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

$\displaystyle \textbf{Question 1: } \text{Given below are two regression lines for the same dataset. }$
$\displaystyle L_{1}:2x-5y=2 \quad L_{2}:3y=2x+1 \text{Identify the regression line of } x \text{ on } y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{We know that the regression line of }x\text{ on }y\text{ is } \text{represented by }x=a+by$
$\displaystyle \text{ and it gives the most } \text{probable value of }x\text{ for the given value of }y.$
$\displaystyle \text{Here, }x\text{ is dependent and }y\text{ is independent variable. } \text{So, the line }L_{1}$
$\displaystyle \text{ is the regression line of }x\text{ on }y.$
$\\$

$\displaystyle \textbf{Question 2: } \text{If correlation coefficient } r=0, \text{ then regression lines are }$
$\displaystyle \text{(a) parallel to each other} \quad \text{(b) not mutually perpendicular}$
$\displaystyle \text{(c) parallel to coordinate axis} \quad \text{(d) overlapping lines}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) If the correlation coefficient }r_{xy}=0,$
$\displaystyle \text{then the lines of regression are parallel to the coordinate axes.}$
$\\$

$\displaystyle \textbf{Question 3: } \text{For a given bivariant distribution, the mean of variable } x=4 \text{ and the mean}$
$\displaystyle \text{of variable } y=6. \text{ Find the point of intersection of two regression lines. }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, the mean of variable }x=4$
$\displaystyle \text{i.e. }\bar{x}=4\text{ and mean of variable }y=6$
$\displaystyle \text{i.e. }\bar{y}=6$
$\displaystyle \text{We know that point of intersection of two regression lines}$
$\displaystyle \text{is }(\bar{x},\bar{y})\text{ i.e. }(4,6).$
$\\$

$\displaystyle \textbf{Question 4: } \text{If the two regression coefficients } b_{xy} \text{ and } b_{yx} \text{ are } -0.8 \text{ and }$
$\displaystyle -0.2 \text{ respectively, then the correlation coefficient } (r) \text{ will be }$
$\displaystyle \text{(a) } 0.16 \quad \text{(b) } -0.16 \quad \text{(c) } 0.4 \quad \text{(d) } -0.4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) Given that }b_{xy}=-0.8\text{ and }b_{yx}=-0.2$
$\displaystyle \text{We know that }r=\pm\sqrt{b_{xy}\cdot b_{yx}}$
$\displaystyle \Rightarrow r=\pm\sqrt{(-0.8)\times(-0.2)}$
$\displaystyle =\pm\sqrt{0.16}$
$\displaystyle \Rightarrow r=-0.4$
$\\$

$\displaystyle \textbf{Question 5: } \text{The line of regression of } y \text{ on } x \text{ is } 4x-5y+33=0 \text{ and the line}$
$\displaystyle \text{of regression of } x \text{ on } y \text{ is } 20x-9y-107=0, \text{ then the value of } x \text{ when }$
$\displaystyle y=7 \text{ is }$
$\displaystyle \text{(a) } 8.5 \quad \text{(b) } -8.5 \quad \text{(c) } 0.5 \quad \text{(d) } -0.5$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Consider equation}$
$\displaystyle 20x-9y-107=0$
$\displaystyle \Rightarrow 20x=9y+107$
$\displaystyle \Rightarrow x=\frac{9y+107}{20}$
$\displaystyle \Rightarrow x=\frac{9\times7+107}{20}$
$\displaystyle =\frac{170}{20}=\frac{17}{2}=8.5$
$\\$

$\displaystyle \textbf{Question 6: } \text{If } \overline{x}=40, \overline{y}=6, \sigma_{x}=10, \sigma_{y}=1.5 \text{ and } r=0.9 \text{ for the two sets}$
$\displaystyle \text{of data } x \text{ and } y, \text{ then the regression line of } x \text{ on } y \text{ will be }$
$\displaystyle \text{(a) } x-6y-4=0 \quad \text{(b) } x+6y-4=0 \quad \text{(c) } x-6y+4=0 \quad \text{(d) } x+6y+4=0$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) Given, }\bar{x}=40,\ \bar{y}=6,\ \sigma_x=10,\ \sigma_y=15\text{ and }r=0.9$
$\displaystyle \therefore b_{xy}=r\frac{\sigma_x}{\sigma_y}=\frac{0.9\times10}{15}=6$
$\displaystyle \text{Regression line of }x\text{ on }y\text{ is}$
$\displaystyle x-\bar{x}=b_{xy}(y-\bar{y})$
$\displaystyle \Rightarrow x-40=6(y-6)$
$\displaystyle \Rightarrow x-40=6y-36$
$\displaystyle \Rightarrow x-6y-4=0$
$\\$

