OP Malhotra Class 12 Maths Solutions Chapter 27 Linear Regression Ex 27

The availability of S Chand Class 12 Maths Solutions Chapter 27 Linear Regression Ex 27 encourages students to tackle difficult exercises.

S Chand Class 12 ICSE Maths Solutions Chapter 27 Linear Regression Ex 27

Question 1.
Determine the equation of a straight line which best fits the data:

X	10	12	13	15	17	20	25
y	10	22	24	27	29	33	37

Solution:
Let X = x – 15 ;Y = y – 25 and let the line of best fit be Y = aX + b …(1)
The normal equations are; ΣY = aΣX + bn …(2)
and ΣXY = aΣX² + bΣX …(3)
We construct the following table :

X	X = x – 15	Y	Y = y – 25	x2	XY
10	-5	10	– 15	25	75
12	-3	22	-3	9	9
13	-2	24	-1	4	2
15	0	27	2	0	0
17	2	29	4	4	8
20	5	33	8	25	40
25	10	37	12	100	120
	ΣX = 7		Σy = 7	ΣX2 = 167	ΣXY = 254

putting all these values in eqn. (1) and eqn. (2); we have
7 = 7a + 7b …(4)
254 = 167a + 7b …(5)
eqn. (5) – eqn. (4) gives ;
247 = 160a ⇒ a = \(\frac { 247 }{ 160 }\) = 1.54375

∴ from(4); b= 1 – a = -0.54375
∴ from (1); Y = 1.54375X – 0.54375
⇒ y – 25 = 1.54375 (x – 15) – 0.54375
⇒ y = 1.54375x+ 1.3

Question 2.
Given the data

X	1	5	3	2	1	1	7	3
y	6	1	0	0	1	2	1	5

(i) Fit the regression line of y on x and hence predict y, if x = 10.
(ii) Fit the regression line of x on y and hence predict x, if y = 2.5.
Solution:
We construct the table of values is as under :

X	Y	Xy	x2	y2
1	6	6	1	36
5	1	5	25	1
3	0	0	9	0
2	0	0	4	0
1	1	1	1	1
1	2	2	1	4
7	1	7	49	1
3	5	15	9	25
Σx = 23	Σy = 16	Σxy = 36	Σx2 = 99	Σy2 = 68

(i) Thus regression line of y on x is given by

When x = 10 ∴ from (1) ; y = 2.8745 – 3.042 = -0.13

(ii) Thus, regression line of x on y is given by

Question 3.
The two lines of regression for a distribution (x, y) are 3x + 2y = 1 and x + 4y = 9. Find the regression coefficient b and b.
Solution:
Given lines are 3x + 2y = 7 …(1)
and x + 4y = 9 …(2)
Assuming line (1) be the regression line of y on x
∴ 2y = 7 – 3x ⇒ y = \(\frac { 7 }{ 2 }\) – \(\frac { 3 }{ 2 }\)x
∴ b_yx = – \(\frac { 3 }{ 2 }\) < 0

Then line (2) be the regression line of x on y
∴ x = 9 – 4y ⇒ b_yx = – 4 < 0
Now b_yx . b_xy = \(\left(-\frac{3}{2}\right)\) (-4) = 6 > 1

Thus our assumption is wrong.
∴ line (1) be the regression line of x on y
∴ 3x = 7 – 2y ⇒ x = – \(\frac { 2 }{ 3 }\) y + \(\frac { 7 }{ 3 }\)
∴ b_xy = – \(\frac { 2 }{ 3 }\)

Then line (2) be the regression line of y on x
∴ 4y = 9 – x ⇒ y = – \(\frac { x }{ 4 }\) + \(\frac { 9 }{ 4 }\)
∴ b_yx = – \(\frac { 1 }{ 4 }\)
Since b_xy . b_yx = – \(\frac { 2 }{ 3 }\) × \(\left(-\frac{1}{4}\right)\) = \(\frac { 1 }{ 6 }\) < 1
Hence, b_xy = – \(\frac { 2 }{ 3 }\) and b_yx = – \(\frac { 1 }{ 4 }\)

Question 4.
Given two lines of regression x + 3y = 11, 2x + y = 7, find the coefficient of correlation between x and y. Also estimate the value of x when y = 4.
Solution:
Given lines are; x + 3 = 11 … (1) and 2x + y = 7 …(2)
Assuming line (1) as regression line on x on y
∴ 2x = 7 – y ⇒ x = \(\frac { 7 }{ 2 }\) – \(\frac { 1 }{ 2 }\) y
∴ b_xy = – \(\frac { 1 }{ 2 }\)
Now b_xyb_yx = \(\left(-\frac{1}{2}\right)\)\(\left(-\frac{1}{3}\right)\) = \(\frac { 1 }{ 6 }\) < 1
So our assumption is true.

Since r and regression coefficients have same sign
∴ r = -0.4082 [∵ b_xyb_yx < 0]
When y = 4 ∴ from (2); 2x + 4 = 7 ⇒ x = \(\frac { 3 }{ 2 }\)

Question 5.
(i) Out of the two regression lines x + 2y – 5 = 0,2x + 3y = 8, find the line of regression of on x.
(ii) Out of the following two regression lines, find the line of regression of x on y.
3x + 12y = 9,9x + 3y = 46.
Solution:
(i) Given lines are x + 2y – 5 = 0 …(1)
and 2x + 3y = 8 …(2)
Assuming line (1) as regression line of y on x
∴ 2x = 5 – x ⇒ y = \(\frac { 5 }{ 2 }\) – \(\frac { x }{ 2 }\)
∴ b_yx = – \(\frac { 1 }{ 2 }\) < 0
Then line (2) as regression line of x on y
∴ 2x = 8 – 3y ⇒ x = 4 – \(\frac { 3 }{ 2 }\) y
∴ b_xy = – \(\frac { 3 }{ 2 }\)
Since b_xy . b_yx = \(\frac { 1 }{ 2 }\) × \(\left(-\frac{3}{2}\right)\) = \(\frac { 3 }{ 4 }\) < 1
Therefore our assumption is true.
∴ x + 2y – 5 = 0 be the regression line of y on x.

(ii) Given lines are 3x + 12y = 9 …(1)
and 9x + 3y = 46 …(2)
Assuming line (1) as regression line of y on x
∴ 12y = 9 – 3x ⇒ y = \(\frac { 3 }{ 4 }\) – \(\frac { 1 }{ 4 }\)x
∴ b_yx = – \(\frac { 1 }{ 4 }\) < 0

Then line (2) be the regression line of x on y
∴ 9x = 46 – 3y ⇒ x = \(\frac { 46 }{ 9 }\) – \(\frac { y }{ 3 }\)
∴ b_xy = – \(\frac { 1 }{ 3 }\) < 0
Here b_yx . b_xy = \(\left(-\frac{1}{4}\right)\) \(\left(-\frac{1}{3}\right)\) = \(\frac { 1 }{ 12 }\) < 1
Thus our assumption is true ∴ 9x + 3y = 46 be the regression line of x on y.

