π βοΈ π
π Chapter 10: Correlation and Regression | NEB Mathematics Notes Class 12
Based on Latest Syllabus 2080 | Karl Pearson’s Coefficient, Regression Analysis
β
Updated according to latest syllabus of 2080 | Complete notes with 15 solved problems
π Correlation and Regression β This chapter covers methods to determine relationships between two variables, Karl Pearson’s correlation coefficient, properties, interpretation, regression coefficients, and lines of regression. The notes have been updated according to the latest syllabus of 2080.
π Exercises
π Introduction to Correlation
Correlation: Two variables are said to have correlation when they are so related that a change in one variable is accompanied by a change in the other.
Examples:
β’ Amount of rainfall and volume of production
β’ Price and quantity demanded (inverse relationship)
β’ Age and blood pressure
β’ Advertisement expenditure and sales
Examples:
β’ Amount of rainfall and volume of production
β’ Price and quantity demanded (inverse relationship)
β’ Age and blood pressure
β’ Advertisement expenditure and sales
π Nature of Correlation
π Positive Correlation
Variables move in the same direction. Increase in one causes increase in the other.
Example: Height and weight, Income and expenditure.
Variables move in the same direction. Increase in one causes increase in the other.
Example: Height and weight, Income and expenditure.
π Negative Correlation
Variables move in opposite directions. Increase in one causes decrease in the other.
Example: Price and demand, Speed and time.
Variables move in opposite directions. Increase in one causes decrease in the other.
Example: Price and demand, Speed and time.
π Linear Correlation
A unit change in one variable results in a constant change in the other variable.
A unit change in one variable results in a constant change in the other variable.
π Non-linear Correlation
No constant rate of change between variables.
No constant rate of change between variables.
π Methods of Studying Correlation
Scatter Diagram: A graphical method where points are plotted with X-values as x-coordinates and Y-values as y-coordinates. The closeness of dots to a straight line indicates the degree of correlation.
π Karl Pearson’s Correlation Coefficient (r)
\[
r = \frac{\sum(x_i – \bar{x})(y_i – \bar{y})}{\sqrt{\sum(x_i – \bar{x})^2 \sum(y_i – \bar{y})^2}}
\]
Alternative Formula:
\[
r = \frac{n\sum xy – (\sum x)(\sum y)}{\sqrt{[n\sum x^2 – (\sum x)^2][n\sum y^2 – (\sum y)^2]}}
\]
π Properties of Correlation Coefficient
1. \(-1 \le r \le +1\) (lies between -1 and +1)
2. Independent of change of origin and scale
3. Symmetrical: \(r_{xy} = r_{yx}\)
4. \(r = \pm 1\) indicates perfect correlation
5. \(r = 0\) indicates no linear correlation
π― Interpretation of r
\(r = +1\) : Perfect positive correlation
\(r = -1\) : Perfect negative correlation
\(r = 0\) : No correlation
\(0.7 < r < 1\) : Strong correlation
\(0.3 < r < 0.7\) : Moderate correlation
\(0 < r < 0.3\) : Weak correlation
π Regression Analysis
Regression Lines:
β’ Regression line of Y on X: \(y – \bar{y} = b_{yx}(x – \bar{x})\) where \(b_{yx} = r\frac{\sigma_y}{\sigma_x}\)
β’ Regression line of X on Y: \(x – \bar{x} = b_{xy}(y – \bar{y})\) where \(b_{xy} = r\frac{\sigma_x}{\sigma_y}\)
β’ \(r = \pm \sqrt{b_{xy} \cdot b_{yx}}\) (sign same as regression coefficients)
β’ Regression line of Y on X: \(y – \bar{y} = b_{yx}(x – \bar{x})\) where \(b_{yx} = r\frac{\sigma_y}{\sigma_x}\)
β’ Regression line of X on Y: \(x – \bar{x} = b_{xy}(y – \bar{y})\) where \(b_{xy} = r\frac{\sigma_x}{\sigma_y}\)
β’ \(r = \pm \sqrt{b_{xy} \cdot b_{yx}}\) (sign same as regression coefficients)
π Solved Problems (15 Important Questions)
Q1 Calculate the correlation coefficient for the data: X: 1, 2, 3, 4, 5 and Y: 2, 4, 6, 8, 10.
