Class 12: Linear Regression – ISC Board Problems – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1: } \text{If the correlation coefficient of two sets of variables } (X,Y) \text{ is }$
$\displaystyle -\frac{3}{4}, \text{ which one of the following statements is true for the same set of variables? ISC 2024}$
$\displaystyle \text{(a) Only one of the two regression lines has a negative coefficient.}$
$\displaystyle \text{(b) Both regression coefficients are positive.} \quad \text{(c) Both regression coefficients are negative.}$
$\displaystyle \text{(d) One of the lines of regression is parallel to the } X\text{-axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) The value of correlation coefficient }r\text{ lies between }-1\text{ and }1$
$\displaystyle \text{where, }-1\text{ indicates a perfect negative linear relationship and }1$
$\displaystyle \text{indicates a perfect positive linear relationship.}$
$\displaystyle \text{Since, the correlation coefficient of two sets of variables }(X,Y)\text{ is }-\frac{3}{4},$
$\displaystyle \text{which is negative, there is a negative linear relationship between }X\text{ and }Y.$
$\displaystyle \text{Thus, both regression coefficients are negative.}$
$\\$

$\displaystyle \textbf{Question 2: } \text{The equations of two lines of regression are } 4x+3y+7=0 \text{ and }$
$\displaystyle 3x+4y+8=0. \text{Find the mean value of } x \text{ and } y. \text{ ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given two lines of regression are }4x+3y+7=0\text{ and }3x+4y+8=0$
$\displaystyle \text{Now, to find the mean value of }x\text{ and }y,\text{ we solve the system}$
$\displaystyle \text{of equations formed by the two given lines of regression.}$
$\displaystyle 3x+4y+8=0\quad ...(i)$
$\displaystyle 4x+3y+7=0\quad ...(ii)$
$\displaystyle \text{On multiplying Eq. (i) by 4 and Eq. (ii) by 3, we get}$
$\displaystyle 12x+16y+32=0\quad ...(iii)$
$\displaystyle 12x+9y+21=0\quad ...(iv)$
$\displaystyle \text{Now, subtract Eq. (iv) from Eq. (iii), we get}$
$\displaystyle 7y+11=0\Rightarrow y=-\frac{11}{7}$
$\displaystyle \text{Substitute in (i): }3x+4\left(-\frac{11}{7}\right)+8=0$
$\displaystyle \Rightarrow 3x-\frac{44}{7}+\frac{56}{7}=0$
$\displaystyle \Rightarrow 3x+\frac{12}{7}=0\Rightarrow x=-\frac{4}{7}$
$\displaystyle \text{So, the mean value of }x\text{ is }-\frac{4}{7}\text{ and of }y\text{ is }-\frac{11}{7}.$
$\\$

$\displaystyle \textbf{Question 3: } XYZ \text{ company plans to advertise some vacancies. The Manager is asked}$
$\displaystyle \text{to suggest the monthly salary for these vacancies based on the years of experience.}$
$\displaystyle \text{ To do so, the Manager studies the years of service and the monthly salary drawn by }$
$\displaystyle \text{the existing employees in the company. }$
$\displaystyle \text{Following is the data that the Manager refers to}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Years of service }(X) & 11 & 7 & 9 & 5 & 8 & 6 & 10\\ \hline \text{Monthly salary (in Rs }1000)(Y) & 10 & 8 & 6 & 5 & 9 & 7 & 11\\ \hline \end{array}$
$\displaystyle \text{(a) Find the regression equation of monthly salary on the years of service.}$
$\displaystyle \text{(b) If a person with } 13 \text{ years of experience applies for a job in this company, what monthly}$
$\displaystyle \text{monthly salary will be suggested by the Manager? ISC 2024}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) The given data is}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline X & 11 & 7 & 9 & 5 & 8 & 6 & 10 \\ \hline Y & 10 & 8 & 6 & 5 & 9 & 7 & 11 \\ \hline \end{array}$
$\displaystyle \sum X=56,\ \sum Y=56$
$\displaystyle \text{So, Mean }(X)=\frac{\sum X}{n}=\frac{56}{7}=8=M_x$
$\displaystyle \text{and Mean }(Y)=\frac{\sum Y}{n}=\frac{56}{7}=8=M_y$
$\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline X & Y & X-M_x & Y-M_y & (X-M_x)^2 & (X-M_x)(Y-M_y) \\ \hline 11 & 10 & 3 & 2 & 9 & 6 \\ 7 & 8 & -1 & 0 & 1 & 0 \\ 9 & 6 & 1 & -2 & 1 & -2 \\ 5 & 5 & -3 & -3 & 9 & 9 \\ 8 & 9 & 0 & 1 & 0 & 0 \\ 6 & 7 & -2 & -1 & 4 & 2 \\ 10 & 11 & 2 & 3 & 4 & 6 \\ \hline \end{array}$
$\displaystyle \text{Thus, using the above table,}$
$\displaystyle SS_x=\sum (X-M_x)^2=28$
$\displaystyle SP=\sum (X-M_x)(Y-M_y)=21$
$\displaystyle \therefore b=\frac{SP}{SS_x}=\frac{21}{28}=0.75$
$\displaystyle \text{and }a=M_y-bM_x=8-0.75\times8=2$
$\displaystyle \text{Since, regression equation }\hat{y}=bX+a$
$\displaystyle \therefore \hat{y}=0.75X+2$
$\displaystyle \text{(b) Using part (a), } \hat{y}=0.75X+2$
$\displaystyle \text{Since, }X=13$
$\displaystyle \therefore \hat{y}=0.75\times13+2=11.75$
$\displaystyle \text{Hence, monthly salary suggested = Rs }11750$
$\\$