Question 6.
For lines of regression 4x – 2y = 3 and 2x – 3y = 5, find
(i) b_xy and b_yx
(ii) P(x, y)
(iii) y when x = 3.
Solution:
Given lines are 4x – 2y = 3 …(1)
and 2x – 3y = 5 …(2)
Assuming line (1) as regression line of x on y
∴ 4x = 2y + 3 ⇒ x = \(\frac { 1 }{ 2 }\)y + \(\frac { 3 }{ 4 }\)
∴ b_xy = \(\frac { 1 }{ 2 }\) > 0
Then line (2) be the regression line of y on x
3y = 2x – 5 ∴ y = \(\frac { 2 }{ 3 }\)x – \(\frac { 5 }{ 3 }\)
∴ b_yx = \(\frac { 2 }{ 3 }\) > 0
Since b_xy . b_yx = \(\frac { 1 }{ 2 }\) × \(\frac { 2 }{ 3 }\) = \(\frac { 1 }{ 3 }\) < 1
Hence our assumption is true.
∴ line (1) be the regression line of x on y and line (2) be the regression line of y on x.
(i) ∴ b_xy = \(\frac { 1 }{ 2 }\) and b_yx = \(\frac { 2 }{ 3 }\)
(ii)
Since ρ has same sign as regression coefficients
∴ ρ = 0.58 [∵ b_xy b_yx > 0]

(iii) When x = 3 ∴ from (2); 3y = 2x – 5 ⇒ 3y = 6 – 5 = 1

Question 7.
Find (i) x and y, (ii) b_yx and b_xy (iii) ρ (x, y) when the two regression lines are 3x + 12y = 19, 9x + 3y = 46.
Solution:
Given lines are
3x + 12y = 19 …(1)
9x + 3y = 46 …(2)
Thus, A.M be the point of intersection of line (1) are (2)
eqn. (2) – 3 × eqn. (1)
3y – 36y = 46 – 57 ⇒ – 33y = -11 ⇒ y = \(\frac { 1 }{ 3 }\)
∴ from(1); 3x + 4 = 19 ⇒ x = 5
Thus x̄ = 5 and ȳ = \(\frac { 1 }{ 3 }\)
Assuming line (1) be the regression line of y on x
∴ 12y = 19 – 3x ⇒ y = – \(\frac { x }{ y }\) + \(\frac { 19 }{ 12 }\) ∴ b_yx = – \(\frac { 1 }{ 4 }\) < 0
Then line (2) be the regression line x on y
∴ 9x = 46 – 3y ⇒ x = – \(\frac { y }{ 3 }\) + \(\frac { 46 }{ 9 }\) ∴ b_xy = – \(\frac { 1 }{ 3 }\) < 0
Here b_yx . b_xy = \(\left(-\frac{1}{4}\right)\) \(\left(-\frac{1}{3}\right)\) = \(\frac { 1 }{ 12 }\) < 1
Thus our assumption is true.
∴ b_xy = – \(\frac { 1 }{ 3 }\) and b_yx = – \(\frac { 1 }{ 4 }\)

Since ρ has the same as regression coefficients.
∴ ρ = – 0.2886 [∵ b_xy,b_yx < 0]

Question 8.
If 4x – 5y + 33 = 0 and 20x – 9y – 107 = 0 are two lines of regression, find (i) the mean values of x and y, (ii) the regression coefficients b and b, (iii) the correlation coefficient between x and y, (iv) the standard deviation of y, if the variance of x is 9 , (v) the value of y for x = 3, (vi) the value of x for y = 2.
Solution:
Given lines of regression are
4x – 5y + 33 = 0 …(1)
20x – 9y – 107 = 0 …(2)

(i) Clearly A.M be the point of intersection of lines (1) and (2).
eqn. (2) – 5 × eqn. (1); We have
-9y + 25y – 107 – 165 = 0 ⇒ 16y – 272 = 0 ⇒ y = \(\frac { 272 }{ 16 }\) = 17
∴ from (1); 4x – 85 + 33 = 0 ⇒ 4x = 52 ⇒ x = 13
Thus x̄ = 13 and ȳ = 17

(ii) Assuming line (1) as regression line of y on x.
∴ 5y = 4x + 33 ⇒ y = \(\frac { 4 }{ 5 }\)x + \(\frac { 33 }{ 5 }\) ∴ b_yx = \(\frac { 4 }{ 5 }\) > 0
Then line (2) be the regression line of x on y
∴ 20x = 9y + 107 ⇒ x = \(\frac { 9 }{ 20 }\)y + \(\frac { 107 }{ 20 }\) ∴ b_xy = \(\frac { 9 }{ 20 }\) > 0
Here, b_xy . b_yx = \(\frac { 9 }{ 20 }\) × \(\frac { 4 }{ 5 }\) = \(\frac { 9 }{ 25 }\) < 1
Therefore our assumption is true.
∴ b_yx = \(\frac { 4 }{ 5 }\) and b_xy = \(\frac { 9 }{ 20 }\)

(iii) Since | ρ | = \(\sqrt{b_{y x} b_{x y}}\) = \(\sqrt{\frac{4}{3} \times \frac{9}{20}}\) = \(\sqrt{\frac{9}{25}}\) = \(\frac { 3 }{ 5 }\) = 0.6 ⇒ ρ = ± 0.6.
But ρ and both regression coeff’s have same sign.
∴ ρ = 0.6 [∵ b_xy . b_yx > 0]

(iv) Now b_yx = \(\rho \frac{\sigma_y}{\sigma_x}\) ⇒ \(\frac { 4 }{ 5 }\) = \(\frac { 3 }{ 5 }\) × \(\frac{\sigma_y}{3}\) [∵ \(\sigma_x^2\) = 9 ⇒ \(\sigma_x\) = 3]
∴\(\sigma_y\) = 3

(v) when x = 3 ∴ from (1); y = \(\frac { 4 }{ 5 }\) × 3 + \(\frac { 33 }{ 5 }\) = \(\frac { 45 }{ 5 }\) = 9

(vi) When y = 2 ∴ from (2); x = \(\frac { 9 }{ 20 }\) × 2 + \(\frac { 107 }{ 20 }\) = \(\frac { 125 }{ 20 }\) = \(\frac { 25 }{ 4 }\) = 6.25

Question 9.
Find the regression coefficient of y on x for the following data:
Σx = 24, Σy = 44, Σxy = 306, Σx² = 164, Σy² = 574, n = 4.
Solution:
Given Σx = 4; Σy = 44; Σxy = 306; Σx² = 164, Σy² = 574, n = 4.