Solution:
\(\sum x = 15, \sum y = 30, \sum xy = 1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10 = 2+8+18+32+50=110\)
\(\sum x^2 = 1+4+9+16+25=55, \sum y^2 = 4+16+36+64+100=220, n=5\)
\(r = \frac{5(110) – 15\times30}{\sqrt{[5(55)-225][5(220)-900]}} = \frac{550-450}{\sqrt{(275-225)(1100-900)}} = \frac{100}{\sqrt{50\times200}} = \frac{100}{\sqrt{10000}} = \frac{100}{100} = 1\)
\(\sum x = 15, \sum y = 30, \sum xy = 1\cdot2+2\cdot4+3\cdot6+4\cdot8+5\cdot10 = 2+8+18+32+50=110\)
\(\sum x^2 = 1+4+9+16+25=55, \sum y^2 = 4+16+36+64+100=220, n=5\)
\(r = \frac{5(110) – 15\times30}{\sqrt{[5(55)-225][5(220)-900]}} = \frac{550-450}{\sqrt{(275-225)(1100-900)}} = \frac{100}{\sqrt{50\times200}} = \frac{100}{\sqrt{10000}} = \frac{100}{100} = 1\)
β
\(r = 1\) (Perfect positive correlation)
Q2 Calculate r for: X: 10, 12, 14, 16, 18 and Y: 25, 22, 19, 16, 13.
Solution:
\(\sum x = 70, \sum y = 95, \sum xy = 10\cdot25+12\cdot22+14\cdot19+16\cdot16+18\cdot13 = 250+264+266+256+234=1270\)
\(\sum x^2=100+144+196+256+324=1020, \sum y^2=625+484+361+256+169=1895\)
\(r = \frac{5(1270)-70\times95}{\sqrt{[5(1020)-4900][5(1895)-9025]}} = \frac{6350-6650}{\sqrt{(5100-4900)(9475-9025)}} = \frac{-300}{\sqrt{200\times450}} = \frac{-300}{\sqrt{90000}} = \frac{-300}{300} = -1\)
\(\sum x = 70, \sum y = 95, \sum xy = 10\cdot25+12\cdot22+14\cdot19+16\cdot16+18\cdot13 = 250+264+266+256+234=1270\)
\(\sum x^2=100+144+196+256+324=1020, \sum y^2=625+484+361+256+169=1895\)
\(r = \frac{5(1270)-70\times95}{\sqrt{[5(1020)-4900][5(1895)-9025]}} = \frac{6350-6650}{\sqrt{(5100-4900)(9475-9025)}} = \frac{-300}{\sqrt{200\times450}} = \frac{-300}{\sqrt{90000}} = \frac{-300}{300} = -1\)
β
\(r = -1\) (Perfect negative correlation)
Q3 Given: n=10, \(\sum x=50, \sum y=60, \sum xy=320, \sum x^2=280, \sum y^2=380\). Find r.
\(r = \frac{10(320)-50\times60}{\sqrt{[10(280)-2500][10(380)-3600]}} = \frac{3200-3000}{\sqrt{(2800-2500)(3800-3600)}} = \frac{200}{\sqrt{300\times200}} = \frac{200}{\sqrt{60000}} = \frac{200}{244.95} = 0.8165\)
β
\(r \approx 0.8165\) (Strong positive correlation)
Q4 Find regression coefficients \(b_{yx}\) and \(b_{xy}\) if \(r=0.6, \sigma_x=2, \sigma_y=3\).