$\displaystyle \textbf{Question 7: } \text{If the regression line of } x \text{ on } y \text{ is } 9x+3y-46=0 \text{ and } y \text{ on } x \text{ is}$
$\displaystyle 3x+12y-7=0, \text{ then the correlation coefficient } r \text{ is equal to }$
$\displaystyle \text{(a) } -\frac{1}{12} \quad \text{(b) } \frac{1}{12} \quad \text{(c) } -\frac{1}{2\sqrt{3}} \quad \text{(d) } \frac{1}{2\sqrt{3}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) Given, regression line of }x\text{ on }y\text{ is}$
$\displaystyle 9x+3y-46=0$
$\displaystyle \Rightarrow x=-\frac{3}{9}y+\frac{46}{9}$
$\displaystyle \therefore b_{xy}=-\frac{3}{9}=-\frac{1}{3}$
$\displaystyle \text{and the regression line of }y\text{ on }x\text{ is }3x+12y-7=0$
$\displaystyle \Rightarrow y=-\frac{3}{12}x+\frac{7}{12}$
$\displaystyle \therefore b_{yx}=-\frac{3}{12}=-\frac{1}{4}$
$\displaystyle \therefore \text{Correlation coefficient }r=-\sqrt{b_{xy}\times b_{yx}}$
$\displaystyle =-\sqrt{\left(-\frac{1}{3}\right)\left(-\frac{1}{4}\right)}$
$\displaystyle =-\sqrt{\frac{1}{12}}=-\frac{1}{2\sqrt{3}}$
$\\$

$\displaystyle \textbf{Question 8: } \text{If the lines of regression are parallel to coordinate axes, then the }$
$\displaystyle \text{coefficient of correlation is }$
$\displaystyle \text{(a) } 1 \quad \text{(b) } 0 \quad \text{(c) } -1 \quad \text{(d) } \frac{1}{2}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(b) If the lines of regression are parallel to coordinate } \text{axes, then }r_{xy}\text{ will be zero.}$
$\\$

$\displaystyle \textbf{Question 9: } \text{If } \sigma_x=3, \sigma_y=4 \text{ and } b_{xy}=\frac{1}{3}, \text{ then find the value of correlation}$
$\displaystyle \text{coefficient } (r).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given that }b_{xy}=\frac{1}{3},\ \sigma_x=3\text{ and }\sigma_y=4$
$\displaystyle \text{Now, }b_{xy}=r\times\frac{\sigma_x}{\sigma_y}$
$\displaystyle \Rightarrow \frac{1}{3}=r\times\frac{3}{4}$
$\displaystyle \Rightarrow r=\frac{1}{3}\times\frac{4}{3}$
$\displaystyle \Rightarrow r=\frac{4}{9}$
$\\$

$\displaystyle \textbf{Question 10: } \text{The mean and standard deviation of the two variables } x \text{ and } y$
$\displaystyle \text{are given as } \overline{x}=6, \overline{y}=8, \sigma_x=4 \text{ and } \sigma_y=12. \text{ The correlation coefficient is given as }$
$\displaystyle r=\frac{2}{3}. \text{ Find the regression line of } x \text{ on } y.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }\bar{x}=6,\ \bar{y}=8,\ r=\frac{2}{3},\ \sigma_x=4,\ \sigma_y=12$
$\displaystyle \text{Regression line }x\text{ on }y\text{ is}$
$\displaystyle x-\bar{x}=b_{xy}(y-\bar{y})$
$\displaystyle x-6=b_{xy}(y-8)$
$\displaystyle \text{Now, }b_{xy}=r\frac{\sigma_x}{\sigma_y}=\frac{2}{3}\times\frac{4}{12}=\frac{2}{9}$
$\displaystyle \Rightarrow (x-6)=\frac{2}{9}(y-8)$
$\displaystyle \Rightarrow 9x-54=2y-16$
$\displaystyle \Rightarrow 9x-2y=-16+54$
$\displaystyle \Rightarrow 9x-2y=38$
$\\$

