OP Malhotra Class 11 Maths Solutions Chapter 29 Correlation Analysis Ex 29(a)

Continuous practice using ISC Class 11 Maths Solutions S Chand Chapter 29 Correlation Analysis Ex 29(a) can lead to a stronger grasp of mathematical concepts.

S Chand Class 11 ICSE Maths Solutions Chapter 29 Correlation Analysis Ex 29(a)

Question 1.
A physicist is experimenting with the resistance in a circuit she is using. She measures and records the resulting current.

Resistance (ohms)	5	10	15	20	25	30	40
Current (amps)	10	4.9	3.2	2.4	1.9	1.7	1.0

(i) Draw a scatter graph of her results.
(ii) Estimate the current for a resistance of 40 ohms.
(iii) Estimate the resistance for a current of 7.5 amps.
Solution:
(i) Plot the points (5, 10), (10, 4.9), (15, 3.2), (20, 2.4), (25, 1.9), (30, 1.7) and (50, 1.0) on graph paper.

(ii) Clearly from scatter diagram, the corresponding current for a resistance of 40 ohms be 1.3 amps.
(iii) Clearly from scatter diagram, the corresponding value of resistance for current of 7.5 amps be 6.5 ohms.

Question 2.
In a small survey the heights of eight boys were measured and their shoe sizes were recorded.

Height (cm)	172	182	164	190	167	169	175	185
Shoe size	8 1/2	10 1/2	7	13	8 1/2	8	10	12

Draw a scatter graphs and use it to find out whether there is a relationship between these sets of data.
Solution:

The plotted points are approximately lie along a straight line suggesting that the shoe size of a boy is related to his height.
TYPE 2. (Based on first formula : \(r=\frac{\Sigma d_x d_y}{\sqrt{\left(\Sigma d_x^2\right)\left(\sum d_y^2\right)}}\), where d_x = x – \(\bar{x}\), d_y = y – \(\bar{y}\))

Question 3.
Calculate Karl Pearson’s coefficient of correlation between the values of X and Y for the following data. Comment on the value of r.

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

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

X	Y	X – X̄ X̄ = 3	Y = ȳ ȳ = 5	(X – X̄ ) (Y – ȳ)	(X – X̄)2	(Y – ȳ)2
1	7	-2	2	-4	4	4
2	6	-1	1	-1	1	1
3	5	0	0	0	0	0
4	4	1	-1	1	1	1
5	3	2	-2	-4	4	4
		Σ(X – X̄) = 0		Σ (X – X̄ ) (Y – ȳ) = – 10	Σ(X – X̄)2 = 10	Σ(Y – ȳ)2 = 10

Here \(\bar{X}\) = \(\frac{1+2+3+4+5}{5}\) = \(\frac{15}{5}\) = 3
and \(\bar{Y}\) = \(\frac{7+6+5+4+3}{5}\) = \(\frac{25}{5}\) = 5
Thus coefficient of correlation

since r = – 1, which shows perfect negative correlation between X and Y.

Question 4.

X	1	2	3	4	5	6	7	8	9
Y	12	11	13	15	14	17	16	19	18

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

X	Y	dX = X – X̄	dy = Y – ȳ	d2x	d2y	dxdy
1	12	-4	-3	16	9	12
2	11	-3	-4	9	16	12
3	13	-2	-2	4	4	4
4	15	-1	0	1	0	0
5	14	0	-1	0	1	0
6	17	1	2	1	4	2
7	16	2	1	4	1	2
8	19	3	4	9	16	12
9	18	4	3	16	9	12
		ΣX = 45	ΣY = 135	Σd2x = 60	Σd2y = 60	Σdxdy = 56

Here \(\bar{X}\) = \(\frac{\Sigma X}{n}\) = \(\frac{45}{9}\) = 5
and \(\bar{Y}\) = \(\frac{\Sigma Y}{n}\) = \(\frac{135}{9}\) = 15
Karl Pearson’s coeff. of correlation \(r=\frac{\Sigma d_{\mathrm{X}} d_{\mathrm{Y}}}{\sqrt{\Sigma d_{\mathrm{X}}^2} \sqrt{\Sigma d_{\mathrm{Y}}^2}}\) = \(\frac{56}{\sqrt{60} \sqrt{60}}=\frac{56}{60}\) = \(\frac{14}{15}\) = 0.933
which shows that their is a high positive correlation between X and Y.

Question 5.

