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

\(\bar{X}\) = 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

\(\bar{X} =\frac{\Sigma X}{N}=\frac{55}{10}=5.5 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{55}{10}=5.5 \)
\(b_{y x} =\frac{N \Sigma X Y-\Sigma X \Sigma Y}{N \Sigma X^2-(\Sigma X)^2} \)
\(=\frac{10(350)-(55)(55)}{10(385)-55^2}=\frac{3500-3025}{3850-3025} \)
\(=\frac{475}{825}=0.576\)
Regression line of Y on X is.
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 5.5 = 0.576(X - 5.5)
Y = 0.576 X - 3.168 + 5.5
Y = 0.576 X + 2.332
At  X = 6, Y = 0.576(6) + 2.332
= 3.456 + 2.332
= 5.788

X	Y	X2	Y2	XY
1	9	1	81	9
2	8	4	64	16
3	10	9	100	30
4	12	16	144	48
5	11	25	121	55
6	13	36	169	78
7	14	49	196	98
8	16	64	256	128
9	15	81	225	135
ΣX = 45	ΣY = 108	ΣX2 = 285	ΣY2 = 1356	ΣXY = 597

\(r =\frac{N \Sigma X Y-\Sigma X \Sigma Y}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{9(597)-(45)(108)}{\sqrt{9(285)-(45)^2} \sqrt{9(1356)-(108)^2}} \)
\(=\frac{5373-4860}{\sqrt{2565-2025} \sqrt{12204-11664}} \)
\(=\frac{513}{\sqrt{540 \times 540}}=\frac{513}{540}=0.95\)

Let the regression equation of Y on X be
3X + 2Y = 26
2Y = –3X + 26
Y = \(\frac{1}{2}\)(-3X + 26)
Y = -1.5X + 13
r\(\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } \) = -1.5
Implies by x = r\(\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } \) = -1.5
Let the regression equation of X on Y be
6X + 3Y = 31
X = \(\frac{1}{6}\)(-3Y + 31) = -0.5Y + 5.17
r\(\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } \) = -0.5
Implies b_xy = r\(\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } \) = -0.5
r = ±\(\sqrt { { b }_{ xy }.{ b }_{ yx } } \)
=\(-\sqrt { (-1.5).(-0.5) } \) (Since both the regression coefficient are negative r is negative)
\(\therefore\) r = -8.66

Let us consider X for price P and Y for stock S. Then the mean and SD for P is considered as \(\bar { X } \) = 100 and σ_x = 8 respectively and the mean and SD of S is considered as \(\bar { Y } \) = 103 and σ_y = 4. The correlation coefficient between the series is r(X, Y) = 0.4
Let the regression line X on Y be
\(X-\bar { X } =r\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } (Y-\bar { Y } )\)
X-100 = (0.4)\(\frac{8}{4}\)(Y-103)
X–100 = 0.8(Y–103 )
X–0.8Y–17.6 = 0 (or)  X = 0.8Y+17.6
The regression line Y on X be \(Y-\bar { Y } =r\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } (X-\bar { X } )\)
Y-103 = (0.4)\(\frac{4}{8}\)(X-100)
Y–103 = 0.2 (X–100 )
Y–103 = 0.2 X–20
Y = 0.2 X + 83 (or) 0.2 X–Y + 83 = 0

Let R_x is considered for the ranks of Statistics and R_y is considered for the ranks of mathematics.