$\displaystyle \textbf{Question 11: } \text{For } 5 \text{ observation of pairs } (x,y) \text{ of variables } X \text{ and } Y, \text{ the following}$
$\displaystyle \text{results are obtained } \sum x=15, \sum y=25, \sum x^{2}=55, \sum y^{2}=135 \text{ and }$
$\displaystyle \sum xy=83. \text{ Calculate the value of } b_{xy} \text{ and } b_{yx}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given for 5 observations,}$
$\displaystyle \sum x=15,\ \sum y=25,\ \sum x^{2}=55,\ \sum y^{2}=135,\ \sum xy=83\text{ and }n=5$
$\displaystyle \therefore \bar{x}=\frac{\sum x}{n}=\frac{15}{5}=3,\quad \bar{y}=\frac{\sum y}{n}=\frac{25}{5}=5$
$\displaystyle \text{Now, }b_{xy}=\frac{\sum xy-\frac{1}{n}\sum x\sum y}{\sum y^{2}-\frac{1}{n}(\sum y)^{2}}$
$\displaystyle =\frac{83-\frac{1}{5}(15)(25)}{135-\frac{1}{5}(25)^{2}}$
$\displaystyle =\frac{83-75}{135-125}=\frac{8}{10}=0.8$
$\displaystyle \text{Now, }b_{yx}=\frac{\sum xy-\frac{1}{n}\sum x\sum y}{\sum x^{2}-\frac{1}{n}(\sum x)^{2}}$
$\displaystyle =\frac{83-\frac{1}{5}(15)(25)}{55-\frac{1}{5}(15)^{2}}$
$\displaystyle =\frac{83-75}{55-45}=\frac{8}{10}=0.8$
$\\$

$\displaystyle \textbf{Question 12: } \text{Shown below is a data set that includes the average temperature and}$
$\displaystyle \text{the number of ice cream cones sold per day by an ice cream truck in a small town}$
$\displaystyle \text{over the course of six days. }$
$\displaystyle \begin{array}{|c|c|} \hline \text{Temperature }({}^{\circ}F)\ x & \text{Cones sold } y\\ \hline 65 & 180\\ 75 & 250\\ 60 & 160\\ 70 & 210\\ 80 & 300\\ 70 & 220\\ \hline \end{array}$
$\displaystyle \text{Find the line of regression using average temperature as the independent variable.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let us construct the table}$
$\displaystyle \begin{array}{|c|c|c|c|} \hline \text{Temperature }(x) & \text{Cones sold }(y) & x^{2} & xy \\ \hline 65 & 180 & 4225 & 11700 \\ 75 & 250 & 5625 & 18750 \\ 60 & 160 & 3600 & 9600 \\ 70 & 210 & 4900 & 14700 \\ 80 & 300 & 6400 & 24000 \\ 70 & 220 & 4900 & 15400 \\ \hline \Sigma x=420 & \Sigma y=1320 & \Sigma x^{2}=29650 & \Sigma xy=94150 \\ \hline \end{array}$
$\displaystyle \text{Here, }\bar{x}=\frac{\Sigma x}{n}=\frac{420}{6}=70\text{ and }\bar{y}=\frac{\Sigma y}{n}=\frac{1320}{6}=220$
$\displaystyle \text{Regression coefficient of }y\text{ on }x,$
$\displaystyle b_{yx}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma x^{2}-\frac{1}{n}(\Sigma x)^{2}}$
$\displaystyle =\frac{94150-\frac{420\times1320}{6}}{29650-\frac{(420)^{2}}{6}}$
$\displaystyle =\frac{94150-92400}{29650-29400}=\frac{1750}{250}=7$
$\displaystyle \text{Equation of line of regression of }y\text{ on }x\text{ is}$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \Rightarrow y-220=7(x-70)$
$\displaystyle \Rightarrow y-220=7x-490$
$\displaystyle \Rightarrow y=7x-270$
$\\$

$\displaystyle \textbf{Question 13: } \text{The line of regression of marks in Maths } (X) \text{ and marks in English } (Y) \text{ for a}$
$\displaystyle \text{class of } 50 \text{ students is } 3y-5x+180=0. \text{ The average score in English is } 44 \text{ and variance of}$
$\displaystyle \text{marks in Maths is } \frac{9}{16} \text{th of the variance of marks in English. Find the average score in Maths.}$
$\displaystyle \text{Also, find out the coefficient of correlation between marks in Maths and English. }$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }h=50$
$\displaystyle \text{Regression line of }x\text{ on }y\text{ is}$
$\displaystyle 3y-5x+180=0\quad ...(i)$
$\displaystyle \Rightarrow 5x=3y+180$
$\displaystyle \Rightarrow x=\frac{3}{5}y+36$
$\displaystyle \therefore b_{xy}=\frac{3}{5}$
$\displaystyle \text{Let variance of marks in Maths }=\sigma_x^{2}\text{ and}$
$\displaystyle \text{variance of marks in English }=\sigma_y^{2}$
$\displaystyle \text{From given condition, we have}$
$\displaystyle \sigma_x^{2}=\frac{9}{16}\sigma_y^{2}\Rightarrow \frac{\sigma_x}{\sigma_y}=\frac{3}{4}$
$\displaystyle \text{We have, }b_{xy}=r\frac{\sigma_x}{\sigma_y}$
$\displaystyle \Rightarrow \frac{3}{5}=r\times\frac{3}{4}$
$\displaystyle \Rightarrow r=\frac{4}{5}$
$\displaystyle \text{Given, }\bar{y}=44$
$\displaystyle \text{On substituting }y=44\text{ in Eq. (i), we get}$
$\displaystyle 3(44)-5x+180=0$
$\displaystyle \Rightarrow 132-5x+180=0$
$\displaystyle \Rightarrow 5x=312\Rightarrow x=\frac{312}{5}=62.4$
$\displaystyle \therefore \bar{x}=62.4$
$\displaystyle \text{Mean marks in Maths }=62.4$
$\\$