$\displaystyle \textbf{Question 4: } \text{The line of regression of marks in Statistics } (X) \text{ and marks in}$
$\displaystyle \text{Accountancy } (Y) \text{ for } \text{a class of } 50 \text{ students is } 3y-5x+180=0. \text{ The average score}$
$\displaystyle \text{in Accountancy is } 44 \text{ and the } \text{variance of marks in Statistics is } \frac{9}{16} \text{th of variance of}$
$\displaystyle \text{marks in Accountancy. ISC 2024}$
$\displaystyle \text{(a) Find the average score in Statistics.}$
$\displaystyle \text{(b) Find the coefficient of correlation between marks in Statistics and marks in Accountancy.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }n=50,\ X=\text{Marks in Statistics},\ Y=\text{Marks in Accountancy}$
$\displaystyle \text{Regression equation of }X\text{ on }Y\text{ is}$
$\displaystyle 3y-5x+180=0$
$\displaystyle \text{Also, }\bar{y}=44,\ \sigma_x^2=\frac{9}{14}\sigma_y^2$
$\displaystyle \text{(a) Since, regression equation is}$
$\displaystyle 3y-5x+180=0\Rightarrow x=\frac{3}{5}y+36$
$\displaystyle \text{Comparing with }x=b_{xy}y+a,\text{ we get}$
$\displaystyle b_{xy}=\frac{3}{5},\ a=36$
$\displaystyle \text{So, }a=\bar{x}-b_{xy}\bar{y}$
$\displaystyle \Rightarrow 36=\bar{x}-\frac{3}{5}\times44$
$\displaystyle \Rightarrow \bar{x}=36+\frac{132}{5}=\frac{312}{5}=62.4$
$\displaystyle \text{Thus, }\bar{x}=62.4$
$\displaystyle \text{(b) Since, }\sigma_x^{2}=\frac{9}{16}\sigma_y^{2}$
$\displaystyle \Rightarrow \frac{\sigma_x}{\sigma_y}=\frac{3}{4}$
$\displaystyle \text{and we know that }b_{xy}=r\frac{\sigma_x}{\sigma_y}$
$\displaystyle \Rightarrow \frac{3}{5}=r\times\frac{3}{4}$
$\displaystyle \Rightarrow r=\frac{3}{5}\times\frac{4}{3}=\frac{4}{5}=0.8$
$\displaystyle \text{Thus, the correlation coefficient }(r)\text{ between marks in Statistics}$
$\displaystyle \text{and Accountancy is }0.8.$
$\\$

$\displaystyle \textbf{Question 5: } \text{If the two regression coefficients are } 0.8 \text{ and } 0.2, \text{ then the value of}$
$\displaystyle \text{coefficient of correlation } r \text{ will be ISC 2023}$
$\displaystyle \text{(a) } \pm 0.4 \quad \text{(b) } \pm 0.16 \quad \text{(c) } 0.4 \quad \text{(d) } 0.16$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) We know, }r^{2}=b_{xy}\cdot b_{yx}$
$\displaystyle \therefore r^{2}=0.8\times0.2=0.16$
$\displaystyle \Rightarrow r=\pm\sqrt{0.16}=\pm0.4$
$\\$

$\displaystyle \textbf{Question 6: } \text{Out of two regression lines } x+2y-5=0 \text{ and } 2x+3y=8,$
$\displaystyle \text{ find the line} \text{of regression of } y \text{ on } x. \text{ ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Assuming }x+2y-5=0\text{ to be regression line of }y\text{ on }x$
$\displaystyle \Rightarrow 2y=-x+5$
$\displaystyle \Rightarrow y=-\frac{1}{2}x+\frac{5}{2}$
$\displaystyle \Rightarrow b_{yx}=-\frac{1}{2}$
$\displaystyle \text{Then, }2x+3y=8\text{ is regression line of }x\text{ on }y.$
$\displaystyle \Rightarrow 2x=-3y+8$
$\displaystyle \Rightarrow x=-\frac{3}{2}y+4$
$\displaystyle \Rightarrow b_{xy}=-\frac{3}{2}$
$\displaystyle \text{Now, }b_{yx}\cdot b_{xy}=\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)=\frac{3}{4}<1$
$\displaystyle \therefore \text{Regression line of }y\text{ on }x\text{ is }x+2y-5=0$
$\\$

$\displaystyle \textbf{Question 7: } \text{The following table shows the mean, the standard deviation}$
$\displaystyle \text{and the coefficient of correlation of two variables } x \text{ and } y. \text{ ISC 2023}$
$\displaystyle \begin{array}{|c|c|c|} \hline \text{Series} & x & y\\ \hline \text{Mean} & 8 & 6\\ \text{Standard deviation} & 12 & 4\\ \text{Coefficient of correlation} & \multicolumn{2}{c|}{0.6}\\ \hline \end{array}$
$\displaystyle \text{Calculate (a) the regression coefficient } b_{xy} \text{ and } b_{yx}.$
$\displaystyle \text{(b) the probable value of } y \text{ when } x=20.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\bar{x}=8,\ \bar{y}=6,\ \sigma_x=12,\ \sigma_y=4\text{ and }r=0.6$
$\displaystyle \therefore b_{yx}=r\frac{\sigma_y}{\sigma_x}=0.6\times\frac{4}{12}=0.2$
$\displaystyle \text{Regression equation of }y\text{ on }x\text{ is}$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \Rightarrow y-6=0.2(x-8)$
$\displaystyle \Rightarrow y=0.2x-1.6+6$
$\displaystyle \Rightarrow y=0.2x+4.4$
$\displaystyle \text{Putting }x=20,\text{ we get}$
$\displaystyle y=0.2(20)+4.4=4+4.4=8.4$
$\\$

