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Maths (commerce) Solve the Following Question.(3 Marks)

Linear Regression (p-2) · Maharashtra Board STD 12 Commerce / Arts (MSBSHSE - English Medium). 17 questions.

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Solve the Following Question.(3 Marks)

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17 questions · self-marked practice — reveal the answer and mark yourself.

Question 13 Marks
Identify the regression equations of x on y and y on x from the following equations.
2x + 3y = 6 and 5x + 7y – 12 = 0
Answer
$
\begin{array}{lr}
3 y=2 x+6 & 5 x=-7 y+12 \\
y=\frac{-2}{3} x+2 & x=\frac{-7}{5} y+\frac{12}{5} \\
\therefore b_{y x}=\frac{-2}{3} & \therefore b_{x y}=\frac{-7}{5} \\
b_{y x} \cdot b_{x y}=\frac{-2}{3} \times \frac{-7}{5} & \\
=\frac{14}{15} \in[0,1] & \text {}
\end{array}
$
$\therefore$ Our assumption is correct
$\therefore$ Regression equation of $Y$ on $X$ is $2 x+3 y=6$
$\therefore$ Regression equation of $X$ on $Y$ is $5 x+7 y-12=0$
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Question 23 Marks
In partially destroyed record, the following data are available variance of X = 25. Regression equation of Y on X is 5y – x = 22 and Regression equation of X on Y is 64x – 45y = 22 Find
(i) Mean values of X and Y.
(ii) Standard deviation of Y.
(iii) Coefficient of correlation between X and Y.
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Question 33 Marks
The equation of two regression lines are $x-4 y=5$ and $16 y-x=64$. Find means of $X$ and $\mathrm{Y}$. Also, find the correlation coefficient between $\mathrm{X}$ and $\mathrm{Y}$.
Answer
Since $(\bar{x}, \bar{y})$ is the point of intersection of the regression lines.
$
\begin{aligned}
& x-4 y=5 \ldots .(i) \\
& -x+16 y=64 . \\
& 12 y=69 \\
& y=5.75
\end{aligned}
$
Substituting $y=5.75$ in equation (i)
$
\begin{aligned}
& x-4(5.75)=5 \\
& x-23=5 \\
& x=28 \\
& \therefore \bar{x}=28, \bar{y}=5.75 \\
& x-4 y=5 \\
& x=4 y+5 \\
& \therefore b_{x y}=4 \\
& 16 y-x=64 \\
& 16 y=x+64 \\
& y=\frac{1}{16} x+4 \\
& b_{y x}=\frac{1}{16} \\
& b_{y x} \cdot b_{x y}=\frac{1}{16} \times 4=\frac{1}{4} \in[0,1] \\
& \therefore \text { Our assumption is correct } \\
& \therefore r^2=b_{y x} \cdot b_{x y} \\
& r^2=\frac{1}{4} \\
& r= \pm \frac{1}{2}
\end{aligned}
$
Since $b_{y x}$ and $b_{x y}$ are positive,
$
\therefore r=\frac{1}{2}=0.5
$
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Question 43 Marks
For 50 students of a class, the regression equation of marks in statistics $(\mathrm{X})$ on the marks in Accountancy $(Y)$ is $3 y-5 x+180=0$. The mean marks in accountancy is 44 and the variance of marks in statistics $\left(\frac{9}{16}\right)^{\text {th }}$ of the variance of marks in accountancy. Find the mean in statistics and the correlation coefficient between marks in two subjects.
Answer
Given, $\mathrm{n}=50, y=44$
$
\begin{aligned}
& \sigma_x^2=\frac{9}{16} \sigma_y^2 \\
& \therefore \frac{\sigma_x}{\sigma_x}=\frac{3}{4}
\end{aligned}
$
Since $(\bar{x}, \bar{y})$ is the point intersection of the regression line.
$\therefore(\bar{x}, \bar{y})$ satisfies the regression equation.