Question 10.
For observation of pairs (x, y) of the variables X and Y, the following results are obtained. Σx = 125, Σy = 100, Σx² = 1650, Σy² = 1500, Σxy = 50 and n = 25.
Find the equation of the line of regression of x on y. Estimate the value of x if y = 5.
Solution:
Given Σx = 125, Σy = 100, Σx² = 1650, Σy² = 1500, Σxy = 50 and n = 25

Thus regression line of x on y is given by x – x̄ = b_xy (y – ȳ)
⇒ x – 5 = – \(\frac { 9 }{ 22 }\)(y – 4) ⇒ 22x – 110 = – 9y + 36 ⇒ 22x = -9y = 146 …(1)
when y = 5 ∴ from (1); 22x = – 45 + 146 = 101 ⇒ x = \(\frac { 101 }{ 22 }\) = 4.591

Question 11.
For a bivariate data, you are given the following information :
Find (i) two lines of regression (ii) coefficient of-correlation between x and y.
Solution:
Let u = x – 58 and v = y – 58
∴Σu = 46; Σu² = 3086; Σv = 9; Σv² = 483 ; Σuv = 1095 and n = 7

(i) Thus expression line of y on x be given by
y – ȳ = b_yx (x – x̄) ⇒ y – 59.29 = 0.3721 (x – 64.57)
and Regression line of x on y be given by
x – x̄ = b_xy (y – ȳ) ⇒ x – 64.57 = 2.197 (y – 59.29)

(ii) Since |ρ| = \(\sqrt{b_{x y} \cdot b_{y x}}\) = \(\sqrt{0.3721 \times 2.197}\) = \(\sqrt{0.8175}\) = 0.9042 ⇒ ρ = ± 0.9042
But ρ and regression coeff’s have same sign.
∴ ρ = 0.9042 [∵ b_xy, b_yx > 0]

Question 12.
Find the equation of two lines of regression for the data:

X	1	2	3	4	5
y	7	6	5	4	3

and hence find an estimate of y for x = 3.5 from the appropriate line of regression.
Solution:
We construct the table of values is an under:

x	y	xy	x2
1	7	7	1
2	6	12	4
3	5	15	9
4	4	16	16
5	3	15	25
Σx = 15	Σx = 25	Σx = 65	Σx = 55

Question 13.
Compute Karl Pearson’s coefficient of correlation and interpret the result. Also find the line of best fit in the following table:

X	1	2	3	4	5
y	3	1	2	5	4

Solution:
We construct the table of values is given as under:

Here \(\frac{\Sigma x}{n}=\bar{x}\) ⇒ x̄ = \(\frac { 15 }{ 5 }\) = 3 and ȳ = \(\frac{\Sigma y}{n}\) = \(\frac { 15 }{ 5 }\) = 3
Karl Parson’s coefficient of correlation ρ (x, y) = \(\frac{\Sigma d_x d_y}{\sqrt{\Sigma d_x^2} \sqrt{\Sigma d_y^2}}\) = \(\frac{6}{\sqrt{10} \sqrt{10}}\) = \(\frac{6}{10}\) = 0.6
So there is a substantial relationship between y and x.
Thus the line of best fit be given by y – ȳ = \(\frac{\Sigma d_x d_y}{\Sigma d_x^2}(x-\bar{x})\) ⇒ y – 3 = \(\frac { 6 }{ 10 }\)(x – 3)
⇒ 10y – 30 = 6x – 18 ⇒ 10y = 6x + 12 ⇒ 5y = 3x + 6

Question 14.
The marks for seven candidates in an Intelligence test

Candidate	A	B	C	D	E	F	G
Intelligence test	30	52	60	62	45	32	41
Arithmetic test	41	62	70	78	53	45	57

Calculate Karl Pearson’s coefficient for Correlation and interpret it.
Also find a line of best fit.
A candidate X scored 40 at the Intelligence Test but was absent from the Arithmetic Test.
Estimate his probable score for the latter test.
Solution:
The table of values is given as under :

Question 15.
From the following data find Karl Pearson’s coefficient of Correlation and obtain the two regression lines

X	1	2	3	4	5	6	7	8	9
y	9	8	10	12	11	13	14	16	15

Solution:
We construct the table of values is given as under:

Question 16.
Find the equation of the regression line of y on x, if the observations (x, y) are the following (1,4),(2,8),(3,2),(4,12),(5,10),(6,4),(7,6),(8,6),(9,18).
Solution:
We construct the table of values is as under :
Here n = 9

X	Y	Xy	X2
1	4	4	1
2	8	16	4
3	2	6	9
4	12	48	16
5	10	50	25
6	4	24	36
7	6	42	49
8	6	48	64
9	18	162	81
Σx = 45	Σy = 70	Σxy = 400	x2 = 285

Question 17.
(i) Consider the observations (1,2),(2,4),(3,8),(4,7),(5,10,(6,5),(7,14),(8,16),(9,2), (10,20) of the corresponding values of x and y. Use the least square line of regression to predict.
(a) The value of y when that of x is 6.5.
(b) The value of x when that of y is 9.
(ii) Find the coefficient of correlation between x and y.
Solution:
(i) Let the regression line of y on x be given by
y = ax + b …(1)
and normal eqns. are ; Σy = aΣx = bn ….(2)
Σxy = aΣx² + bΣx ….(3)
The table of values is given as under:

x	y	xy	x2	y2
1	2	2	1	4
2	4	8	4	16
3	8	24	9	64
4	7	28	16	49
5	10	50	25	100
6	5	30	36	25
7	14	98	49	196
8	16	128	64	256
9	2	18	81	4
10	20	200	100	400
Σx = 55	Σy = 88	Σxy = 586	Σx2 = 385	Σy2 = 1114

putting the values in eqns. (2) and (3); we have
88 = 55a + 10b …(4)
586 = 385a + 55b ….(5)
eqn. (5) – 7 × eqn. (4) gives;
586 – 616 = 55b – 70b ⇒ – 30 = – 15b ⇒ b = 2
∴ from (4); 88 = 55a + 20 ⇒ 55a = 68 ⇒ a = \(\frac { 68 }{ 55 }\)
∴ from (1); y = \(\frac { 68 }{ 55 }\) x + 2
when x = 65; y = \(\frac { 68 }{ 55 }\) × 6.5 + 2 = 10.56 and b_yx = \(\frac { 68 }{ 55 }\) > 0
Let the regression line of x on y be ; x = cy + d …(6)
and normal eqns. are ; Σx = cΣy + 10d …(7)
Σxy = cΣy² + dΣy …(8)
putting the values in eqn. (7) and (8); we have
55 = 88c + 10d …(9)
586 = 1114c + 88d
⇒ 293 = 557c + 44d …(10)
On solving eqn. (9) and eqn. (10); we have
c = 0.3004 ; d = 2.85648
∴ from (6) ; x = (0.3004)y + 2.85648
b_xy = 0.3004
When y = 9 ; x = 5.56
and ρ = \(\sqrt{b_{x y} \cdot b_{y x}}\) = \(\sqrt{\frac{68}{55} \times 0.3004}\) = \(\sqrt{0.3714}\) = 0.6094

