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

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

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

\(r =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{25(520)-(125)(100)}{\sqrt{25(650)-(125)^2} \sqrt{25(436)-(100)^2}} \)
\(=\frac{13000-12500}{\sqrt{625 \times 900}}=\frac{500}{(25)(30)}\)
\(r=\frac{500}{750}=0.667\)

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

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 n = 15, \(\sigma_x=3.01, \sigma_y=3.03 \)
\(\Sigma(X-\bar{X})(Y-\bar{Y})=122 \)
\(r(x, y) =\frac{1}{N} \frac{\Sigma(X-\bar{X})(Y-\bar{Y})}{\sigma_x \sigma_y} \)
\(=\frac{122}{15 \times 3.01 \times 3.03} \)
\(=\frac{122}{136.8045}=0.8918\)

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.

X	Y	Rx	Ry	d = Rx-Ry	d2
30	27	2	1	1	1
45	48	4	5	-1	1
80	73	7	7	0	0
25	29	1	2	-1	1
50	63	6	6	0	0
97	87	8	8	0	0
47	39	5	3	2	4
40	45	3	4	-1	1
					\(\sum\)d2 = 8

\(\rho =1-\frac{6 \sum d^2}{n\left(n^2-1\right)}=1-\frac{6(8)}{8(63)} \)
\(=1-\frac{6}{63}=1-0.095=0.905\)

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 \(\rho =1-\frac{6 \Sigma d^2}{n\left(n^2-1\right)}=1-\frac{6(226)}{10(99)} \)
\(=1-\frac{1356}{990}=1-1.37=-0.37\)

X	Y	Rx	Ry	d=Rx-Ry	d2
300	273	2	1	1	1
450	486	4	5	-1	1
800	734	7	7	0	0
250	297	1	2	-1	1
500	631	6	6	0	0
975	872	8	8	0	0
475	396	5	3	2	4
400	457	3	4	-1	1
					\(\sum\)d2 = 8

\(\rho =1-\frac{6 \Sigma d^2}{n\left(n^2-1\right)} \)
\(=1-\frac{6(8)}{8(63)} \)
= 1-0.0952
= 0.905