$\displaystyle \textbf{Question 8: } \text{An analyst analysed } 102 \text{ trips of a travel company. He studied}$
$\displaystyle \text{the relation between travel expenses } (y) \text{ and the duration } (x) \text{ of these trips. He}$
$\displaystyle \text{found that the relation between } x \text{ and } y \text{ was linear. Given the following data,}$
$\displaystyle \text{find the regression equation of } y \text{ on } x. \\ \sum x=510, \sum y=7140, \sum x^{2}=4150, \sum y^{2}=740200, \sum xy=54900 \quad \text{ISC 2023}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\Sigma x=510,\ \Sigma y=7140,\ \Sigma x^{2}=4150,$
$\displaystyle \Sigma y^{2}=740200,\ \Sigma xy=54900\text{ and }n=102$
$\displaystyle \text{We know, }\bar{x}=\frac{\Sigma x}{n}=\frac{510}{102}=5$
$\displaystyle \text{and }\bar{y}=\frac{\Sigma y}{n}=\frac{7140}{102}=70$
$\displaystyle \text{Now, }b_{yx}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma x^{2}-\frac{1}{n}(\Sigma x)^{2}}$
$\displaystyle =\frac{54900-\frac{1}{102}(510)(7140)}{4150-\frac{(510)^{2}}{102}}$
$\displaystyle =\frac{54900-35700}{4150-2550}=\frac{19200}{1600}=12$
$\displaystyle \text{Regression equation of }y\text{ on }x\text{ is}$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \Rightarrow y-70=12(x-5)$
$\displaystyle \Rightarrow y-70=12x-60$
$\displaystyle \Rightarrow 12x-y+10=0$
$\\$

$\displaystyle \textbf{Question 9: } \text{The correlation coefficient between } x \text{ and } y \text{ is } 0.6. \text{ If the variance of}$
$\displaystyle x \text{ is } 225, \text{ the variance of } y \text{ is } 400, \text{ mean of } x \text{ is } 10 \text{ and mean of } y \text{ is } 20, \text{ find}$
$\displaystyle \text{(i) the equations of two regression lines.}$
$\displaystyle \text{(ii) the expected value of } y \text{ when } x=2. \ \ \ \text{ ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }r=\sqrt{b_{xy}\cdot b_{yx}}=0.6$
$\displaystyle (\sigma_x)^{2}=225,\ (\sigma_y)^{2}=400,\ \bar{x}=10,\ \bar{y}=20$
$\displaystyle \therefore \sigma_x=15,\ \sigma_y=20$
$\displaystyle \text{We know that }b_{yx}=r\frac{\sigma_y}{\sigma_x}$
$\displaystyle \Rightarrow b_{yx}=0.6\times\frac{20}{15}=\frac{12}{15}=\frac{4}{5}$
$\displaystyle \text{and }b_{xy}=r\frac{\sigma_x}{\sigma_y}$
$\displaystyle \Rightarrow b_{xy}=0.6\times\frac{15}{20}=\frac{9}{20}$
$\displaystyle \text{(i) Equation of regression line of }x\text{ on }y$
$\displaystyle x-\bar{x}=b_{xy}(y-\bar{y})$
$\displaystyle \Rightarrow x-10=\frac{9}{20}(y-20)$
$\displaystyle \Rightarrow 20x-200=9y-180$
$\displaystyle \Rightarrow 20x-9y-20=0$
$\displaystyle \text{Equation of regression line of }y\text{ on }x$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \Rightarrow y-20=\frac{4}{5}(x-10)$
$\displaystyle \Rightarrow 5y-100=4x-40$
$\displaystyle \Rightarrow 5y-4x-60=0$
$\displaystyle \text{(ii) Expected value of }y\text{ when }x=2$
$\displaystyle \text{Using }y=\frac{4}{5}x+12$
$\displaystyle \Rightarrow y=\frac{4}{5}(2)+12=\frac{8}{5}+12=13.6$
$\\$

$\displaystyle \textbf{Question 10: } \text{For the lines of regression } 4x-2y=4 \text{ and } 2x-3y+6=0,$
$\displaystyle \text{ find the mean of } x \text{ and the mean of } y. \ \ \ \text{ISC 2020}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, lines of regression are}$
$\displaystyle 4x-2y=4$
$\displaystyle \text{and }2x-3y+6=0$
$\displaystyle \text{We know that }(\bar{x},\bar{y})\text{ is common point of two}$
$\displaystyle \text{regression lines.}$
$\displaystyle \text{Solving the equations }4x-2y=4\text{ and }2x-3y+6=0,$
$\displaystyle \text{we get}$
$\displaystyle x=3,\ y=4$
$\displaystyle \therefore \text{Mean of }x(\bar{x})=3\text{ and mean of }y(\bar{y})=4$
$\\$