Question 18.
Find the regression coefficients b and b of y on x and x on y respectively, if standard deviations of x and y are 4 and 3 respectively and coefficient of correlation between x and y is 0.8.
Solution:
Given \(\sigma_x\) = 4 and \(\sigma_y\) = 3 and ρ = 0.8
∴ b_xy = \(r \frac{\sigma_x}{\sigma_y}\) = 0.8 × \(\frac { 4 }{ 3 }\) = 1.07
and b_yx = \(r \frac{\sigma_y}{\sigma_x}\) = 0.8 × \(\frac { 3 }{ 4 }\) = 0.6

Question 19.
The correlation coefficient between x and y is 0.60 . If the variance of x = 225, the variance of y = 400, mean of x = 10 and mean of y = 20, find the equation of the regression lines of (i) y on x, (ii) x on y.
Solution:
Given
Var (x) = \(\sigma_x^2\) = 225 ⇒ \(\sigma_x\) = 15
Var (y) = \(\sigma_y^2\) = 400 ⇒ \(\sigma_y\) = 20 and x̄ = 10; ȳ = 20
Also coefficient of correlation ρ = 0.6
∴ b_xy = \(\rho \frac{\sigma_x}{\sigma_y}\) = 0.6 × \(\frac { 15 }{ 20 }\) = 0.6 × \(\frac { 3 }{ 4 }\) = 0.45
and b_yx = \(\rho \frac{\sigma_y}{\sigma_x}\) = 0.6 × \(\frac { 20 }{ 15 }\) = 0.6 × \(\frac { 4 }{ 3 }\) = 0.8
Thus regression line of y on x be given by
y – ȳ = b_xy (x – x̄) ⇒ y – 20 = 0.8 (x – 10) ⇒ y = 0.3x + 12
The regression line of x on y is given by
x – x̄ = b_yx (y – ȳ) ⇒ x – 10 = 0.45 (y – 20) ⇒ x = 0.45y + 1

Question 20.
The regression lines of y on x and x on y are respectively given as :
y = x + 5 and 16x = 9y + 95. If \(\sigma_y\) = 4, then find the value of x̄, ȳ \(\sigma_y\) and r_xy. Also, find the estimate of (i) x when y = 12, (ii) y when x = 30.
Solution:
The regression line of y on x be y = x + 15 …(1)
∴ b_yx = 1 > 0
The regression line of x on y be 16x = 9y + 95 …(1)
⇒ x = \(\frac { 9 }{ 16 }\)y + \(\frac { 95 }{ 16 }\) ∴ b_xy = \(\frac { 9 }{ 16 }\) > 0
∴ \(\left|r_{x y}\right|=\sqrt{b_{x y} \cdot b_{y x}}=\sqrt{\frac{9}{16} \times 1}=\frac{3}{4}=0.75 \Rightarrow r_{x y}= \pm 0.75\)
Since r_xy has the same sign as both regression coefficients
∴ r_xy = + 0.75 [∵ b_xy , b_yx > 0]
since b_yx = 1 ⇒ 1 = \(\frac{r \sigma_y}{\sigma_x}\) ⇒ 1 = 0.75 × \(\frac{4}{\sigma_x}\) ⇒ \(\sigma_x\) = 3
x̄ and ȳ can be found out by finding the point of interseccion (1) and (2).
From (1) and (2); we have
16x = 9 (x + 5) + 95 ⇒ 7x = 140 ⇒ x = 20
∴ from (1); y = 20 + 5 = 25
Thus, x̄ = 20 and ȳ = 25
When y = 12 ∴ from (2); x = \(\frac { 9 }{ 16 }\) × 12 + \(\frac { 95 }{ 16 }\) = 12.6875
When x = 30 ∴ from (1); y = 30 + 5 = 35

Question 21.
Karl Pearson’s coefficient of correlation between two variables x and y is 0.28 , their co-variance is + 7.6 . If the variance of x is 9 , find the standard deviation of y-series.
Solution:
Given Karl pearson coeff. of correlation = 0.28

Question 22.
You are given the following data:

Series	X	Y
Mean Standard deviation	6 4	8 12

Coefficient of correlation = 2/3. Find : (i) The regression coefficients b_yx and b_xy, (ii) The lines of regression, (iii) The most likely value of y when x = 10.
Solution:
Given x̄ = 6; ȳ = 8 ; \(\sigma_x\) = 4; \(\sigma_y\) = 6
and r = coeff. of correlation = \(\frac { 2 }{ 3 }\)

(i) b_yx = r\(\frac{\sigma_y}{\sigma_x}\) = \(\frac { 2 }{ 3 }\) × \(\frac { 6 }{ 4 }\) = \(\frac { 12 }{ 12 }\) = 1
b_xy = r\(\frac{\sigma_x}{\sigma_y}\) = \(\frac { 2 }{ 3 }\) × \(\frac { 4 }{ 6 }\) = \(\frac { 8 }{ 18 }\) = \(\frac { 4 }{ 9 }\)

(ii) regression line of x on y is given by
x – x̄ = b_xy (y – ȳ) ⇒ x – 6 = \(\frac { 4 }{ 9 }\) (y – 8) ⇒ 9x – 54 = 4y – 32
⇒ 9x = 4y + 22 ⇒ x = \(\frac { 1 }{ 9 }\) (4y + 22) …(1)
When y = 14 ∴ from (1); we have
x = \(\frac { 1 }{ 9 }\) (4 × 14 + 22) = \(\frac { 1 }{ 9 }\) × 78 = \(\frac { 26 }{ 3 }\) = 8.667

Examples

Question 1.
If the two regression lines of a bivariate distribution are
4x – 5y + 33 = 0 and 20x – 9y – 107 = 0
(i) calculate x and y, the arithmetic mean of x and y respectively.
(ii) estimate the value of x when y = 7.
(iii) find the variance of y when a_x = 3.
Solution:
(i) Given regression lines are 4x – 5y + 33 = 0 …(1)
and 20x – 9y – 107 = 0 ….(2)
A.M be the point of intersection of line (1) and (2).
For this, we solve (1) and (2) simultaneously.
eqn. (2) -5 eqn. (1); we have
-9y + 25y – 107 – 165 = 0
16y = + 272 ⇒ y = \(\frac { 272 }{ 16 }\) = 17
∴ from (1); 4x – 85 + 33 = 0
Thus x̄ = 13 and ȳ = 17

