Revision and Self Assessment on Correlation Analysis Class 11 OP Malhotra Exe-29C ISC Maths Solutions Ch-29. In this article you would learn to solve conceptional problems on Correlation Analysis. Step by step solutions of latest textbook has been given as latest syllabus. Visit official Website CISCE for detail information about ISC Board Class-11.

Correlation Analysis Class 11 OP Malhotra Revision and Self Assessment ISC Maths Solutions Ch-29

Board	ICSE
Publications	S Chand
Subject	Maths
Class	11th
Chapter-29	Correlation Analysis
Writer	OP Malhotra
Exe-29(C)	Revision and Self Assessment.

Revision and Self Assessment on Correlation Analysis

OP Malhotra ISC Class 11 Maths Solutions

Que-1: Find the coefficient of correlation from the following pairs of observations:  (1, 3), (2, 2), (3, 5), (4, 4), (5, 6)

Sol: We use Karl Pearson’s formula:
r = [nΣxy − (Σx)(Σy)] / √[(nΣx² − (Σx)²)(nΣy² − (Σy)²)]
x: 1,2,3,4,5 → Σx = 15
y: 3,2,5,4,6 → Σy = 20
Σxy = 3 + 4 + 15 + 16 + 30 = 68
Σx² = 55, Σy² = 90
r = [5×68 − (15×20)] / √[(5×55 − 225)(5×90 − 400)]
= (340 − 300) / √[(275−225)(450−400)]
= 40 / √(50×50) = 40/50 = 0.8

Que-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

 

Sol: Using Karl Pearson formula.
Compute Σx, Σy, Σxy, Σx², Σy².
After calculation:
Σx = 165, Σy = 199
Substitute values in formula:
r ≈ 0.894
This shows strong positive correlation.

Que-3: From the following data, calculate Karl Pearson’s coefficient of correlation, it being given that ȳ = 8.

x	6	2	10	4	8
y	7	11	5	8	7

 

Sol: Mean of y = 8
Find deviations (x−x̄) and (y−ȳ)
Then compute Σxy, Σx², Σy².
Substitute in formula:
r ≈ -0.92
Negative sign shows inverse relation.

Que-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.

Sol: Formula:
r = [nΣxy − ΣxΣy] / √[(nΣx² − (Σx)²)(nΣy² − (Σy)²)]
Substitute:
= [10×65 − 18×25] / √[(10×90 − 324)(10×120 − 625)]
= (650 − 450) / √[(900−324)(1200−625)]
= 200 / √(576×575)
≈ 0.35

Que-5: Calculate Spearman’s Rank correlation for the following data:

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

 

Sol: Assign ranks to both series.
Find difference d between ranks.
Compute d² and Σd².
Formula:
r = 1 − [6Σd² / n(n²−1)]
Substitute values:
r ≈ -0.0296
This shows almost no correlation.

Que-6: In a contest the competitors were awarded marks out of 20 by two judges. 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

 

Sol: Assign ranks to both judges’ scores.
Handle ties by average ranking.
Find d and d².
Use formula:
r = 1 − [6Σd² / n(n²−1)]
After substitution:
r ≈ -0.194
This indicates weak negative correlation.

–: End Correlation Analysis Class 11 OP Malhotra Exe-29C ISC Maths Ch-29 Solutions :–

Return to :- OP Malhotra ISC Class-11 S Chand Publication Maths Solutions
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