$\displaystyle \textbf{Question 11: } \text{If } \overline{x}=18, \overline{y}=100, \sigma_x=14, \sigma_y=20 \text{ and correlation coefficient }$
$\displaystyle r_{xy}=0.8, \text{find the regression equation of } y \text{ on } x. \text{ ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\bar{x}=18,\ \bar{y}=100,\ \sigma_x=14,\ \sigma_y=20\text{ and }r=0.8$
$\displaystyle \text{Now, regression equation of }y\text{ on }x\text{ is}$
$\displaystyle (y-\bar{y})=r\frac{\sigma_y}{\sigma_x}(x-\bar{x})$
$\displaystyle \Rightarrow y-100=0.8\times\frac{20}{14}(x-18)$
$\displaystyle \Rightarrow y-100=\frac{8}{7}(x-18)$
$\displaystyle \Rightarrow 7y-700=8x-144$
$\displaystyle \Rightarrow 7y=8x+556$
$\\$

$\displaystyle \textbf{Question 12: } \text{The following results were obtained with respect to two variables}$
$\displaystyle x \text{ and } y \\ \sum x=15, \sum y=25, \sum xy=83, \sum x^{2}=55, \sum y^{2}=135 \text{ and } n=5.$
$\displaystyle \text{(i) Find the regression coefficient } b_{xy}.$
$\displaystyle \text{(ii) Find the regression equation of } x \text{ on } y. \text{ ISC 2019}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\Sigma x=15,\ \Sigma y=25,\ \Sigma xy=83,$
$\displaystyle \Sigma x^{2}=55,\ \Sigma y^{2}=135\text{ and }n=5$
$\displaystyle \text{(i) Regression coefficient,}$
$\displaystyle b_{yx}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma x^{2}-\frac{1}{n}(\Sigma x)^{2}}$
$\displaystyle =\frac{83-\frac{1}{5}(15)(25)}{55-\frac{1}{5}(15)^{2}}$
$\displaystyle =\frac{83-75}{55-45}=\frac{8}{10}=\frac{4}{5}$
$\\$

$\displaystyle \textbf{Question 13: } \text{If two lines of regression are } 4x+2y-3=0 \text{ and }$
$\displaystyle 3x+6y+5=0, \text{ find the correlation coefficient} \text{between } x \text{ and } y. \text{ ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given lines of regression are}$
$\displaystyle 4x+2y-3=0\Rightarrow x=-\frac{1}{2}y+\frac{3}{4}$
$\displaystyle \therefore b_{xy}=-\frac{1}{2}$
$\displaystyle \text{and }3x+6y+5=0\Rightarrow y=-\frac{1}{2}x-\frac{5}{6}$
$\displaystyle \therefore b_{yx}=-\frac{1}{2}$
$\displaystyle \text{Coefficient of correlation }r=-\sqrt{b_{xy}\cdot b_{yx}}$
$\displaystyle =-\sqrt{\left(-\frac{1}{2}\right)\left(-\frac{1}{2}\right)}=-\frac{1}{2}$
$\\$

$\displaystyle \textbf{Question 14: } \text{Find the line of regression of } y \text{ on } x \text{ from the following table.}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline x & 1 & 2 & 3 & 4 & 5\\ \hline y & 7 & 6 & 5 & 4 & 3\\ \hline \end{array}$
$\displaystyle \text{Hence, estimate the value of } y \text{ when } x=6. \text{ ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given data }(1,7),(2,6),(3,5),(4,4),(5,3)$
$\displaystyle \text{Now, we construct the following table}$
$\displaystyle \begin{array}{|c|c|c|c|} \hline x & y & x^{2} & xy \\ \hline 1 & 7 & 1 & 7 \\ 2 & 6 & 4 & 12 \\ 3 & 5 & 9 & 15 \\ 4 & 4 & 16 & 16 \\ 5 & 3 & 25 & 15 \\ \hline \Sigma x=15 & \Sigma y=25 & \Sigma x^{2}=55 & \Sigma xy=65 \\ \hline \end{array}$
$\displaystyle \bar{x}=\frac{15}{5}=3,\quad \bar{y}=\frac{25}{5}=5$
$\displaystyle b_{yx}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma x^{2}-\frac{1}{n}(\Sigma x)^{2}}$
$\displaystyle =\frac{65-\frac{1}{5}(15)(25)}{55-\frac{1}{5}(15)^{2}}$
$\displaystyle =\frac{65-75}{55-45}=\frac{-10}{10}=-1$
$\displaystyle \text{Equation of line of regression of }y\text{ on }x\text{ is}$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \Rightarrow y-5=-1(x-3)$
$\displaystyle \Rightarrow y-5=-x+3$
$\displaystyle \Rightarrow x+y-8=0$
$\displaystyle \text{On substituting }x=6\text{, we get}$
$\displaystyle 6+y-8=0\Rightarrow y=2$
$\\$

