt { 90(640) } } =\frac { 120 }{ \sqrt { 57600 } } =\frac { 120 }{ 240 } \) = 0.5

We know that correlation coefficient
r= \(\frac { N\Sigma XY-(\Sigma X)(\Sigma Y) }{ \sqrt { N\Sigma X^{ 2 }-(\Sigma X)^{ 2 }\sqrt { N\Sigma { Y }^{ 2 }-(\Sigma Y)^{ 2 } } } } \)
= \(\frac { 9(597)-(45\times 108) }{ \sqrt { 9(285)-(45)^{ 2 }\times \sqrt { 9(1356)-(108^{ 2 }) } } } \)
r = +0.95

Let A=69, B=112, dx=x-69, dy=y-112.
Given \(\sum\)dxdy=2176, \(\bar{X}\)=74.5, \(\bar{Y}\)=125.5
\({ \sigma }_{ x }\)=13.07, \({ \sigma }_{ y }\)=15.85, n=8, A=69, B=112
we know, \(\bar{x}\)=A+\(\frac { \sum { dx } }{ n } \)
\(\Rightarrow\)74.5=69+\(\frac { \sum { dx } }{ 8 } \) \(\sum\)dx=(74.5-69)(8)
\(\Rightarrow\)\(\sum\)dx =(5.5)(8)=4
Also, \(\bar{y}\)=B+\(\frac { \sum { dy } }{ n } \)
\(\Rightarrow\)125.5=112+\(\frac { \sum { dy } }{ 8 } \)
\(\Rightarrow\)(125.5-112)8=\(\sum\)dy
\(\Rightarrow\)\(\sum\)dy=108
The correlation co-efficient between x and y
r(x, y)=\(\frac { n\sum { dxdy } -(\sum { dx } )(\sum { dy } ) }{ \sqrt { n\sum { d{ x }^{ 2 } } }-(\Sigma dx)^{ 2 } \sqrt { n\sum { d{ y }^{ 2 }-{ (\sum { dy } ) }^{ 2 } } } } \) ...(1) 
\({ \sigma }_{ x }=\sqrt { \frac { { \sum { dx } }^{ 2 } }{ n } -{ \left( \frac { \sum { dx } }{ n } \right) }^{ 2 } } =\sqrt { \frac { { \sum { dx } }^{ 2 } }{ n } -\frac { { (\sum { dx } ) }^{ 2 } }{ { n }^{ 2 } } } \)
13.07= \(\frac { \sqrt { n\sum { d{ x }^{ 2 }-{ (\sum { dx } ) }^{ 2 } } } }{ n } \)  
13.07(8)=\(\sqrt { n\sum { { dx }^{ 2 }-({ \sum { dx } ) }^{ 2 } } } \) 
\(\sqrt { n\sum { { dx }^{ 2 }-({ \sum { dx } ) }^{ 2 } } } \)=104.56 .....(2)
SImilarly, \(\sqrt { n\sum { { dy }^{ 2 }-({ \sum { dy } ) }^{ 2 } } } \)=8(15.85)-126.8   ..(3)
Substuting (2) and (3) in (1) we get,
r=\(\frac { 8(2176)-44(108) }{ (104.56)(126.8) } =\frac { 12656 }{ 13258.208 } \)
=0.9545

X	Y	X2	Y2	XY
4	2	16	4	8
2	3	4	9	6
3	2	9	4	6
4	4	16	16	16
2	4	4	16	8
15	15	49	49	44

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 are given by
b_xy= \(r.\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } \)and b_yx=\(r.\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } \)
where r is the co-efficient of correlation.

\(r=\sqrt { { b }_{ xy }.{ b }_{ yx } } \)

X	Y	RX	RY	d=RX-RY	d2
46	36	1	1	0	0
54	40	2	2	0	0
56	44	3	4	-1	1
57	54	4	6	-2	4
58	42	5	3	2	4
60	58	6	7	-1	1
62	53	7	5	2	4
					\(\sum\)d2=14

Rank correlation \(\rho =1-\frac { 6\sum { { d }^{ 2 } } }{ N({ N }^{ 2 }-1) } =1-\frac { 6(14) }{ 7(49-1) } \)
=1-\(\frac { 6(14) }{ 7(48) } =1-\frac { 84 }{ 336 } \) =1-0.25
\(\rho\)=0.75

\(\bar{X}\) = 6, \(\bar{Y}\) = 8, \(\sigma _x=5\)
\(\sigma_y=\frac{40}{3}, \quad r=\frac{8}{15} \)
\(b_{y x}=r \frac{\sigma_x}{\sigma_y}=\frac{8}{15}\left(\frac{40}{3 \times 5}\right)=\frac{64}{45}=1.422\)
Regression line of Y on X is
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 8 = 1.422(X - 6)
Y = 1.422 X - 8.532 + 8
Y = 1.422 X - 0.532 .
If  X = Rs 100,
Y = 142.2 - 0.532 = Rs 141.67

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

X	Y	RX	RY	d=RX-RY	d2
40	45	1	4	-3	9
46	46	2	5	-3	9
54	50	3	6	-3	9
60	43	4	3	1	1
70	40	5	1	4	16
80	75	6	11	-5	25
82	55	7	7	0	0
85	72	8	10	-2	4
87	65	9	8	1	1
90	42	10	2	8	64
95	70	11	9	2	4
					Σd2 =142

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

X	Y	x2	y2	xy
14	31	196	961	434
19	36	361	1296	684
24	48	576	2304	1152
21	37	441	1369	777
26	50	676	2500	1300
22	45	225	1089	495
15	33	225	1089	495
20	41	400	1681	820
19	39	361	1521	741
\(\sum\)X = 180	\(\sum\)Y = 360	\(\sum\)x2 = 3720	\(\sum\)y2 = 14746	\(\sum\)xy = 7393

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

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.

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

X	Y	X2	Y2	XY
23	18	529	324	414
27	22	729	484	594
28	23	784	529	644
29	24	841	576	696
30	25	900	625	750
31	26	961	676	806
33	28	1089	784	924
35	29	1225	841	1015
36	30	1296	900	1080
39	32	1521	1024	1248
\(\Sigma X\) = 311	\(\Sigma Y\) = 257	\(\Sigma X^2\) = 9875	\(\Sigma Y^2\) = 6763	\(\Sigma XY\) = 8171

\(r =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(8171)-(311)(257)}{\sqrt{10(9875)-(311)^2} \sqrt{10(6763)-(257)^2}} \)
\(=\frac{81710-79927}{\sqrt{98750-96721} \times \sqrt{67630-66049}} \)
\(=\frac{1783}{\sqrt{2029 \times 1581}} \)
\(r =\frac{1783}{1791.05}=0.9955\)