$
\begin{aligned}
& 3 \bar{y}-5 \bar{x}+180=0 \\
& 3(44)-5 \bar{x}+180=0 \\
& \therefore 5 \bar{x}=132+180 \\
& \bar{x}=\frac{312}{5}=62.4
\end{aligned}
$
$\therefore$ Mean marks in statistics is 62.4
Regression equation of $\mathrm{X}$ on $\mathrm{Y}$ is $3 \mathrm{y}-5 \mathrm{x}+180=0$
$
\begin{aligned}
& \therefore 5 x=3 y+180 \\
& \therefore x=\frac{3}{5}+36 \\
& \therefore b_{x y}=\frac{3}{5}
\end{aligned}
$
Also, $b_{x y}=r \frac{\sigma_x}{\sigma_y}$ $\frac{3}{5}=r \frac{3}{4}$
$
r=\frac{4}{5}=0.8
$
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Question 53 Marks
In a partially destroyed laboratory record of an analysis of regression data, the following data are legible:
Variance of X = 9
Regression equations:
8x – 10y + 66 = 0 And 40x – 18y = 214.
Find on the basis of the above information
(i) The mean values of X and Y.
(ii) Correlation coefficient between X and Y.
(iii) Standard deviation of Y.
Answer
Given, ${\sigma_x}^2=9, \sigma_x=3$
(i) $(\bar{x}, \bar{y})$ is the point of intersection of the regression lines
$
\begin{aligned}
& 40 \mathrm{x}-50 \mathrm{y}=-330 \ldots \ldots . .(\mathrm{i}) \\
& 40 \mathrm{x}-50 \mathrm{y}=+214 \ldots \ldots . .(\mathrm{i}) \\
& -32 \mathrm{y}=-544 \\
& \mathrm{y}=17 \\
& \therefore \bar{y}=17 \\
& 8 \mathrm{x}-10(17)+66=0 \\
& 8 \mathrm{x}=104 \\
& \mathrm{x}=13 \\
& \therefore \bar{x}=13
\end{aligned}
$
(ii)$
\begin{array}{ll}
8 x-10 y+66=0, & 40 x-18 y=214 \\
10 y=8 x+66, & 40 x=18 y+214 \\
y=\frac{8 x}{10}+\frac{66}{10} & x=\frac{18}{40}+\frac{214}{40} \\
\therefore b_{y x}=\frac{8}{10}=\frac{4}{5} & b_{x y}=\frac{18}{40}+\frac{9}{20} \\
b_{y x} \cdot b_{x y}=\frac{4}{5} \times \frac{9}{20}=\frac{9}{25} & \text { which belongs to }[0,1]
\end{array}
$
$\therefore$ Our assumption is correct
$
r^2=b_{y x} \cdot b_{x y}
$
$r^2=\frac{9}{25}$
$
r= \pm \frac{3}{5}
$
Since $b_{y x}$ and $b_{x y}$ are positive $\therefore r=\frac{3}{5}$
(iii)
$
\begin{aligned}
& b_{y x}=r \frac{\sigma_y}{\sigma_x} \\
& \frac{4}{5}=\frac{3}{5} \times \frac{\sigma_y}{3} \\
& \sigma_y=4
\end{aligned}
$
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Question 63 Marks
From the two regression equations find $\mathrm{r}, \bar{x}$ and $\bar{y}$.
$4 y=9 x+15$ and $25 x=4 y+17$
Answer
Given $4 y=9 x+15$ and $25 x=4 y+17$
$
\begin{aligned}
& y=\frac{9 x}{4}+\frac{15}{4} \\
& x=\frac{4 y}{25}+\frac{17}{25} \\
& b_{y x}=\frac{9}{4} b_{x y}=\frac{4}{25} \\
& b_{y x} \cdot b_{x y}=\frac{9}{4} \times \frac{4}{25}
\end{aligned}
$
$=\frac{9}{25} \quad$ which belongs to $[0,1]$
$\therefore$ Our assumption is correct
$
\therefore r^2=b_{y x} \cdot b_{x y}
$
$r^2=\frac{9}{25} \quad$
$
r= \pm \frac{3}{5}
$
Since $b_{y x}$ and $b_{x y}$ are positive.