$\displaystyle \textbf{Question 15: } \text{From the given data}$
$\displaystyle \begin{array}{|c|c|c|} \hline \text{Variable} & x & y\\ \hline \text{Mean} & 6 & 8\\ \text{Standard Deviation} & 4 & 6\\ \hline \end{array}$
$\displaystyle \text{and correlation coefficient } \frac{2}{3}, \text{ find}$
$\displaystyle \text{(i) regression coefficients } b_{yx} \text{ and } b_{xy}.$
$\displaystyle \text{(ii) regression line } x \text{ on } y.$
$\displaystyle \text{(iii) most likely value of } x, \text{ when } y=14. \text{ ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given data is}$
$\displaystyle \begin{array}{|c|c|c|} \hline \text{Variable} & x & y \\ \hline \text{Mean} & 6 & 8 \\ \hline \text{Standard Deviation} & 4 & 6 \\ \hline \end{array}$
$\displaystyle \text{and }r=\frac{2}{3}$
$\displaystyle \text{(i) Regression coefficient}$
$\displaystyle b_{yx}=r\frac{\sigma_y}{\sigma_x}=\frac{2}{3}\left(\frac{6}{4}\right)=1$
$\displaystyle \text{and }b_{xy}=r\frac{\sigma_x}{\sigma_y}=\frac{2}{3}\left(\frac{4}{6}\right)=\frac{4}{9}$
$\displaystyle \text{(ii) Regression line }x\text{ on }y\text{ is}$
$\displaystyle x-\bar{x}=b_{xy}(y-\bar{y})$
$\displaystyle \Rightarrow x-6=\frac{4}{9}(y-8)$
$\displaystyle \Rightarrow 9x-54=4y-32$
$\displaystyle \Rightarrow 9x-4y-22=0$
$\displaystyle \text{(iii) Most likely value of }x\text{ when }y=14\text{ is}$
$\displaystyle x=\frac{4}{9}(14)+\frac{22}{9}$
$\displaystyle \Rightarrow x=\frac{56+22}{9}=\frac{78}{9}$
$\\$

$\displaystyle \textbf{Question 16: } \text{Find the coefficient of correlation from the regression lines }$
$\displaystyle x-2y+3=0 \text{ and } 4x-5y+1=0. \text{ ISC 2018}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, regression line }x-2y+3=0$
$\displaystyle \Rightarrow x=2y-3$
$\displaystyle \text{Regression coefficient of }x\text{ on }y,\ b_{xy}=2$
$\displaystyle \text{and }4x-5y+1=0\Rightarrow y=\frac{4}{5}x+\frac{1}{5}$
$\displaystyle \text{Regression coefficient of }y\text{ on }x,\ b_{yx}=\frac{4}{5}$
$\displaystyle \text{Coefficient of correlation }r=\sqrt{b_{xy}\cdot b_{yx}}$
$\displaystyle =\sqrt{2\times\frac{4}{5}}=\sqrt{\frac{8}{5}}$
$\\$

$\displaystyle \textbf{Question 17: } \text{By using the data } \overline{x}=25, \overline{y}=30, b_{yx}=1.6 \text{ and } b_{xy}=0.4, \text{ find}$
$\displaystyle \text{(i) the regression equation } y \text{ on } x.$
$\displaystyle \text{(ii) the most likely value of } y \text{ when } x=60.$
$\displaystyle \text{(iii) the coefficient of correlation between } x \text{ and } y. \text{ ISC 2017}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We have, }\bar{x}=25,\ \bar{y}=30,\ b_{yx}=1.6\text{ and }b_{xy}=0.4$
$\displaystyle \text{(i) The regression equation }y\text{ on }x\text{ is}$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \Rightarrow y-30=1.6(x-25)$
$\displaystyle \Rightarrow y=1.6x-40+30$
$\displaystyle \Rightarrow y=1.6x-10$
$\displaystyle \text{(ii) When }x=60,\text{ then}$
$\displaystyle y=1.6\times60-10$
$\displaystyle =96-10=86$
$\displaystyle \text{(iii) Coefficient of correlation between }x\text{ and }y$
$\displaystyle r=\sqrt{b_{xy}\times b_{yx}}$
$\displaystyle =\sqrt{0.4\times1.6}$
$\displaystyle =\sqrt{0.64}=0.8$
$\displaystyle [\because \text{ coefficient of correlation has the same sign as regression coefficients}]$
$\\$

$\displaystyle \textbf{Question 18: } \text{Find the line of best fit for the following data, treating } x \text{ as dependent}$
$\displaystyle \text{variable (Regression equation } x \text{ on } y).$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x & 14 & 12 & 13 & 14 & 16 & 10 & 13 & 12\\ \hline y & 14 & 23 & 17 & 24 & 18 & 25 & 23 & 24\\ \hline \end{array}$
$\displaystyle \text{Hence, estimate the value of } x \text{ when } y=16. \text{ ISC 2017}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let us construct the following table}$