	X series	Y series
Number of pairs of observation	15	15
Arithmetic mean	25	18
Standard deviation	3.01	3.03
Sum of the squares of deviation from the mean	136	138
Sum of the product of the deviations of x and y	–
series from their respective means	122

Solution:
Given sum of the squares of deviation from the mean of series \(\mathrm{X}=d_{\mathrm{X}}^2=\Sigma(\mathrm{X}-\overline{\mathrm{X}})=136\)
\(d_{\mathrm{Y}}^2=\Sigma(\mathrm{Y}-\overline{\mathrm{Y}})^2=138\)
\(d_{\mathrm{X}} d_{\mathrm{Y}}=\Sigma(\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})=122\)
∴ \(r=\frac{\Sigma d_{\mathrm{X}} d_{\mathrm{Y}}}{\sqrt{\Sigma d_{\mathrm{X}}^2} \sqrt{\Sigma d_{\mathrm{Y}}^2}}=\frac{122}{\sqrt{136} \sqrt{138}}=0.89\)
So there is high positive correlation between X and Y.

Question 6.
Calculate the Pearson’s coefficient of correlation between the ages of husband and wife.

Age of husband	35	34	40	43	56	20	38
Age of wife	32	30	31	32	53	20	33

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

Age of husband x	Age of wife y	dx = x – x̄ x̄ = 38	dy = y – ȳ ȳ = 33	dxdy	dx2	dy2
35	32	-3	-1	3	9	1
34	30	-4	-3	12	16	9
40	31	2	-2	-4	4	4
43	32	5	-1	-5	25	1
56	53	18	20	360	324	400
20	20	-18	-13	234	324	169
38	33	0	0	0	0	0
Σx = 266	Σy = 231			Σdxdy = 600	Σdx2 = 702	Σdy2 = 584

\(\bar{x}\) = \(\frac{\Sigma x}{n}\) = \(\frac{266}{7}\) = 38
and \(\bar{y}\) = \(\frac{\Sigma y}{n}\) = \(\frac{231}{7}\) = 33
\(r=\frac{\Sigma d_x d_y}{\sqrt{\Sigma d_x^2} \sqrt{\Sigma d_y^2}}=\frac{600}{\sqrt{702} \sqrt{584}}=0.937\)

Question 7.
Given r = 0.8, Σxy = 60, σ_y = 2.5 and Σx² = 90, find the number of items. x and y are deviations from their respective mean.
Solution:
Given r = 0.8 ; Σxy = Σ(x – \(\bar{x}\))(y – \(\bar{y}\))=60 ; σ_y = 2.5 ; Σx² = Σ(x – \(\bar{x}\))² = 90

Calculate Karl Pearson’s coefficient of correlation between the values of x and y for the following data.

Question 8.
(1,2), (2,4), (3,8), (4,7), (5,10), (6,5), (7,14), (8,16), (9,2), (10,20)
Solution:
We construct 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

Question 9.

X	-3	-2	-1	0	1	2	3
Y	9	4	1	0	1	4	9

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

X	Y	XY	X2	Y2
-3	9	-27	9	81
-2	4	-8	4	16
-1	1	-1	1	1
0	0	0	0	0
1	1	1	1	1
2	4	8	4	16
3	9	27	9	81
ΣX = 0	ΣY = 28	ΣXY = 0	ΣX2 = 28	ΣY2 = 196

Question 10.
n = 50, Σx = 75, Σy = 80, Σx² = 150, Σy² = 140, Σxy = 120.
Solution:
Given n = 50 ; Σx = 75 ; Σy = 80 ; Σx² = 150 ; Σy² = 140 and Σxy = 120

Question 11.
n = 10, Σx = 55, Σy = 40, Σx² = 385, Σy² = 192 and Σ(x + y)² = 947.
Solution:
Given n = 10 ; Σx = 55 ; Σy = 40 ; Σx² = 385, Σy² = 192
and Σ(x + y)² = 947
⇒ 948 = Σ(x² + y² + 2xy)
⇒ 947 = Σx² + Σy² + 2Σxy
⇒ 947 = 385 + 192 + 2Σxy
⇒ 947 = 577 + 2Σxy
⇒ 2Σxy = 370
⇒ Σxy = 185

Where u = X – A or \(\frac{\mathbf{X}-\mathbf{A}}{h}\), v = Y – B or \(\frac{\mathbf{Y}-\mathbf{B}}{k}\), A and B being assumed means.

Question 12.