$
\therefore r=\frac{3}{5}=0.6
$
$(\bar{x}, \bar{y})$ is the point of intersection of the regression lines
$
\begin{aligned}
& 9 \mathrm{x}-4 \mathrm{y}=-15 \\
& 25 \mathrm{x}-4 \mathrm{y}=17 \\
& -16 \mathrm{x}=-32 \\
& \mathrm{x}=2 \\
& \therefore \bar{x}=2
\end{aligned}
$
Substituting $x=2$ in equation (i)
$
\begin{aligned}
& 9(2)-4 y=-15 \\
& 18+15=4 y \\
& 33=4 y \\
& y=33 / 4=8.25 \\
& \therefore \bar{y}=8.25
\end{aligned}
$
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Question 73 Marks
Given the following information about the production and demand of a commodity obtain the two regression lines:
  PRODUCTION (X)DEMAND (Y)
MEN 85 90
S.D. 5 6
Coefficient of correlation between X and Y is 0.6. Also estimate the problem when demand is 100.
Answer
Given $\bar{x}=85, \bar{y}=90, \sigma_{\mathrm{x}}=5, \sigma_{\mathrm{y}}=6$ and $\mathrm{r}=0.6$
$
\begin{aligned}
& \text { bxy }=\frac{r \sigma_x}{\sigma_y}=0.6 \times \frac{5}{6}=0.5 \\
& \text { byx }=\frac{r \sigma_y}{\sigma_x}=0.6 \times \frac{6}{5}=0.72
\end{aligned}
$
Regression equation of $\mathrm{X}$ on $\mathrm{Y}$ is
$
\begin{aligned}
& (X-\bar{x})=b_{x y}(Y-\bar{y}) \\
& (X-85)=0.5(y-90) \\
& (X-85)=0.5 y-45 \\
& X=0.5 y+40
\end{aligned}
$
When $\mathrm{y}=100$,
$
\begin{aligned}
& x=0.5(100)+40 \\
& =50+40 \\
& =90
\end{aligned}
$
Regression equation of $\mathrm{Y}$ on $\mathrm{X}$ is
$
\begin{aligned}
& (Y-\bar{y})=b_{y x}(X-\bar{x}) \\
& (Y-90)=0.72(x-85) \\
& (Y-90)=0.72 x-61.2 \\
& Y=0.72 x+28.8
\end{aligned}
$
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Question 83 Marks
You are given the following information about advertising expenditure and sales.
Advertisement expenditure (Rs.in lakh)Sales (Rs.in lakh)
Arithmetic mean1090
Standard deviation312
Correlation coefficient between X and Y is 0.8
(i) Obtain two regression equations.
(ii) What is the likely sales when the advertising budget is ₹ 15 lakh?
(iii) What should be the advertising budget if the company wants to attain sales target of ₹ 120 lakh?
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Question 93 Marks
From the data of 7 pairs of observations on $\mathrm{X}$ and $\mathrm{Y}$ following results are obtained.
$
\begin{aligned}
& \Sigma\left(x_i-70\right)=-35, \Sigma\left(y_i-60\right)=-7, \Sigma\left(x_i-70\right)^2=2989, \Sigma\left(y_i-60\right)^2=476, \Sigma\left(x_i-70\right)\left(y_i-60\right)= \\
& 1064 \text { [Given } v 0.7884=0.8879] \\
& \text { Obtain }
\end{aligned}
$(i) The line of regression of $Y$ on $X$.
(ii) The line of regression of $X$ on $Y$.
(iii) The correlation coefficient between $\mathrm{X}$ and $\mathrm{Y}$.
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Question 103 Marks
From the data of 20 pairs of observation on X and Y, following result are obtained $\bar{x}=199$,
$
\begin{aligned}
& \bar{y}=94, \sum\left(x_i-\bar{x}\right)^2=1200, \sum\left(y_i-\bar{y}\right)^2=300 \\
& \sum\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=-250 \\
& \text { Find }
\end{aligned}
$
(i) The line of regression of $Y$ on $X$.
(ii) The line of regression of $X$ on $Y$.
(iii) Correlation coefficient between $\mathrm{X}$ on $\mathrm{Y}$.
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Question 113 Marks
For certain bivariate data the following information are available
XY
A.M.1317
S.D.32
Correlation coefficient between x and y is 0.6, estimate x when y = 15 and estimate y when x = 10.