(ii) Assuming 4x – 5y + 33 = 0 be the regression line of y on x.
5y = 4x + 33 ⇒ y = \(\frac { 4 }{ 5 }\)x + \(\frac { 33 }{ 5 }\) ∴ b_yx = \(\frac { 4 }{ 5 }\) > 0
Then line (2) be the regression line of x on y
∴ 20x = 9y + 107 ⇒ x = \(\frac { 9 }{ 20 }\)x + \(\frac { 107 }{ 20 }\) ∴ b_xy = \(\frac { 9 }{ 20 }\) > 0
Here b_xy . b_yx = \(\frac { 9 }{ 20 }\) × \(\frac { 4 }{ 5 }\) = \(\frac { 9 }{ 25 }\) < 1
Hence our assumption is true.

(ii) When y = 7 ∴ from (2); x = \(\frac { 9 }{ 20 }\) × 7 + \(\frac { 107 }{ 20 }\) = \(\frac { 170 }{ 20 }\) = \(\frac { 17 }{ 2 }\) = 8.5

(iii) since b_xy = \(r \frac{\sigma_x}{\sigma_y}\) ….(3)
Now r² = b_xy . b_yx
\(\Rightarrow|r|=\sqrt{b_{x y} \cdot b_{y x}}=\sqrt{\frac{9}{25}}=\frac{3}{5} \quad \Rightarrow r= \pm \frac{3}{5}\)
Since r has same sign as that of regression coefficients.
∴ r = + \(\frac { 3 }{ 5 }\) [∵ b_xy . b_yx > 0]
∴ from (3); \(\frac { 9 }{ 20 }\) = \(\frac { 3 }{ 5 }\) × \(\frac{3}{\sigma_y}\) ⇒ \(\sigma_y\) = \(\frac { 9 }{ 5 }\) × \(\frac { 20 }{ 9 }\) = 4
∴ Var(y) = \(\sigma_y^2\) = 4² = 16

Question 2.
If the regression equation of x on y is given by mx – y + 10 = 0 and the equation of y on x is given by 2x + 5y = 14, determine the value of ‘m’ if the coefficient of correlation between x and y is \(\frac{1}{\sqrt{10}}\)
Solution:
Given regression line of x on y be given by
mx – y + 10 = 0 ⇒ mx = y – 10 ⇒ x = \(\frac { y }{ m }\) – \(\frac { 10 }{ m }\) ∴ b_xy = \(\frac { 1 }{ m }\)
and Regression line of y on x be given by
-2x + 5y = 14 ⇒ 5y = 14 + 2x ⇒ y = \(\frac { 2 }{ 5 }\)x + \(\frac { 14 }{ 5 }\) ∴ b_yx = \(\frac { 2 }{ 5 }\) > 0
Also given ρ = \(\frac{1}{\sqrt{10}}\)
Since \(|\rho|=\sqrt{b_{x y} \cdot b_{y x}}=\sqrt{\frac{1}{m} \times \frac{2}{5}} \Rightarrow \rho= \pm \sqrt{\frac{2}{5 m}}\)
Since ρ and both regression coeff’s have same sign
∴ ρ = \(\sqrt{\frac{2}{5 m}}\) [∵ b_xy and b_yx > 0]
Thus \(\sqrt{\frac{2}{5 m}}=\frac{1}{\sqrt{10}} \quad \Rightarrow \frac{2}{5 m}=\frac{1}{10} \quad \Rightarrow 5 m=20 \quad \Rightarrow m=4\)

Question 3.
Find the equations of the two lines of regression for the following observations:
(3,6),(4,5),(5,4),(6,3),(7,2)
Find an estimate of y for x = 2.5
Solution:
We construct the table of values is as under :

x	y	xy	x2	y2
3	6	18	9	36
4	5	20	16	26
5	4	20	25	16
6	3	18	36	9
7	2	14	49	4
Σx = 25	Σy = 20	Σxy = 90	Σx2 = 135	Σy2 = 90

The regression line of y on x be given by
y – ȳ = b_yx (x – x̄) ⇒ y – 4 = -1 (x – 5) ⇒ x + y = 9 …(1)
The regression line of x on y is given by
x – x̄ = b_xy (y – ȳ) ⇒ x – 5 = -1 (y – 4) ⇒ x = – y + 9
When x = 2.5
∴ from (1); y = 9 – 2.5 = 6.5

Question 4.
Two regression lines are represented by 2x + 3y – 10 = 0 and 4x + y – 5 = 0. Find the line of regression of y on x.
Solution:
Let us assume, 2x + 3y – 10 = 0 be the regression line of y on x.
∴ 3y = 10 – 2x ⇒ y = –\(\frac { 2x }{ 3 }\) + \(\frac { 10 }{ 3 }\) ⇒ b_yx = –\(\frac { 2 }{ 3 }\)
Then 4x + y – 5 be the line of regression of x on y
Then 4x = -y + 5 ⇒ x = – \(\frac { 1 }{ 4 }\)y + \(\frac { 5 }{ 4 }\) ∴ b_xy = –\(\frac { 1 }{ 4 }\)
Now b_yxb_xy = –\(\frac { 2 }{ 3 }\) × \(\left(-\frac{1}{4}\right)\) = \(\frac { 1 }{ 6 }\) < 1
∴ 2x + 3y – 10 = 0 be the regression line of y on x.

Question 5.
From the equations of the two regression lines, 4x + 3y + 1 = 0 and 3x + 4y + 8 = 0, find : (i) mean of x and y (ii) regression coefficients (iii) coefficient of correlation.
Solution:
(i) Given regression lines re
4x + 3y + 7 = 0 ….(1)
3x + 4y + 8 = 0 …(2)
Since (x̄, ȳ) be the common point on both lines (1) and (2)
On solving (1) and (2); we have

Question 6.
The following results were obtained with respect to two variables x and y.
Σx = 30, Σy = 42, Σxy = 199, Σx² = 184, Σy² = 318, n = 6
Find the following :
(i) the regression coefficients
(ii) correlation coefficient between x and y
(iii) regression equation of y on x
(iv) the likely value of y when x =10.
Solution:
Given Σx = 30, Σy = 42, Σxy = 199, Σx² = 184, Σy² = 318, n = 6

(ii) correlation coefficient between x and y = | r | = \(\sqrt{b_{x y} b_{y x}}\)
| r | = \(\sqrt{(-0.3235) \times(-0.4583)}\) = 0.385
Since the sign of b_xy, b_yx and r should be same
∴ r = – 0.385