$\displaystyle \textbf{Question 14: } \text{Find the equation of the regression line of } y \text{ on } x, \text{ if the observations } (x,y)$
$\displaystyle \text{are as follows } (1,4),(2,8),(3,2),(4,12),(5,10),(6,14),(7,16),(8,6),(9,18).$
$\displaystyle \text{Also, find the estimated value of } y \text{ when } x=14.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let us construct the following table for given data}$
$\displaystyle \begin{array}{|c|c|c|c|c|} \hline x & y & x^{2} & y^{2} & xy \\ \hline 1 & 4 & 1 & 16 & 4 \\ 2 & 8 & 4 & 64 & 16 \\ 3 & 2 & 9 & 4 & 6 \\ 4 & 12 & 16 & 144 & 48 \\ 5 & 10 & 25 & 100 & 50 \\ 6 & 14 & 36 & 196 & 84 \\ 7 & 16 & 49 & 256 & 112 \\ 8 & 6 & 64 & 36 & 48 \\ 9 & 18 & 81 & 324 & 162 \\ \hline \Sigma x=45 & \Sigma y=90 & \Sigma x^{2}=285 & \Sigma y^{2}=1140 & \Sigma xy=530 \\ \hline \end{array}$
$\displaystyle \text{Here, }n=9$
$\displaystyle \therefore \bar{x}=\frac{\Sigma x}{n}=\frac{45}{9}=5,\quad \bar{y}=\frac{\Sigma y}{n}=\frac{90}{9}=10$
$\displaystyle \text{Regression coefficient of }y\text{ on }x$
$\displaystyle b_{yx}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma x^{2}-\frac{1}{n}(\Sigma x)^{2}}$
$\displaystyle =\frac{530-\frac{1}{9}(45)(90)}{285-\frac{1}{9}(45)^{2}}$
$\displaystyle =\frac{530-450}{285-225}=\frac{80}{60}=\frac{4}{3}$
$\displaystyle \text{Now, regression equation of }y\text{ on }x\text{ is}$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle \Rightarrow y-10=\frac{4}{3}(x-5)$
$\displaystyle \Rightarrow 3y-30=4x-20$
$\displaystyle \Rightarrow 4x-3y+10=0$
$\displaystyle \text{Now, estimate value of }y\text{ when }x=14$
$\displaystyle 4(14)-3y+10=0$
$\displaystyle \Rightarrow 56-3y+10=0$
$\displaystyle \Rightarrow 3y=66\Rightarrow y=22$
$\\$