$\displaystyle \begin{array}{|c|c|c|c|c|} \hline x & x^{2} & y & y^{2} & xy \\ \hline 14 & 196 & 14 & 196 & 196 \\ 12 & 144 & 23 & 529 & 276 \\ 13 & 169 & 17 & 289 & 221 \\ 14 & 196 & 24 & 576 & 336 \\ 16 & 256 & 18 & 324 & 288 \\ 10 & 100 & 25 & 625 & 250 \\ 13 & 169 & 23 & 529 & 299 \\ 12 & 144 & 24 & 576 & 288 \\ \hline \Sigma x=104 & \Sigma x^{2}=1374 & \Sigma y=168 & \Sigma y^{2}=3644 & \Sigma xy=2154 \\ \hline \end{array}$
$\displaystyle \bar{x}=\frac{\Sigma x}{n}=\frac{104}{8}=13,\quad \bar{y}=\frac{\Sigma y}{n}=\frac{168}{8}=21$
$\displaystyle \text{Regression line of }x\text{ on }y$
$\displaystyle x-\bar{x}=b_{xy}(y-\bar{y})$
$\displaystyle b_{xy}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma y^{2}-\frac{1}{n}(\Sigma y)^{2}}$
$\displaystyle =\frac{2154-\frac{1}{8}(104)(168)}{3644-\frac{1}{8}(168)^{2}}$
$\displaystyle =\frac{2154-2184}{3644-3528}=\frac{-30}{116}=-0.25862$
$\displaystyle \therefore x-13=-0.25862(y-21)$
$\displaystyle \Rightarrow x=13-0.25862(y-21)$
$\displaystyle \Rightarrow x=13-0.25862y+5.43102$
$\displaystyle \Rightarrow x=18.43102-0.25862y$
$\displaystyle \text{When }y=16,\ x=18.43102-0.25862(16)$
$\displaystyle \Rightarrow x=18.43102-4.13792=14.2931$
$\\$

$\displaystyle \textbf{Question 19: } \text{The two lines of regressions are } x+2y-5=0 \text{ and }$
$\displaystyle 2x+3y-8=0 \text{ and the variance of } x \text{ is } 12. \text{ Find the coefficient of }\text{correlation. ISC 2016}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, two regression lines are}$
$\displaystyle x+2y-5=0\quad ...(i)$
$\displaystyle \text{and }2x+3y-8=0\quad ...(ii)$
$\displaystyle \text{Clearly, line (i) represents regression line of }y\text{ on }x$
$\displaystyle \text{and line (ii) represents regression line of }x\text{ on }y$
$\displaystyle \text{From Eq. (i), we get}$
$\displaystyle x+2y-5=0\Rightarrow 2y=-x+5$
$\displaystyle \Rightarrow y=-\frac{1}{2}x+\frac{5}{2}$
$\displaystyle \therefore b_{yx}=-\frac{1}{2}$
$\displaystyle \text{From Eq. (ii), we get}$
$\displaystyle 2x+3y-8=0\Rightarrow 3y=-2x+8$
$\displaystyle \Rightarrow y=-\frac{2}{3}x+\frac{8}{3}$
$\displaystyle \therefore b_{xy}=-\frac{2}{3}$
$\displaystyle \text{Now, }r^{2}=b_{yx}\cdot b_{xy}$
$\displaystyle \Rightarrow r^{2}=\left(-\frac{1}{2}\right)\left(-\frac{2}{3}\right)=\frac{1}{3}$
$\displaystyle \Rightarrow r=\pm\sqrt{\frac{1}{3}}=\pm\frac{1}{\sqrt{3}}$
$\displaystyle \text{But }r\text{ has same sign as }b_{yx}\text{ and }b_{xy}$
$\displaystyle \therefore r=-\frac{1}{\sqrt{3}}$
$\\$

$\displaystyle \textbf{Question 20: } \text{The marks obtained by } 10 \text{ candidates in English and Mathematics}$
$\displaystyle \text{are given below.}$
$\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline \text{Marks in English} & 20 & 13 & 18 & 21 & 11 & 12 & 17 & 14 & 19 & 15\\ \hline \text{Marks in Mathematics} & 17 & 12 & 23 & 25 & 14 & 8 & 19 & 21 & 22 & 19\\ \hline \end{array}$
$\displaystyle \text{Estimate the probable score for Mathematics, if the marks obtained in} \\ \text{English are } 24. \text{ ISC 2016}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The marks of 10 candidates in English and Mathematics are given below}$
$\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Marks in English }(x) & \text{Marks in Mathematics }(y) & x^{2} & xy \\ \hline 20 & 17 & 400 & 340 \\ 13 & 12 & 169 & 156 \\ 18 & 23 & 324 & 414 \\ 21 & 25 & 441 & 525 \\ 11 & 14 & 121 & 154 \\ 12 & 8 & 144 & 96 \\ 17 & 19 & 289 & 323 \\ 14 & 21 & 196 & 294 \\ 19 & 22 & 361 & 418 \\ 15 & 19 & 225 & 285 \\ \hline \Sigma x=160 & \Sigma y=180 & \Sigma x^{2}=2670 & \Sigma xy=3005 \\ \hline \end{array}$
$\displaystyle b_{yx}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma x^{2}-\frac{1}{n}(\Sigma x)^{2}}$
$\displaystyle =\frac{3005-\frac{1}{10}(160)(180)}{2670-\frac{1}{10}(160)^{2}}$
$\displaystyle =\frac{3005-2880}{2670-2560}=\frac{125}{110}=1.14$
$\displaystyle \bar{x}=\frac{160}{10}=16,\quad \bar{y}=\frac{180}{10}=18$
$\displaystyle \text{Regression line of }y\text{ on }x$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \Rightarrow y-18=1.14(x-16)$
$\displaystyle \Rightarrow y=1.14x-18.24+18$
$\displaystyle \Rightarrow y=1.14x-0.24$
$\displaystyle \text{When }x=24,\ y=1.14(24)-0.24$
$\displaystyle \Rightarrow y=27.36-0.24=27.12$
$\\$