(iii) The regression equation of y on x is given by

Question 7.
Two regression lines are represented by 4x + 10y = 9 and 6x + 3y = 4. Find the line of regression of y on x.
Solution:
(i) Let us assume, 4x + 10y = 4 be the regression ineof y on x.
Thus, 10y = -4x + 4 ⇒ y= – \(\frac { 2 }{ 5 }\)x + \(\frac { 2 }{ 5 }\) ⇒ b_yx = –\(\frac { 2 }{ 5 }\)
Then 6x + 3y = 4 be the regression line of x on

(ii) 6x = 3 – 3y ⇒ x = \(\frac { y }{ 2 }\) + \(\frac { 2 }{ 3 }\) ⇒ b_xy = –\(\frac { 1 }{ 2 }\)
Now, b_xy, b_yx = \(\left(-\frac{2}{5}\right)\left(-\frac{1}{2}\right)=\frac{1}{5}<1\)
Thus, our assumption is true and hence 4x + 10y = 4 be the regression line of y on x.

Question 8.
For the given lines of regression, 3x – 2y = 5 and x – 4y = 7, find :
(i) regression coefficients b and b
(ii) coefficient of correlation r(x, y)
Solution:
(i) Given regression lines are ; 3x – 2y = 5
and x – 4y = 7
Assuming line (1) as regression line of x on y.
∴ 3x = 2y + 5 ⇒ x = \(\frac { 2 }{ 3 }\)y + \(\frac { 5 }{ 3 }\) ∴ b_xy = \(\frac { 2 }{ 3 }\) > 0
Then line (2) be the regression line of x on y
∴ 4y = x – 7 ⇒ y = \(\frac { x }{ 4 }\) – \(\frac { 7 }{ 4 }\) ∴ b_yx = \(\frac { 1 }{ 4 }\) > 0
Here b_xy . b_yx = \(\frac { 2 }{ 3 }\) × \(\frac { 1 }{ 4 }\) = \(\frac { 1 }{ 6 }\) < 1
Hence our assumption is true.

(ii) Since | r | = \(\sqrt{b_{x y} \times b_{y x}}=\sqrt{\frac{2}{3} \times \frac{1}{4}}=\frac{1}{\sqrt{6}} \Rightarrow r= \pm \frac{1}{\sqrt{6}}\)
Since r and both regression coefficients have same sign.
∴ r = \(\frac{1}{\sqrt{6}}\) [ ∵ b_xy and b_yx > 0]

Question 9.
The coefficient of correlation between the values denoted by X and Y is 0.5 . The mean of X is 3 and that 7 is 5 . Their standard deviations are 5 and 4 respectively. Find
(i) the two lines of regression,
(ii) the expected value of Y, when X is given 14,
(iii) the expected value of X, when Y is given 9.
Solution:
Here given r = 0.5; x̄ = 3 and ȳ = 5 ; \(\sigma_{\mathrm{X}}\) = 5 ; \(\sigma_{\mathrm{Y}}\) = 4
∴ b_YX\(r \frac{\sigma_{\mathrm{Y}}}{\sigma_{\mathrm{X}}}\) = 0.5 × \(\frac { 4 }{ 5 }\) = 0.4
and b_XY\(r \frac{\sigma_{\mathrm{X}}}{\sigma_{\mathrm{Y}}}\) = 0.5 × \(\frac { 5 }{ 4 }\) =\(\frac { 2.5 }{ 4 }\) = 0.625

(i) The regression line of Y on X is given by
Y – ȳ = b_YX(X – x̄) ⇒ Y – 5 = 0.4 (X – 3) ⇒ Y = 0.4X + 3.8 …(1)
The regression line of X on Y is given by
X – x̄ = b_XY (Y – ȳ) ⇒ X – 3 = 0.625 (Y – 5) ⇒ X = 0.625Y – 0.125 …(2)

(ii) When X = 14 ∴ from eqn. (1); Y = 0.4 × 14 + 3.8 = 5.6 + 3.8 = 9.4
(iii) When Y = 9 ∴ from eqn. (2); X = 0.625 × 9 – 0.125 = 5.5

Question 10.
The two lines of regressions are 4x + 2y – 3 = 0 and 3x + 6y + 5 = 0. Find the correlation coefficient between x and y.
Solution:
Given regression lines are ;
4x + 2y – 3 = 0 …(1)
and 3x + 6y + 5 = 0 ….(2)
Assuming line (1) as regression line of x on y.
∴ 4x = 3 – 2y ⇒ x = – \(\frac { y }{ 2 }\) + \(\frac { 3 }{ 4 }\) ∴ b_xy = – \(\frac { 1 }{ 2 }\) < 0
Then line (2) be the regression line of y on x.
∴ 6y = -3x – 5 ⇒ y = – \(\frac { x }{ 2 }\) – \(\frac { 5 }{ 6 }\) ∴ b_yx = – \(\frac { 1 }{ 2 }\) < 0
since |r|= \(\sqrt{b_{x y} \cdot b_{y x}}\) = \(\sqrt{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)}\) = \(\frac{1}{2}\)
but r and both regression coefficients have same sign.
∴ r = – \(\frac{1}{2}\) [∵ b_xy , b_yx < 0]

Question 11.
Given that the observations are (9,-4),(10,-3),(11,-1),(12,0),(13,1),(14,3),(15,5),(16,8). Find the two lines of regression and estimate the value of y when x = 13.5.
Solution:
We construct the table of values is as under :

x	y	xy	x2	y2
9	-4	-36	80	16
10	-3	-30	100	9
11	-1	-11	121	1
12	0	0	144	0
13	1	13	169	1
14	3	42	196	9
15	5	75	225	25
16	8	128	256	64
Σx = 100	Σy = 9	Σxy = 181	Σx2 = 1296	Σy2 = 125