Answer
$\begin{aligned} & \text { Given, } \bar{x}=13, \bar{y}=17, \sigma_{\mathrm{x}}=3, \sigma_{\mathrm{y}}=2, \mathrm{r}=0.6 \\ & \mathrm{~b}_{\mathrm{yx}}=\frac{r \sigma_y}{\sigma_x}=0.6 \times \frac{2}{3}=0.4 \\ & \mathrm{~b}_{\mathrm{xy}}=\frac{r_{\sigma_x}}{\sigma_y}=0.6 \times \frac{3}{2}=0.9 \\ & \text { Regression equation of } \mathrm{Y} \text { on } \mathrm{X} \\ & (\mathrm{Y}-\bar{y})=\mathrm{b}_{\mathrm{yx}}(\mathrm{X}-\bar{x}) \\ & \mathrm{Y}-17=0.4(\mathrm{x}-13) \\ & \mathrm{Y}=0.4 \mathrm{x}+11.8 \\ & \text { When } \mathrm{x}=10 \\ & \mathrm{Y}=0.4(10)+11.8 \\ & =4+11.8 \\ & =15.8 \\ & \text { Regression equation of } \mathrm{X} \text { on } \mathrm{Y} \\ & (\mathrm{X}-\bar{x})=\mathrm{b}_{\mathrm{xy}}(\mathrm{Y}-\bar{y}) \\ & (\mathrm{X}-13)=0.9(\mathrm{y}-17) \\ & \mathrm{X}-13=0.9 \mathrm{y}-15.3 \\ & \mathrm{X}=0.9 \mathrm{y}-2.3 \\ & \text { When } \mathrm{y}=15 \\ & \mathrm{X}=0.9(15)-2.3 \\ & =13.5-2.3 \\ & =11.2\end{aligned}$
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Question 123 Marks
For bivariate data.
$
\bar{x}=53, \bar{x}=28, b_{y x}=-1.2, b_{x y}=-0.3
$
Find,
(i) Correlation coefficient between $\mathrm{X}$ and $\mathrm{Y}$.
(ii) Estimate $Y$ for $X=50$
(iii) Estimate $X$ for $Y=25$
Answer
$
\begin{aligned}
& \text { (i) } r^2=b_{y x} \cdot b_{x y} \\
& r^2=(-1.2)(-0.3) \\
& r^2=0.36 \\
& r= \pm 0.6
\end{aligned}
$
Since, $b_{y x}$ and bxy are negative, $r=-0.6$
(ii) Regression equation of $\mathrm{Y}$ on $\mathrm{X}$ is
$
\begin{aligned}
& (Y-\bar{y})=b_{y x}(X-\bar{x}) \\
& Y-28=-1.2(50-53) \\
& Y-28=-1.2(-3) \\
& Y-28=3.6 \\
& Y=31.6
\end{aligned}
$
(iii) Regression equation of $\mathrm{X}$ on $\mathrm{Y}$ is
$
\begin{aligned}
& (X-\bar{x})=b_{x y}(Y-\bar{y}) \\
& (X-53)=-0.3(25-28) \\
& X-53=-0.3(-3) \\
& X-53=0.9 \\
& X=53.9
\end{aligned}
$
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Question 133 Marks
Compute the appropriate regression equation for the following data.
X [ Independent variable]2456811
Y[Dependent variable]181210875
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Question 143 Marks
The following sample gave the number of hours of study (X) per day for an examination and marks (Y) obtained by 12 students.
X 3 3 3 4 4 5 5 5 6 6 7 8
Y 45 60 55 60 75 70 80 75 90 80 75 85
Obtain the line of regression of marks on hours of study.
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Question 153 Marks
Find the following data, find the regression equation of Y on X, and estimate Y when X = 10.
X 1 2 3 4 5 6
Y 2 4 7 6 5 6
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Question 163 Marks
For the following data, find the regression line of Y on X
X123
Y216
Hence find the most likely value of y when x = 4
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Question 173 Marks
The HRD manager of the company wants to find a measure which he can use to fix the monthly income of persons applying for the job in the production department. As an experimental project. He collected data of 7 persons from that department referring to years of service and their monthly incomes.
Years of service (X)117958610
Monthly Income (Rs. 1000's) (Y)108659711
(i) Find the regression equation of income on years of service.
(ii) What initial start would you recommend for a person applying for the job after having served in a similar capacity in another company for 13 years?
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