X	16	18	21	20	22	26	27	15
Y	22	25	24	26	25	30	33	14

Solution:
Let Assumed mean for serics X be 20 i.e. A = 20 and for serics Y be 25 i.e. B = 25, Here n = 8 We construct the table of values is an under:

X	Y	u = X – 20	V = Y – 25	uv	u2	v2
16	22	-4	-3	12	16	9
18	25	-2	0	0	4	0
21	24	1	-1	-1	1	1
20	26	0	1	0	0	1
22	25	2	0	0	4	0
26	30	6	5	30	36	25
27	33	7	8	56	49	64
15	14	-5	-11	55	25	121
		Σu = 5	Σv = -1	uv = 152	Σu2 = 135	Σv2 = 221

Thus using formula, we have

Question 13.

X	1	2	4	5	7	8	10
Y	2	6	8	10	14	16	20

Solution:

X	Y	u = X – A A = 5	u = Y – B B = 10	uv	u2	v2
1	2	-4	-8	32	16	64
2	6	-3	-4	12	9	16
4	8	-1	-2	2	1	4
5	10	0	0	0	0	0
7	14	2	4	8	4	16
8	16	3	6	18	9	36
10	20	5	10	50	25	100
		Σu = 2	Σv = 6	Σuv = 122	Σu2 = 64	Σv2 = 236

So there is a positive and perfect correlation between X an Y.

Question 14.
Calculate Karl Pearson’s correlation coefficient between the marks in English and Hindi obtained by 10 students.

Marks in English	10	25	13	25	22	11	12	25	21	20
Marks in Hindi	12	22	16	15	18	18	17	23	24	17

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

X	Y	u = X – A A = 20	v = Y – 17	uv	u2	v2
10	12	-10	-5	50	100	25
25	22	+5	5	25	25	25
13	16	-7	-1	7	49	1
25	15	+5	-2	-10	25	4
22	18	2	1	2	4	1
11	18	-9	1	-9	81	1
12	17	-8	0	0	64	0
25	23	5	6	30	25	36
21	24	1	7	7	1	49
20	17	0	0	0	0	0
		Σu = -16	Σv = 12	Σuv = 102	Σu2 = 374	Σv2 = 142

Question 15.
Show that the coefficient of correlation ρ between two variables x and y is given by \(\rho=\frac{\sigma_x^2+\sigma_y^2–\sigma_{x-y}^2}{2 \sigma_y \sigma_x}\) where \(\sigma_x^2, \sigma_y^2\) and \(\sigma_{x-y}^2\) are the variances of x, y and x-y respectively.
Solution:

Question 16.
A computer expert while calculating correlation coefficient between X and Y from 25 pairs of observations obtained the following results :
n = 25, ΣX = 125, ΣX² = 650, ΣY = 100, ΣY² = 460, ΣXY = 508
It was, however, later discovered at the time of checking that he had copied down two pairs as while the correct values were Obtain the correct value of correlation coefficient.
Solution:
Given n = 25, ΣX = 125 ; ΣX² = 650 ; ΣY² = 460 ; ΣY = 100 ; ΣXY = 508
Corrected ΣX = Given ΣX – (Sum of incorrect values) + (Sum of correct values) = 125 – (6 + 8) + (8 + 6) = 125
Corrected ΣX² = Given ΣX² – (6² + 8²) + (8² + 6²) = 650 – 100 + 100 = 650
Corrected ΣXY = Given ΣXY – (6 × 14 + 8 × 6) + (8 × 12 + 6 × 8) = 508 – (84 + 48) + (96 + 48) = 520
Corrected ΣY = 100 – (14 + 6) + (12 + 8) = 100
Corrected ΣY² = 460 – (14² + 6²) +(12² + 8²) = 436

Question 17.
A computer obtained the data: n = 30, Σx = 120, Σy = 90, Σx² = 600, Σy² = 250 and Σxy = 56. Later on, it was found that pairs are wrong while the correct values are . Find the correct values of ρ(X, Y).
Solution:
Given n = 30; Σx = 120; Σy = 90; Σx² = 600 ; Σy² = 250 and Σxy = 356
Corrected Σx = Given Σx – (Sum of incorrect values) + (Sum of correct values) = 120 – (8 + 12) + (8 + 10) = 118
Corrected Σx² = 600 – (8² + 12²) + (8² + 10²) = 556
Corrected Σy = 90 – (10 + 7) + (12 + 8) = 93
Corrected Σy² = 250 – (10² + 7²) + (12² + 8²) = 309
Corrected Σxy = 356 – (8 × 10 + 12 × 7) + (8 × 12 + 10 × 8) = 368

Question 18.
Show that Pearson’s coefficient of correlation lies between -1 and +1 , i.e., -1 ≤ r ≤ 1 or | r | ≤ 1.
Solution: