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Maths (commerce) Solve the Following Question.(5 Marks)

Linear Regression (p-2) · Maharashtra Board STD 12 Commerce / Arts (MSBSHSE - English Medium). 7 questions.

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Solve the Following Question.(5 Marks)

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

Question 15 Marks
The data obtained on X, the length of time in weeks that a promotional project has been in progress at a small business, and Y the percentage increase in weekly sales over the period just prior to the beginning of the campaign.
X123413123424
Y101018201115121517191316
Find the equation of regression line to predict percentage increase in sales if the company has been in progress for 1.5 weeks.
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Question 25 Marks
If the two regression lines for a bivariate data are $2 x=y+15(x$ on $y)$ and $4 y-3 x+25(y$ on $x$ ) find
(i) $\bar{x}$
(ii) $\bar{y}$
(iii) $b_{y x}$
(iv) $b_{x y}$
(v) $r$ [Given $\sqrt{ } 0.375=0.61$ ]
Answer
Since $(\bar{x}, \bar{y})$ is the point of intersection of the regression line
$
\begin{aligned}
& 2 x=y+15 \\
& 4 y=3 x+25 \\
& 2 x-y=15 \ldots \ldots \text { (i) } \\
& 3 x-4 y=-25 . . . . . .(i i)
\end{aligned}
$
Multiplying equation (i) by 4
$
\begin{aligned}
& 8 x-4 y=60 \\
& 3 x-4 y=-25
\end{aligned}
$
on Subtracting,
$
\begin{aligned}
& 5 x=85 \\
& \therefore x=17
\end{aligned}
$
Substituting $x$ in equation (i)
$
\begin{aligned}
& 2(17)-y=15 \\
& 34-15=y \\
& \therefore y=15
\end{aligned}
$
$\begin{array}{lr}\bar{x}=17, \bar{y}=19 & \\ 2 x=y+15 & 4 y=3 x+25 \\ x=\frac{1}{2} y+7.5 & y=\frac{3}{4} x+6.25 \\ b_{x y}=\frac{1}{2} & b_{y x}=\frac{3}{4} \\ b_{x y} \cdot b_{y x}=\frac{1}{2} \times \frac{3}{4}=\frac{3}{8} \in[0,1] & \end{array}$
$\therefore$Our assumption is correct
$
\begin{aligned}
r^2 & =b_{x y} \cdot b_{y x} \\
r^2 & =\frac{3}{8} \quad \text {} \\
& =0.375 \\
r & = \pm \sqrt{0.375}= \pm 0.61
\end{aligned}
$
Since $b_{y x}$ and $b_{x y}$ are positive, $\therefore r=0.61$
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Question 35 Marks
The two regression lines between height (X) in includes and weight (Y) in kgs of girls are 4y – 15x + 500 = 0 and 20x – 3y – 900 = 0. Find the mean height and weight of the group. Also, estimate the weight of a girl whose height is 70 inches.
Answer
Since $(\bar{x}, \bar{y})$ is the point intersection of the regression lines
$
\begin{aligned}
& 15 x-4 y=500 \ldots \\
& 20 x-3 y=900 \ldots \\
& 60 x-16 y-2000 \\
& 60 x-9 y=2700 \\
& \text { on subtracting, } \\
& -7 y=-700 \\
& y=100
\end{aligned}
$
on subtracting,
$
\begin{aligned}
& -7 y=-700 \\
& y=100
\end{aligned}
$
$
\begin{aligned}
& \text { Substituting } y=100 \text { in equation (i) } \\
& 15 x-4(100)=500 \\
& 15 x=900 \\
& x=60
\end{aligned}
$
$
\begin{array}{ll}
\therefore \bar{x}=60, \bar{y}=100 & \\
4 y=15 x-500 & 20 x=3 y+100 \\
y=\frac{15}{4} x-125 & x=\frac{3}{20} y+45 \\
b_{y x}=\frac{15}{4} & b_{x y}=\frac{3}{20}
\end{array}
$
$
\begin{aligned}
& b_{y x} \cdot b_{x y}=\frac{15}{4} \times \frac{3}{20} \text { } \\
& =\frac{9}{16} \in[0,1]
\end{aligned}
$
$\therefore$ Our assumption is correct