Thus regression line of y on x be givne by

Question 12.
The two lines of regression are x + 2y – 5 = 0 and 2x + 3y – 8 = 0 and the variance of x is 12 . Find the variance of y and the coefficient of correlation.
Solution:
The given regression lines are
x + 2y – 5 = 0 …(1)
2x + 3y – 8 = 0 …(2)
Assuming line (1) as regression line of y on x.
∴ 2y = 5 – x ⇒ y = –\(\frac{1}{2}\)x + \(\frac { 5 }{ 2 }\) ∴ b_yx = – \(\frac { 1 }{ 2 }\) < 0
Then line (2) be the regression line of x on y.
∴ 2x = 8 – 3y ⇒ x = –\(\frac{3}{2}\)y + 4 ∴ b_xy = – \(\frac { 3 }{ 2 }\) < 0
Here b_xy , b_yx = – \(\frac { 3 }{ 2 }\) × \(\left(-\frac{1}{2}\right)\) = \(\frac { 3 }{ 4 }\) < 1
Thus our supposition is true.
Since | r | = \(\sqrt{b_{x y} \cdot b_{y x}}\) = \(\sqrt{\left(-\frac{3}{2}\right) \times\left(-\frac{1}{2}\right)}\) = \(\sqrt{\frac{3}{4}}\) = 0.8660 ⇒ r = ± 0.8660
But r and both regression coefficient have same sign
∴ r = -0.8660 [∵ b_xy and b_yx < 0]
given \(\sigma_x^2\) = 12 ⇒ \(\sigma_x\) = \(\sqrt{12}\)
Now b_yx = \(r \frac{\sigma_y}{\sigma_x}\) ⇒ – \(\frac { 1 }{ 2 }\) = – 0.8660 × \(\frac{\sigma_y}{\sqrt{12}}\)
⇒ \(\sigma_y\) = + \(\frac{1}{2}\) × \(\frac{\sqrt{12}}{0.8660}\) = 2
∴ Var(y) = \(\sigma_y^2\) = \(2^2\) = 4

Question 13.
A textbook publisher finds that the production costs to each book are Rs. 45 and that the fixed costs are Rs.90,000. If each book can be sold for Rs. 75, then what is the break even point ?
(a) 2000 copies
(b) 2500 copies
(c) 3000 copies
(d) 3500 copies
Solution:
Let the number of books produced by publisher be x.
∴ Cost of x books @ Rs. 45 = V(x) = Rs.45x
Thus cost function C(x) = F + V(x) = 90000 + 45x
and Revenue function R(x) = 75x
For break even point, C(x) = R(x)
⇒ 90000 + 45x = 75x
⇒ 30x = 90000 ⇒ x = 3000 ∴ Ans. (c)

Question 14.
The total revenue received from the sale of x units of a product is given by R(x) = 13 x² + 26x + 15. Find the marginal revenue when x = 7.
(a) 200
(b) 208
(c) 280
(d) 182
Solution:
Given R(x) = 13 x² + 26x + 15
∴ MR = \(\frac{d}{d x}\)R(x) = 26x + 26
When x = 7; MR = 26 x 7 + 26 = 208
∴ Ans. (b)

Question 15.
The demand function of a monopolist is given by p = 1500 -2x – x². Find the marginal revenue (MR) when x = 10.
(a) 820
(b) 1200
(c) 800
(d) None of these
Solution:
Given demand function = p(x)
=1500 – 2x – x²
∴ revenue function = R(x) = xp(x)
= 1500 x – 2x² – x³
∴ MR = \(\frac{d}{d x}\)R(x) = 1500 – 4x – 3x²
When x = 10 ; MR = 1500 – 40 – 300
= 1160 ∴ Ans. (d)

Question 16.
A manufacturer can sell x items (x 0) at a price Rs. (330 – x) each. The cost of producing x items is Rs. (x² + 10x + 12). How many items should he sell to make the maximum profit ?
(a) 100
(b) 120
(c) 90
(d) 80
Solution:
Given Revenue function =R(x) = Rs.(x) (330 – x)
and Cost function = C(x) = Rs. (x² + 10x + 12)
∴ P(x) = R(x) – C(x)
= (330x – x²) – (x² + 10x + 12)
⇒ P(x) = -2 x² + 320x – 12
\(\frac{d \mathrm{P}}{d x}\) = -4x + 320 and \(\frac{d^2 \mathrm{P}}{d x^2}\) = -4
For maxima/minima ;
\(\frac{d \mathrm{P}}{d x}\) = 0 ⇒ -4x + 320 = 0 ⇒ x = 80
∴ Maximum profit = -2 (80)² + 320 × 80 – 12
= – 1280o + 25600 – 12 = 12782
∴ Ans. (d)

Question 17.
Find the total cost of producing 600 pens, if the marginal cost C'(x) is given by C'(x) = \(\frac{x}{600}\) + 2.50, when the output is x.
(a) Rs. 2000
(b) Rs. 1500
(c) Rs. 1800
(d) None of these
Solution:
Given Marginal cost = MC = C'(x) = \(\frac { x }{ 600 }\) + 2.50
∴Total cost of producing 600 pens =C(600)

Question 18.
Find the value of coefficient of correlation if the two regression coefficients are 0.4 and 0.9 .
(a) 0.6
(b) ±0.6
(c) – 0.6
(d) 0.13
Solution:
Let b_xy = 0.4 > 0 and b_yx = 0.9 > 0 ∴ r > 0
⇒ r = \(\sqrt{b_{x y} b_{y x}}\) = \(\sqrt{0.4 \times 0.9}\) = \(\sqrt{0.36}\) = 0.6 ∴ Ans. (a)

Question 19.
Find the regression coefficient b_yx, if \(\sigma_x\) = 11, \(\sigma_y\) = 8 and ρ(x, y) = 0.66.
(a) 0.9075
(b) \(\frac { 8 }{ 11 }\)
(c) 0.48
(d) 0.84
Solution:
We know that,
b_yx = r\(\frac{\sigma_y}{\sigma_x}\) = 0.66 + \(\frac { 8 }{ 11 }\) = 0.48 ∴ Ans. (c)

Question 20.
Find x̄ and ȳ, the arithmetic means of x and y respectively if the two regression lines of a bivariance distribution are x + 2y = 5 and 2x + 3y = 8.
(a) x̄ = 2, ȳ = 1
(b) x̄ = 1, ȳ = 2
(c) x̄ = \(\frac { 1 }{ 2 }\), ȳ = 1
(d) None of these
Solution:
Given x̄ + 2ȳ = 5 ….(1)
2 x̄ + 3 ȳ = 8 …(2)
eqn. (2)-2 × eqn. (1); we have
– ȳ = 8 – 10 ⇒ ȳ = 2
∴ from (1); x̄ = 5 – 4 = 1 ∴ Ans. (b)

Question 21.
Which of the following statements is false ?
(a) Correlation coefficient is the geometric mean between the two regression coefficients.
(b) If one of the regression coefficients is greater than unity numerically, the other must be less than unity numericaliy.
(c) The correlation coefficient and the two regression coefficients have the same sign, i.e., b_yx and b_xy have the same sign as that of r.
(d) If the two regression coefficients are 0.4 and 0.9 , then r = ±0.6.
Solution:
Statement (d) is false.
Since both regression coefficients are positive.
∴ r > 0
∴ Ans. (d)