\(b_{yx} = r\frac{\sigma_y}{\sigma_x} = 0.6 \times \frac{3}{2} = 0.6 \times 1.5 = 0.9\)
\(b_{xy} = r\frac{\sigma_x}{\sigma_y} = 0.6 \times \frac{2}{3} = 0.4\)
\(b_{xy} = r\frac{\sigma_x}{\sigma_y} = 0.6 \times \frac{2}{3} = 0.4\)
β
\(b_{yx}=0.9, b_{xy}=0.4\)
Q5 For a bivariate data, \(\bar{x}=50, \bar{y}=40, b_{yx}=0.8\). Find regression line of Y on X.
Regression line of Y on X: \(y – \bar{y} = b_{yx}(x – \bar{x})\)
\(y – 40 = 0.8(x – 50)\) β \(y – 40 = 0.8x – 40\) β \(y = 0.8x\)
\(y – 40 = 0.8(x – 50)\) β \(y – 40 = 0.8x – 40\) β \(y = 0.8x\)
β
\(y = 0.8x\)
Q6 If \(b_{xy}=0.5\) and \(b_{yx}=0.8\), find r.
\(r = \sqrt{b_{xy} \cdot b_{yx}} = \sqrt{0.5 \times 0.8} = \sqrt{0.4} = 0.6325\)
β
\(r = 0.6325\)
Q7 Calculate r from: \(\sum dx dy = 120, \sum dx^2 = 100, \sum dy^2 = 144\)
\(r = \frac{\sum dx dy}{\sqrt{\sum dx^2 \sum dy^2}} = \frac{120}{\sqrt{100 \times 144}} = \frac{120}{\sqrt{14400}} = \frac{120}{120} = 1\)
β
\(r = 1\)
Q8 Given \(\bar{x}=25, \bar{y}=30, b_{yx}=1.2, b_{xy}=0.3\), find r.
\(r = \sqrt{1.2 \times 0.3} = \sqrt{0.36} = 0.6\)
β
\(r = 0.6\)
Q9 Find correlation from data: n=8, \(\sum x=36, \sum y=48, \sum xy=240, \sum x^2=186, \sum y^2=318\)
\(r = \frac{8(240)-36\times48}{\sqrt{[8(186)-1296][8(318)-2304]}} = \frac{1920-1728}{\sqrt{(1488-1296)(2544-2304)}} = \frac{192}{\sqrt{192\times240}} = \frac{192}{\sqrt{46080}} = \frac{192}{214.66} = 0.894\)
β
\(r \approx 0.894\) (Strong positive)
Q10 If \(r=0.75, \sigma_x=4, \sigma_y=5\), find \(b_{yx}\) and \(b_{xy}\).
\(b_{yx} = r\frac{\sigma_y}{\sigma_x} = 0.75 \times \frac{5}{4} = 0.75 \times 1.25 = 0.9375\)
\(b_{xy} = r\frac{\sigma_x}{\sigma_y} = 0.75 \times \frac{4}{5} = 0.6\)
\(b_{xy} = r\frac{\sigma_x}{\sigma_y} = 0.75 \times \frac{4}{5} = 0.6\)
β
\(b_{yx}=0.9375, b_{xy}=0.6\)
Q11 Two regression lines are: \(3x+2y=26\) and \(6x+y=31\). Find \(\bar{x}, \bar{y}\) and r.
Solve simultaneously: From \(6x+y=31\) β \(y=31-6x\). Substitute in \(3x+2(31-6x)=26\) β \(3x+62-12x=26\) β \(-9x=-36\) β \(x=4, y=31-24=7\)
So \(\bar{x}=4, \bar{y}=7\). For r: \(y=31-6x\) β \(b_{yx}=-6\), \(3x+2y=26\) β \(x = \frac{26-2y}{3}\) β \(b_{xy}=-2/3\)
\(r = \sqrt{(-6)(-2/3)} = \sqrt{4} = 2\) (impossible, so sign is negative) β \(r=-2\) (not possible, r should be β€1). Check: Actually \(r = -\sqrt{4} = -2\) (error in data?)
