\(r =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{10(-115)-(50)(-30)}{\sqrt{10(290)-(50)^2} \sqrt{10(300)-(-30)^2}} \)
\(=\frac{-1150+1500}{\sqrt{400 \times 2100}}=\frac{350}{916.52}=0.382\)

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

\(\bar{X}\) = 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

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

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

Solving the two regression equations we get mean values of X and Y
2X–Y = –1 ... (1)
3X–2Y = –7 ... (2)
Solving equation (1) and equation (2) We get X = 5 and Y = 11
Therefore the regression line passing through the means \(\bar { X } \) = 5 and \(\bar { Y } \) = 11
The regression equation of Y on X is 3X–2Y = –7
2Y = 3X + 7
Y = \(\frac{1}{2}\)(3X+7)
Y = \(\frac { 3 }{ 2 } X+\frac { 7 }{ 2 } \)
∴ b_yx = \(\frac{3}{2}\)(>1)
The regression equation of X on Y is
2X–Y = –1
2X = Y–1
X = \(\frac{1}{2}\)(Y-1)
X = \(\frac{1}{2}\)Y - \(\frac{1}{2}\)
∴ b_yx = \(\frac{1}{2}\)
The regression coefficients are positive
r = ±\(\sqrt { { b }_{ xy }.{ b }_{ yx } } =\pm \sqrt { \frac { 3 }{ 2 } \times \frac { 1 }{ 2 } } \)
= \(\sqrt { \frac { 3 }{ 2 } \times \frac { 1 }{ 2 } } \)
= \(\sqrt { \frac { 3 }{ 4 } } \)
= 0.866
∴ r = 0.866

Applicant	X	Y	Rx	Ry	d=Rx-Ry	d2
A	20	30	2	1	1	1
B	50	60	8	8	0	0
C	28	50	5	6	-1	1
D	25	40	3	3	0	0
E	70	85	9	10	-1	1
F	90	90	11	11	0	0
G	76	56	10	7	3	9
H	45	82	7	9	-2	4
I	30	42	6	4	2	4
J	19	31	1	2	-1	1
K	26	49	4	5	-1	1
						\(\sum\)d2 = 22

\(\rho =1-\frac{6 \sum d^2}{n\left(n^2-1\right)} \)
\(=1-\frac{6(22)}{11(120)}=1-\frac{1}{10}=1-0.1=0.9\)