Question 22.
If the two lines of regression are 3x – 2y – 10 = 0 and 24x – 25y + 145 = 0, then the value of r is
(a) \(\frac { 5 }{ 4 }\)
(b) \(\frac { 4 }{ 5 }\)
(c) \(\frac { 16 }{ 25 }\)
(d) None of these
Solution:
Assuming 3x – 2y – 10 = 0 be the line of regression of x on y
∴ 3x = 2y + 10
⇒ x = \(\frac { 2 }{ 3 }\)y + \(\frac { 10 }{ 3 }\) ⇒ b_xy = \(\frac { 2 }{ 3 }\)
Then 24x – 25y + 145 = 0 be the regression line of y on x be 25y = 24x + 145
⇒ y = \(\frac { 24 }{ 25 }\)x + \(\frac { 145 }{ 25 }\) ⇒ b_yx = \(\frac { 24 }{ 25 }\)
∴ b_xy . b_yx = \(\frac { 2 }{ 3 }\) × \(\frac { 24 }{ 25 }\) = \(\frac { 16 }{ 25 }\) < 1
Hence our assumption is correct.
∴ r = \(\sqrt{b_{x y} b_{y x}}\) = \(\sqrt{\frac{16}{25}}\) = \(\frac { 4 }{ 5 }\) = 0.8
∴ Ans. (b)

Question 23.
The marginal revenue function (MR) when output is $x$ units is given by MR = 6 – 2x – 3x².
(a) 6 – x – x²
(b) 6 + x + x²
(c) – 6 – x- x²
(d) 6 – x + x²
Solution:
Given MR = \(\frac{d \mathrm{R}(x)}{d x}\) = 6 – 2x = 3x²

Question 24.
The total cost C (x) associated with the producing of x units of an item is given by C(x) = 0.005x³ – 0.02x² + 30x + 5000. Find the marginal cost when 3 units are produced.
Solution:
Given cost function C (x)
= 0.005x³ – 0.02x² + 30x + 5000
∴ \(\frac{d \mathrm{C}}{d x}\) = MC = 0.015 x² – 0.04x + 30
When x = 3;
MC = 0.015 × 9 – 0.04 × 3 + 30
= 0.135 – 0.12 + 30 = 30.015

Question 25.
Given the total cost function for x units of a commodity are
C(x) = \(\frac { 1 }{ 4 }\)x³ + x² – 6x,
find the (i) marginal cost (ii) average cost (iii) the slope of marginal cost function and also (iv) the marginal cost when 80 units of the commodity are sold.
Solution:
Given C(x) = \(\frac{x^3}{4}\) + x² – 6x

Question 26.
The average cost function associated with producing and, marketing x units of an item is given by AC =2x – 11 + \(\frac{50}{x}\). Find (i) the total cost function and (ii) marginal cost function.
Solution:
Given AC (x) = 2x – 11 + \(\frac{50}{x}\)
⇒ \(\frac{\mathrm{C}(x)}{x}\) = 2x – 11 + \(\frac{50}{x}\)
⇒ C(x) = 2x² – 11x + 50
∴ MC (x) = \(\frac{d}{d x}\) C(x) = 4x – 11

Question 27.
For the first year the fixed cost for setting up a new company for producing of ceiling fans the investment is Rs. 30,24,000. The variable cost for producing a fan is Rs. 120. The revenue from the sale of fan is expected to be R. 3000 per fan. Find (i) the revenue function, (ii) the break even point (iii) Find the number of fan produced for which company will incur loss.
Solution:
Fixed cost = F = Rs. 3024000
(i) Let the no. of fans produced by company be x units.
∴ V(x) = 120x [∵ cost price of each fan = Rs. 120]
∴ C(x) = F + V(x) = 3024000 + 120x
and R(x) = revenue function = 3000x

(ii) For break even point, C(x) = R(x)
3024000 + 120x = 3000x
⇒ 2280x = 3024000
⇒ x = \(\frac{3024000}{2880}\) = 1050
at breakeven point, there will be no profit and no loss.

(iii) Thus the company incur loss if it produced less then 1050 fans.

Question 28.
The cost function of a firm is given by C(x) = 300 – 10x² + \(\frac{x^3}{3}\). Calculate the output at which the marginal cost is minimum.
Solution:
Given C(x) = 300x – 10x² + \(\frac{x^3}{3}\)
∴ MC(x) = \(\frac{d}{d x}\)C(x) = 300 – 20x + x²
∴ \(\frac{d}{d x}\) MC = -20 + 2x
and \(\frac{d^2}{d x^2}\) MC = 0
⇒ – 20 + 2x = 0 ⇒ x = 10
at x = 10; \(\frac{d^2}{d x^2}\) MC = 2 > 0
Thus, marginal cost (MC) is minimum when output x = 10 units.

Question 29.
If the two regression coefficients are 0.3 and 1.2, the find the value of coefficient of correlation.
Solution:
Since both regression coefficients are 0.3 and 1.2 i.e. positive
∴ correlation coefficient r > 0
and r = \(\sqrt{0.3 \times 1.2}\) = \(\sqrt{0.36}\) = 0.6

Question 30.
Find the two regression coefficients if standard deviation of x and y are 5 and 4 respectively and coefficient of correlation between x and y is 0.5 .
Solution:
Given \(\sigma_x\) = 5 ;\(\sigma_y\) = 4 and r = 0.5
∴ b_yx = r\(\frac{\sigma_y}{\sigma_x}\) = 0.5 × \(\frac{4}{5}\) = 0.4
and b_xy = r\(\frac{\sigma_x}{\sigma_y}\) = 0.5 × \(\frac{5}{4}\)
= \(\frac{25}{40}\) = \(\frac{5}{8}\) = 0.625

Question 31.
The equation of two regression lines are 4x + 3y + 2 = 0 and 3x + 4y – 9 = 0. Find the mean of x and y.
Solution:
Let x̄ and ȳ be the mean of x and y and x̄ , ȳ must satisfies given regression lines. Thus, 4 x̄ + 3 ȳ + 2 = 0
and 3 x̄ + 4 ȳ – 9 = 0
3 × eqn. (1) – 4 × eqn. (2); we have
-7 ȳ + 42 = 0 ⇒ ȳ = 6
∴ from (1); 4 x̄ + 18 + 2 = 0 ⇒ x̄ = -5

Question 32.
Two lines of regression are 4x + 2y – 3 = 0 and 3x + 6y + 5 = 0. Find the correlation coefficient between x and y.
Solution:
Let us assume 4x + 2y – 3 = 0 be the regression line of x on y.
⇒ 4x = 3 – 2y
⇒ x = –\(\frac{y}{2}\) + \(\frac{3}{4}\) ⇒ b_xy = – \(\frac{1}{2}\) < 0
Then 3x + 6y + 5 = 0 be the regression line of y on x
i.e. 6y = -3x – 5 ⇒ y = –\(\frac{x}{2}\) – \(\frac{5}{6}\)