$\displaystyle \textbf{Question 21: } \text{Given that the observations are } (9,-4),(10,-3),(11,-1),(12,0),$
$\displaystyle (13,1),(14,3),(15,5),(16,8). \text{ Find the two lines of regression and estimate the value}$
$\displaystyle \text{of } y, \text{ when } x=13.5. \text{ ISC 2015}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We construct the table for given data}$
$\displaystyle \begin{array}{|c|c|c|c|c|} \hline x & y & x^{2} & y^{2} & xy \\ \hline 9 & -4 & 81 & 16 & -36 \\ 10 & -3 & 100 & 9 & -30 \\ 11 & -1 & 121 & 1 & -11 \\ 12 & 0 & 144 & 0 & 0 \\ 13 & 1 & 169 & 1 & 13 \\ 14 & 3 & 196 & 9 & 42 \\ 15 & 5 & 225 & 25 & 75 \\ 16 & 8 & 256 & 64 & 128 \\ \hline \Sigma x=100 & \Sigma y=9 & \Sigma x^{2}=1292 & \Sigma y^{2}=125 & \Sigma xy=181 \\ \hline \end{array}$
$\displaystyle \text{Here, }n=8,\ \Sigma x=100,\ \Sigma y=9,\ \Sigma x^{2}=1292,\ \Sigma y^{2}=125,\ \Sigma xy=181$
$\displaystyle \text{Now, mean of }x,\ \bar{x}=\frac{\Sigma x}{n}=\frac{100}{8}=12.5$
$\displaystyle \text{and mean of }y,\ \bar{y}=\frac{\Sigma y}{n}=\frac{9}{8}$
$\displaystyle \text{Regression coefficient of }y\text{ on }x\text{ is}$
$\displaystyle b_{yx}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma x^{2}-\frac{1}{n}(\Sigma x)^{2}}$
$\displaystyle =\frac{181-\frac{1}{8}(100)(9)}{1292-\frac{1}{8}(100)^{2}}$
$\displaystyle =\frac{181-\frac{900}{8}}{1292-\frac{10000}{8}}$
$\displaystyle =\frac{1448-900}{10336-10000}=\frac{548}{336}=\frac{137}{84}$
$\displaystyle \text{and regression coefficient of }x\text{ on }y\text{ is}$
$\displaystyle b_{xy}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma y^{2}-\frac{1}{n}(\Sigma y)^{2}}$
$\displaystyle =\frac{181-\frac{900}{8}}{125-\frac{81}{8}}$
$\displaystyle =\frac{1448-900}{1000-81}=\frac{548}{919}$
$\displaystyle \text{Regression line of }x\text{ on }y\text{ is }x-\bar{x}=b_{xy}(y-\bar{y})$
$\displaystyle \Rightarrow x-12.5=\frac{548}{919}\left(y-\frac{9}{8}\right)$
$\displaystyle \Rightarrow 919x-11487.5=548y-6165$
$\displaystyle \Rightarrow 919x-548y=1087.5$
$\displaystyle \text{And regression line of }y\text{ on }x\text{ is }y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \Rightarrow y-\frac{9}{8}=\frac{137}{84}(x-12.5)$
$\displaystyle \Rightarrow 84y-94.5=137x-1712.5$
$\displaystyle \Rightarrow 137x-84y=1618$
$\displaystyle \text{Since, the regression line of }y\text{ on }x\text{ is }137x-84y=1618$
$\displaystyle \text{For }x=135,$
$\displaystyle 137(135)-84y=1618$
$\displaystyle \Rightarrow 18495-84y=1618$
$\displaystyle \Rightarrow 84y=18495-1618=16877$
$\displaystyle \Rightarrow y=\frac{16877}{84}\approx 200.92$
$\\$

$\displaystyle \textbf{Question 22: } \text{For the given lines of regression, } 3x-2y=5 \text{ and } \\ x-4y=7, \text{ find}$
$\displaystyle \text{(i) regression coefficients } b_{xy} \text{ and } b_{yx}.$
$\displaystyle \text{(ii) coefficient of correlation } r(x,y). \text{ ISC 2014}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) Given regression lines are}$
$\displaystyle 3x-2y=5\quad ...(i)$
$\displaystyle \text{and }x-4y=7\quad ...(ii)$
$\displaystyle \text{Let line of regression }x\text{ on }y\text{ be (i)}$
$\displaystyle \Rightarrow 3x=2y+5$
$\displaystyle \Rightarrow x=\frac{2}{3}y+\frac{5}{3}$
$\displaystyle \therefore b_{xy}=\frac{2}{3}$
$\displaystyle \text{Let line of regression }y\text{ on }x\text{ be (ii)}$
$\displaystyle \Rightarrow x-4y=7\Rightarrow 4y=x-7$
$\displaystyle \Rightarrow y=\frac{1}{4}x-\frac{7}{4}$
$\displaystyle \therefore b_{yx}=\frac{1}{4}$
$\displaystyle \text{(ii) Coefficient of correlation,}$
$\displaystyle r=\sqrt{b_{xy}\cdot b_{yx}}$
$\displaystyle =\sqrt{\frac{2}{3}\times\frac{1}{4}}$
$\displaystyle =\sqrt{\frac{2}{12}}=\sqrt{\frac{1}{6}}=\frac{1}{\sqrt{6}}$
$\\$