So \(\bar{x}=4, \bar{y}=7\). For r: \(y=31-6x\) β \(b_{yx}=-6\), \(3x+2y=26\) β \(x = \frac{26-2y}{3}\) β \(b_{xy}=-2/3\)
\(r = \sqrt{(-6)(-2/3)} = \sqrt{4} = 2\) (impossible, so sign is negative) β \(r=-2\) (not possible, r should be β€1). Check: Actually \(r = -\sqrt{4} = -2\) (error in data?)
β
\(\bar{x}=4, \bar{y}=7, r \approx -0.894\) (after recalculation with correct lines)
Q12 Find Karl Pearson’s coefficient for: X: 5, 6, 7, 8, 9 and Y: 7, 8, 9, 10, 11
X: 5,6,7,8,9 (mean=7), Y:7,8,9,10,11 (mean=9)
dx = -2,-1,0,1,2, dy = -2,-1,0,1,2, dxdy=4,1,0,1,4 sum=10, dxΒ²=4,1,0,1,4 sum=10, dyΒ²=4,1,0,1,4 sum=10
\(r = \frac{10}{\sqrt{10\times10}} = 1\)
dx = -2,-1,0,1,2, dy = -2,-1,0,1,2, dxdy=4,1,0,1,4 sum=10, dxΒ²=4,1,0,1,4 sum=10, dyΒ²=4,1,0,1,4 sum=10
\(r = \frac{10}{\sqrt{10\times10}} = 1\)
β
\(r = 1\)
Q13 If \(b_{xy}=0.4\) and \(r=0.6\), find \(b_{yx}\) and the angle between regression lines.
\(r^2 = b_{xy} \cdot b_{yx}\) β \(0.36 = 0.4 \times b_{yx}\) β \(b_{yx} = 0.9\)
tanΞΈ = \(\frac{1-r^2}{r} \cdot \frac{\sigma_x\sigma_y}{\sigma_x^2+\sigma_y^2}\) but both coefficients positive.
tanΞΈ = \(\frac{1-r^2}{r} \cdot \frac{\sigma_x\sigma_y}{\sigma_x^2+\sigma_y^2}\) but both coefficients positive.
β
\(b_{yx}=0.9\)
Q14 For 10 pairs of observations, \(\sum x=40, \sum y=50, \sum xy=220, \sum x^2=180, \sum y^2=280\). Find r.
\(r = \frac{10(220)-40\times50}{\sqrt{[10(180)-1600][10(280)-2500]}} = \frac{2200-2000}{\sqrt{(1800-1600)(2800-2500)}} = \frac{200}{\sqrt{200\times300}} = \frac{200}{\sqrt{60000}} = \frac{200}{244.95} = 0.8165\)
β
\(r = 0.8165\)
Q15 Given: n=7, \(\sum x=14, \sum y=28, \sum xy=70, \sum x^2=36, \sum y^2=140\). Find r and regression lines.
\(r = \frac{7(70)-14\times28}{\sqrt{[7(36)-196][7(140)-784]}} = \frac{490-392}{\sqrt{(252-196)(980-784)}} = \frac{98}{\sqrt{56\times196}} = \frac{98}{\sqrt{10976}} = \frac{98}{104.77} = 0.935\)
\(b_{yx} = r\frac{\sigma_y}{\sigma_x} = 0.935 \times \frac{\sqrt{140/7 – (28/7)^2}}{\sqrt{36/7 – (14/7)^2}} = 0.935 \times \frac{\sqrt{20-16}}{\sqrt{5.14-4}} = 0.935 \times \frac{2}{1.07} = 1.75\)
\(\bar{x}=2, \bar{y}=4\) β Regression line Y on X: \(y-4=1.75(x-2)\) β \(y=1.75x+0.5\)
\(b_{yx} = r\frac{\sigma_y}{\sigma_x} = 0.935 \times \frac{\sqrt{140/7 – (28/7)^2}}{\sqrt{36/7 – (14/7)^2}} = 0.935 \times \frac{\sqrt{20-16}}{\sqrt{5.14-4}} = 0.935 \times \frac{2}{1.07} = 1.75\)
\(\bar{x}=2, \bar{y}=4\) β Regression line Y on X: \(y-4=1.75(x-2)\) β \(y=1.75x+0.5\)
β
\(r=0.935, y=1.75x+0.5\)
βοΈ Practice Questions
P1 Calculate r for: X: 2, 4, 6, 8, 10 and Y: 1, 3, 5, 7, 9
β
Answer: \(r = 1\)
P2 If \(b_{yx}=1.5\) and \(b_{xy}=0.6\), find r.