$\displaystyle \textbf{Question 15: } \text{Find the regression coefficients } b_{yx}, b_{xy} \text{ and correlation coefficient } r$
$\displaystyle \text{for the following data } (2,8),(6,8),(4,5),(7,6),(5,2).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, data }(2,8),(6,8),(4,5),(7,6),(5,2)$
$\displaystyle \text{We construct the following table}$
$\displaystyle \begin{array}{|c|c|c|c|c|} \hline x & y & x^{2} & y^{2} & xy \\ \hline 2 & 8 & 4 & 64 & 16 \\ 6 & 8 & 36 & 64 & 48 \\ 4 & 5 & 16 & 25 & 20 \\ 7 & 6 & 49 & 36 & 42 \\ 5 & 2 & 25 & 4 & 10 \\ \hline \Sigma x=24 & \Sigma y=29 & \Sigma x^{2}=130 & \Sigma y^{2}=193 & \Sigma xy=136 \\ \hline \end{array}$
$\displaystyle n=5$
$\displaystyle b_{yx}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma x^{2}-\frac{1}{n}(\Sigma x)^{2}}$
$\displaystyle =\frac{136-\frac{1}{5}(24)(29)}{130-\frac{1}{5}(24)^{2}}$
$\displaystyle =\frac{136-139.2}{130-115.2}=\frac{-3.2}{14.8}=-\frac{8}{37}$
$\displaystyle b_{xy}=\frac{\Sigma xy-\frac{1}{n}\Sigma x\Sigma y}{\Sigma y^{2}-\frac{1}{n}(\Sigma y)^{2}}$
$\displaystyle =\frac{136-139.2}{193-\frac{1}{5}(29)^{2}}$
$\displaystyle =\frac{-3.2}{193-168.2}=\frac{-3.2}{24.8}=-\frac{4}{31}$
$\displaystyle r=-\sqrt{b_{xy}\cdot b_{yx}}$
$\displaystyle =-\sqrt{\left(-\frac{8}{37}\right)\left(-\frac{4}{31}\right)}$
$\displaystyle =-\sqrt{\frac{32}{1147}}\approx -0.167$
$\\$

$\displaystyle \textbf{Question 16: } \text{Given that the observations are } (9,4),(10,-3),(11,-1),(13,1),$
$\displaystyle (14,3),(15,5),(16,8), \text{ find the two lines of regression. Estimate the value of } y \text{ when } x=13.5.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given data }(9,-4),(10,-3),(11,-1),(13,1),(14,3),(15,5),(16,8)$
$\displaystyle \text{Now, we construct the following table}$
$\displaystyle \begin{array}{|c|c|c|c|c|} \hline x & y & x^{2} & y^{2} & xy \\ \hline 9 & -4 & 81 & 16 & -36 \\ 10 & -3 & 100 & 9 & -30 \\ 11 & -1 & 121 & 1 & -11 \\ 13 & 1 & 169 & 1 & 13 \\ 14 & 3 & 196 & 9 & 42 \\ 15 & 5 & 225 & 25 & 75 \\ 16 & 8 & 256 & 64 & 128 \\ \hline \Sigma x=88 & \Sigma y=9 & \Sigma x^{2}=1148 & \Sigma y^{2}=125 & \Sigma xy=181 \\ \hline \end{array}$
$\displaystyle \text{Here, }n=7,\ \Sigma x=88,\ \Sigma y=9$
$\displaystyle \therefore \bar{x}=\frac{\Sigma x}{n}=\frac{88}{7}\text{ and }\bar{y}=\frac{\Sigma y}{n}=\frac{9}{7}$
$\displaystyle \text{Now, }b_{yx}=\frac{\Sigma xy-\frac{\Sigma x\Sigma y}{n}}{\Sigma x^{2}-\frac{(\Sigma x)^{2}}{n}}$
$\displaystyle =\frac{181-\frac{88\times9}{7}}{1148-\frac{(88)^{2}}{7}}$
$\displaystyle =\frac{1267-792}{8036-7744}=\frac{475}{292}$
$\displaystyle \therefore \text{Equation of line of regression of }y\text{ on }x\text{ is}$
$\displaystyle y-\bar{y}=b_{yx}(x-\bar{x})$
$\displaystyle y-\frac{9}{7}=\frac{475}{292}\left(x-\frac{88}{7}\right)$
$\displaystyle \text{Now, }b_{xy}=\frac{\Sigma xy-\frac{\Sigma x\Sigma y}{n}}{\Sigma y^{2}-\frac{(\Sigma y)^{2}}{n}}$
$\displaystyle =\frac{181-\frac{88\times9}{7}}{125-\frac{81}{7}}$
$\displaystyle =\frac{1267-792}{875-81}=\frac{475}{794}$
$\displaystyle \text{Equation of line of regression of }x\text{ on }y\text{ is}$
$\displaystyle x-\bar{x}=b_{xy}(y-\bar{y})$
$\displaystyle \Rightarrow x-\frac{88}{7}=\frac{475}{794}\left(y-\frac{9}{7}\right)$
$\displaystyle \text{Now, the line of regression of }y\text{ on }x\text{ is}$
$\displaystyle y-\frac{9}{7}=\frac{475}{292}\left(x-\frac{88}{7}\right)$
$\displaystyle \text{Given, }x=13.5$
$\displaystyle \text{We get}$
$\displaystyle y=\frac{475}{292}\left(13.5-\frac{88}{7}\right)+\frac{9}{7}$
$\displaystyle y=1.63(13.5-12.57)+1.28$
$\displaystyle y=1.52+1.28$
$\displaystyle y=2.8$
$\\$