OP Malhotra Class 11 Maths Solutions Chapter 29 Correlation Analysis Chapter Test

Regular engagement with ISC Class 11 Maths Solutions S Chand Chapter 29 Correlation Analysis Chapter Test can boost students’ confidence in the subject.

S Chand Class 11 ICSE Maths Solutions Chapter 29 Correlation Analysis Chapter Test

Question 1.
Find the coefficient of correlation from the following pairs of observations :
(1,3), (2,2), (3,5), (4,4), (5,6)
Solution:
The table of values is given as under :

\(\bar{X}\) = \(\frac{\Sigma \mathrm{X}}{n}\) = \(\frac{15}{5}\) = 3;
\(\bar{Y}\) = \(\frac{\Sigma \mathrm{Y}}{n}\) = \(\frac{20}{5}\) = 4;
∴ r = \(\frac{\Sigma d \mathrm{X} d \mathrm{Y}}{\sqrt{\Sigma d \mathrm{X}^2} \sqrt{\Sigma d \mathrm{Y}^2}}\) = \(\frac{8}{\sqrt{10} \sqrt{10}}\) = \(\frac{4}{5}\) = 0.8

Question 2.
Find the Karl Pearson’s coefficient of correlation between x and y for the following data :

x	16	18	21	20	22	26	27	15
y	22	25	24	26	25	30	33	14

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

Question 3.
From the following data, calculate the Karl Pearson’s coefficient of correlation, it being given that \(\bar{y}\) = 8.

x	6	2	10	4	8
Y	?	11	5	8	7

Solution:
Given \(\bar{X}\) = 6 and \(\bar{Y}\) = 8 = \(\frac{9+11+f+8+7}{5}\) ⇒ 40 = 35 + f ⇒ f = 5

By formula, we have
r = \(\frac{\Sigma(\mathrm{X}-\overline{\mathrm{X}})(\mathrm{Y}-\overline{\mathrm{Y}})}{\sqrt{\Sigma(\mathrm{X}-\overline{\mathrm{X}})^2} \sqrt{\Sigma(\mathrm{Y}-\overline{\mathrm{Y}})^2}}\) = \(\frac{-26}{\sqrt{40} \sqrt{20}}\) = \(\frac{-26}{\sqrt{800}}\) = \(\frac{-26}{20 \sqrt{2}}\) = -0.9192

Question 4.
Calculate Karl Pearson’s coefficient between the values of x and y if Σx = 18, Σx² = 90, n = 10, Σy = 25, Σy² = 120, Σxy = 65.
Solution:
Given Σx = 18, Σx² = 90 ; n = 10 ; Σy = 25 ; Σy² = 120 and Σxy = 65
Karl Pearson coeff. of correlation =

Question 5.
A psychologist selected a random sample of 22 students. He grouped them in 11 pairs so that the students in each pair have nearly equal scores in an intelligence test. In each pair, one student was taught by method A and the other by method B and examined after the course. The marks obtained by them after the course are as follows :

Pairs	1	2	3	4	5	6	7	8	9	10	11
Method A	24	29	19	14	30	19	27	30	20	28	11
Method B	37	35	16	26	23	27	19	20	16	11	21

Calculate Spearman’s Rank correlation.
Solution:
We construct the table of values is as under :

Pairs	Method A	Rank	Method B	Rank	d = R1 – R2	d2
1	24	6	37	1	5	25
2	29	3	35	2	1	1
3	19	8.5	16	9.5	-1	1
4	14	10	26	4	6	36
5	30	1.5	23	5	-3.5	12.25
6	19	8.5	27	3	5.5	30.25
7	27	5	19	8	-3	9
8	30	1.5	20	7	-5.5	30.25
9	20	7	16	9.5	-2.5	6.25
10	28	4	11	11	-7	49
11	11	11	21	6	5	25
						Σd2 = 225

Question 6.
In a contest the competitors were awarded marks out of 20 by two judges. The scores of the 10 competitors are given below. Calculate spearman’s rank correlation.

Pairs	A	B	C	D	E	F	G	H	I	J
Judge A	2	11	11	18	6	5	8	16	13	15
Judge B	6	11	16	9	14	20	4	3	13	17

Solution:
The table of values is given as under :