Rx	Ry	d = Rx-RY	d2
1	1	0	0
2	4	-2	4
3	2	1	1
4	5	-1	1
5	3	2	4
6	9	-3	9
7	7	0	0
8	10	-2	4
9	6	3	9
10	8	2	4
			Σd2 = 36

The rank correlation is given by
\(\rho =1-\frac { 6\Sigma { d }^{ 2 } }{ N(N^{ 2 }-1) } =1-\frac { 6(36) }{ 10(10^{ 2 }-1) } \)
= 1 - 0.218
∴ \(\rho \) = 0.782

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

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

x	y	x2	y2	xy
5	1	25	1	5
10	6	100	36	60
5	2	25	4	10
11	8	121	64	88
12	5	144	25	60
4	1	16	1	4
3	4	9	16	12
2	6	4	36	12
7	5	49	25	35
1	2	1	4	2
\(\sum\)x = 60	\(\sum\)y = 40	\(\Sigma X^2=\) 494	\(\Sigma Y^2=\) 212	\(\Sigma XY=\) 288

Correlation Co-efficient r =\(\frac { N\sum { xy-(\sum { x)(\sum { y) } } } }{ \sqrt { N\sum { { x }^{ 2 }-({ \sum { x) } }^{ 2 }\times \sqrt { N{ \sum { y } }^{ 2 }-\left( { \sum { y } }^{ 2 } \right) } } } } \)
= \(\frac { 10(288)-(60)(40) }{ \sqrt { 10(494)-{ (60) }^{ 2 }\sqrt { 10(211)-{ (40) }^{ 2 } } } } \)
= \(\frac { 2880-2400 }{ \sqrt { 1340 } .\sqrt { 520 } } \)
= \(\frac { 480 }{ (36.61)(22.80) } =\frac { 480 }{ 834.71 } \)
r = 0.575

Let X, Y represent the marks in subject A and B respectively,
\(\bar{X}=39.5, \quad \bar{Y}=47.5 \)
\(\sigma_x=10.8 \quad \sigma_y=16.8 \quad r=0.42 . \)
\(b_{y x}=r \frac{\sigma_y}{\sigma_x}=0.42\left(\frac{16.8}{10.8}\right)=0.653\)
Regression equation of Y on X
\(Y-\bar{Y}=b_{y x}(X-\bar{X})\)
Y - 47.5 = 0.653(X - 39.5)
Y = 0.653 X - 25.79 + 47.5
Y = 0.653 X + 21.7z
At X = 52,
Y = 33.956 + 21.71 = 55.67
Hence if the candidate secures 52 marks in subject A, he will secure 55.67 marks in subject B.

R_A R_B d=R_A-R_B d² 1 6 -5 25 2 7 -5 25 3 5 -2 4 4 10 -6 36 5 3 2 4 6 9 -3 9 7 4 3 9 8 1 7 49 9 8 1 1 10 2 8 64       \(\sum\)d² = 226 \(\rho =1-\frac{6 \Sigma d^2}{n\left(n^2-1\right)}=1-\frac{6(226)}{10(99)} \)
\(=1-\frac{1356}{990}=1-1.37=-0.37\)

RX	Ry	d = RX - RY	d2
6	4	2	4
4	1	3	9
3	6	-3	9
1	7	-6	36
2	5	-3	9
7	8	-1	1
9	10	-1	1
8	9	-1	1
10	3	7	49
5	2	3	9
			\(\sum\)d2 = 128

\(\rho =1-\frac{6 \sum d^2}{n\left(n^2-1\right)} \)
\(=1-\frac{6(128)}{10(99)}=1-\frac{768}{990}=1-0.7758=0.224\)

\(r =\frac{\operatorname{cov}(x, y)}{\sigma_x \sigma_y} \)
\(=\frac{-16.5}{\sqrt{2.89 \times 100}}=\frac{-16.5}{17}=-0.97\)

\(r =\frac{11(2827)-(117)(260)}{\sqrt{11(1313)-(117)^2} \sqrt{11(6580)-(260)^2}} \)
\(=\frac{31097-30420}{\sqrt{754 \times 4780}}=0.3566\)

\(r =\frac{N \Sigma X Y-\Sigma X \Sigma Y}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{25(520)-(125)(100)}{\sqrt{25(650)-(125)^2} \sqrt{25(436)-(100)^2}} \)
= 0.667