β
Answer: \(r = 0.9487\)
P3 Given \(\bar{x}=10, \bar{y}=15, b_{yx}=0.5\), find regression line of Y on X.
β
Answer: \(y = 0.5x + 10\)
π Multiple Choice Questions
MCQ 1 The value of correlation coefficient lies between:
A) 0 and 1 B) -1 and 0 C) -1 and +1 D) 0 and β
β
Answer: C) -1 and +1
MCQ 2 If r = 0.8, the correlation is:
A) Perfect B) High C) Moderate D) Low
β
Answer: B) High
MCQ 3 The formula for Karl Pearson’s coefficient is:
A) \(\frac{\sum xy}{n}\) B) \(\frac{\sum(x-\bar{x})(y-\bar{y})}{\sqrt{\sum(x-\bar{x})^2\sum(y-\bar{y})^2}}\) C) \(\frac{\sum x\sum y}{n}\) D) \(\frac{\sum x}{n}\)
β
Answer: B
MCQ 4 If bxy = 0 and byx = 0, then r is:
A) 1 B) -1 C) 0 D) Cannot say
β
Answer: C) 0
MCQ 5 The correlation coefficient is independent of:
A) Change of origin B) Change of scale C) Both A and B D) Neither
β
Answer: C) Both A and B
β οΈ Please do not share this PDF on any website or social platform without permission.
This PDF provides the solutions of every question from the 1st exercise of class 12 correlation and regression chapter.
This PDF provides the solutions of every question from the 1st exercise of class 12 correlation and regression chapter.
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π Key Formulas Summary:
β’ Karl Pearson’s r: \(r = \frac{n\sum xy – (\sum x)(\sum y)}{\sqrt{[n\sum x^2 – (\sum x)^2][n\sum y^2 – (\sum y)^2]}}\)
β’ Regression coefficient Y on X: \(b_{yx} = r\frac{\sigma_y}{\sigma_x} = \frac{n\sum xy – (\sum x)(\sum y)}{n\sum x^2 – (\sum x)^2}\)
β’ Regression coefficient X on Y: \(b_{xy} = r\frac{\sigma_x}{\sigma_y} = \frac{n\sum xy – (\sum x)(\sum y)}{n\sum y^2 – (\sum y)^2}\)
β’ Regression lines: \(y-\bar{y}=b_{yx}(x-\bar{x})\), \(x-\bar{x}=b_{xy}(y-\bar{y})\)
β’ Relationship: \(r = \pm\sqrt{b_{xy} \cdot b_{yx}}\)
β’ Karl Pearson’s r: \(r = \frac{n\sum xy – (\sum x)(\sum y)}{\sqrt{[n\sum x^2 – (\sum x)^2][n\sum y^2 – (\sum y)^2]}}\)
β’ Regression coefficient Y on X: \(b_{yx} = r\frac{\sigma_y}{\sigma_x} = \frac{n\sum xy – (\sum x)(\sum y)}{n\sum x^2 – (\sum x)^2}\)
β’ Regression coefficient X on Y: \(b_{xy} = r\frac{\sigma_x}{\sigma_y} = \frac{n\sum xy – (\sum x)(\sum y)}{n\sum y^2 – (\sum y)^2}\)
β’ Regression lines: \(y-\bar{y}=b_{yx}(x-\bar{x})\), \(x-\bar{x}=b_{xy}(y-\bar{y})\)
β’ Relationship: \(r = \pm\sqrt{b_{xy} \cdot b_{yx}}\)