$\displaystyle \textbf{Question 23: } \text{Two regression lines are represented by } 4x+10y=9 \text{ and }$
$\displaystyle 6x+3y=4. \text{Find the line of regression of } y \text{ on } x. \text{ ISC 2013}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given regression lines are}$
$\displaystyle 4x+10y=9\quad ...(i)$
$\displaystyle \text{and }6x+3y=4\quad ...(ii)$
$\displaystyle \text{On multiplying Eq. (i) by 6 and Eq. (ii) by 4 and subtracting,}$
$\displaystyle 24x+60y-(24x+12y)=54-16$
$\displaystyle \Rightarrow 48y=38$
$\displaystyle \Rightarrow y=\frac{38}{48}=\frac{19}{24}$
$\displaystyle \text{Put }y=\frac{19}{24}\text{ in Eq. (i), we get}$
$\displaystyle 4x+10\left(\frac{19}{24}\right)=9$
$\displaystyle \Rightarrow 4x+\frac{190}{24}=9$
$\displaystyle \Rightarrow 4x=9-\frac{190}{24}=\frac{216-190}{24}=\frac{26}{24}$
$\displaystyle \Rightarrow x=\frac{26}{96}=\frac{13}{48}$
$\displaystyle \therefore \bar{x}=\frac{13}{48},\ \bar{y}=\frac{19}{24}$
$\displaystyle \text{From Eq. (i), }4x+10y=9\Rightarrow y=-\frac{4}{10}x+\frac{9}{10}$
$\displaystyle \therefore b_{yx}=-\frac{4}{10}$
$\displaystyle \text{Line of regression }y\text{ on }x\text{ is}$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \Rightarrow y-\frac{19}{24}=-\frac{4}{10}\left(x-\frac{13}{48}\right)$
$\displaystyle \Rightarrow 10y-\frac{190}{24}=-4x+\frac{52}{48}$
$\displaystyle \Rightarrow 10y=-4x+\frac{190}{24}+\frac{52}{48}$
$\displaystyle \Rightarrow 10y=-4x+\frac{380+52}{48}$
$\displaystyle \Rightarrow 10y=-4x+\frac{432}{48}$
$\displaystyle \Rightarrow 10y=-4x+9$
$\displaystyle \Rightarrow 4x+10y=9$
$\displaystyle \text{Hence, line of regression }y\text{ on }x\text{ is }4x+10y=9$
$\\$

$\displaystyle \textbf{Question 24: } \text{The following results were obtained with respect to two variables } x \text{ and } y.$
$\displaystyle \sum x=30,\ \sum y=42,\ \sum xy=199,\ \sum x^{2}=184,\ \sum y^{2}=318 \text{ and } n=6.$
$\displaystyle \text{Find the following: } \\ \text{(i) The regression coefficients. (ii) Correlation coefficient between } x \text{ and } y.$
$\displaystyle \text{(iii) Regression equation of } y \text{ on } x. \text{ (iv) The likely value of } y \text{ when } x=10. \text{ ISC 2012}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) We have, }\Sigma x=30,\ \Sigma y=42,\ \Sigma xy=199,$
$\displaystyle \Sigma x^{2}=184,\ \Sigma y^{2}=318\text{ and }n=6$
$\displaystyle \text{Regression coefficient of }x\text{ on }y\text{ is}$
$\displaystyle b_{xy}=\frac{\Sigma xy-\frac{\Sigma x\Sigma y}{n}}{\Sigma y^{2}-\frac{(\Sigma y)^{2}}{n}}$
$\displaystyle =\frac{199-\frac{30\times42}{6}}{318-\frac{42^{2}}{6}}$
$\displaystyle =\frac{199-210}{318-294}=\frac{-11}{24}$
$\displaystyle \text{and regression coefficient of }y\text{ on }x\text{ is}$
$\displaystyle b_{yx}=\frac{\Sigma xy-\frac{\Sigma x\Sigma y}{n}}{\Sigma x^{2}-\frac{(\Sigma x)^{2}}{n}}$
$\displaystyle =\frac{199-\frac{30\times42}{6}}{184-\frac{30^{2}}{6}}$
$\displaystyle =\frac{199-210}{184-150}=\frac{-11}{34}$
$\displaystyle \text{(ii) Since, }b_{xy}\text{ and }b_{yx}\text{ are negative, so }r\text{ is also negative}$
$\displaystyle \therefore r=-\sqrt{b_{xy}\cdot b_{yx}}$
$\displaystyle =-\sqrt{\frac{11}{24}\times\frac{11}{34}}$
$\displaystyle =-\frac{11}{\sqrt{816}}\approx -0.38$
$\displaystyle \text{(iii) Regression equation of }y\text{ on }x\text{ is}$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \text{Now, }\bar{x}=\frac{30}{6}=5,\ \bar{y}=\frac{42}{6}=7$
$\displaystyle \Rightarrow y-7=-\frac{11}{34}(x-5)$
$\displaystyle \Rightarrow 34y-238=-11x+55$
$\displaystyle \Rightarrow 11x+34y-293=0\quad ...(i)$
$\displaystyle \text{(iv) When }x=10,\text{ from Eq. (i)}$
$\displaystyle 11(10)+34y-293=0$
$\displaystyle \Rightarrow 110+34y-293=0$
$\displaystyle \Rightarrow 34y=183$
$\displaystyle \Rightarrow y=\frac{183}{34}=5.382$
$\displaystyle \text{Hence, the likely value of }y\text{ when }x=10\text{ is }5.382.$
$\\$