$\therefore$ Regression equation of $\mathrm{Y}$ on $\mathrm{X}$ is
$
Y=\frac{15}{4} x-125
$
When $x=70$
$
\begin{aligned}
& Y=\frac{15}{4} \times 70=-125 \\
& =262.5-125 \\
& =137.5 \mathrm{~kg}
\end{aligned}
$
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Question 45 Marks
Regression equation of two series are $2 x-y-15=0$ and $4 y+25=0$ and $3 x-4 y+25=0$. Find $\bar{x}, \bar{y}$ and regression coefficients, Also find coefficients of correlation. [Given $\sqrt{ } 0.375=$ $0.61]$
Answer
Since $(\bar{x}, \bar{y})$ is the point of intersection of the regression line
$
\begin{aligned}
& 2 x=y+15 \\
& 4 y=3 x+25 \\
& 2 x-y=15 \ldots \ldots (i) \\
& 3 x-4 y=-15 \ldots \ldots (ii)
\end{aligned}
$
Multiply equation (i) by 4
$
\begin{aligned}
& 8 x-4 y=60 \\
& 3 x-4 y=-25
\end{aligned}
$
on subtracting,
$
\begin{aligned}
& 5 x=85 \\
& x=17
\end{aligned}
$
Substituting $x$ in equation (i)
$
\begin{aligned}
& 2(17)-y=15 \\
& 34-15=y \\
& y=15 \\
& \therefore \bar{x}=17, y=19 \\
& 2 x=y+15 \\
& 4 y=3 x+25 \\
& x=\frac{1}{2} y+7.5 \\
& y=\frac{3}{4} x+6.25 \\
&
\end{aligned}
$
$
\begin{aligned}
& b_{x y}=\frac{1}{2} \quad \text {} \\
& b_{x y} \cdot b_{y x}=\frac{1}{2} \times \frac{3}{4}=\frac{3}{8} \in[0,1]
\end{aligned}
$
$\therefore$ Our assumption is correct
$
\begin{aligned}
& r^2=b_{x y} \cdot b_{y x} \\
& r^2=\frac{3}{8}=0.375 \\
& r= \pm \sqrt{ } 0.375= \pm 0.61
\end{aligned}
$
Since, $b_{y x}$ and $b_{x y}$ are positive, $\therefore r=0.61$
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Question 55 Marks
\begin{align}
\text { The two regression equation are } 5 x-6 y+90=0 \text { and } 15 x-8 y-130=0 \text {. Find } \bar{x}, \bar{y}, r \text {. }
\end{align}
Answer
Since $(\bar{x}, y)$ is the point of intersection of the regression lines
$
\begin{aligned}
& 5 x-6 y+90=0 \\
& 15 x-8 y-130=0 \\
& 15 x-18 y+270=0 \\
& 15 x-8 y-130=0
\end{aligned}
$
on subtracting,
$
\begin{aligned}
& -10 y+400=0 \\
& y=40
\end{aligned}
$
Substituting $\mathrm{y}=40$ in equation (i)
$
\begin{array}{ll}
5 x-6(40)+90=0 & \\
5 x=150 & \\
\mathrm{x}=30 & \\
\therefore \bar{x}=30, \bar{y}=40 & \\
\quad 5 x-6 y+90=0 & 15 x-8 y-130=0 \\
6 y=5 x+90 & x=\frac{8}{15} y+\frac{130}{15} \\
y=\frac{5}{6} x+15 & b_{x y}=\frac{8}{15} \\
b_{y x}=\frac{5}{6} & \\
b_{y x} \cdot b_{x y}=\frac{5}{6} \times \frac{8}{15}=\frac{4}{9} \in[0,1] &
\end{array}
$
$
b_{y x} \cdot b_{x y}=\frac{5}{6} \times \frac{8}{15}=\frac{4}{9} \in[0,1]
$
$\therefore$ Our assumption is correct
$
\begin{aligned}
\therefore & r^2=b_{y x} \cdot b_{x y} \\
& =\frac{4}{9} \quad \text { } \\
\therefore \quad r & = \pm \frac{2}{3}
\end{aligned}
$
Since $b_{y x}$ and $b_{x y}$ are positive
$
\therefore r=\frac{2}{3}
$
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Question 65 Marks
The following table gives the aptitude test scores and productivity indices of 10 works selected at workers selected randomly.
Aptitude score (X) 60 62 65 70 72 48 53 73 65 82
Productivity Index (Y) 68 60 62 80 85 40 52 62 60 81
Obtain the two regression equation and estimate
(i) The productivity index of a worker whose test score is 95.
(ii) The test score when productivity index is 75.
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