Solve the following Question.(1 Marks)

Question 11 Mark

Find the line of regression of $\mathrm{X}$ on $\mathrm{Y}$ for the following data: $\mathrm{n}=8, \Sigma\left(\mathrm{x}_i-\mathrm{x}\right)^2=36, \Sigma\left(\mathrm{y}_i-\mathrm{y}\right)^2=44, \Sigma\left(\mathrm{x}_{\mathrm{i}}-\mathrm{x}\right)\left(\mathrm{y}_{\mathrm{i}}-\mathrm{y}\right)=24$

Answer

$b_{x y}=\frac{\sum(x-\bar{x})(y-\bar{y})}{\sum(y-\bar{y})^2}$
$=\frac{24}{44}=\frac{6}{11}$
Regression equation of $\mathrm{X}$ on $\mathrm{Y}$ is
$
\begin{aligned}
& (x-\bar{x})=b_{x y}(y-\bar{y}) \\
& (x-\bar{x})=\frac{6}{11}(y-\bar{y})
\end{aligned}
$

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Question 21 Mark

If for a bivariate data $\bar{x}=10, \bar{y}=12, \mathrm{v}(\mathrm{x})=9, \sigma_{\mathrm{y}}=4$ and $\mathrm{r}=0.6$. Estimate $\mathrm{y}$ when $\mathrm{x}=5$.

Answer

Given, $V(x)=9$
$
\begin{aligned}
& \therefore \sigma_{\mathrm{x}}=3 \\
& \mathrm{~b}_{\mathrm{yx}}=\frac{r \cdot \sigma_y}{\sigma_x} \\
& =0.6 \times \frac{4}{3} \\
& =0.8
\end{aligned}
$
$\therefore$ Regression equation of $\mathrm{Y}$ on $\mathrm{X}$ is
$
\begin{aligned}
& (y-\bar{y})=b_{y x}(x-\bar{x}) \\
& (y-12)=0.8(5-10) \\
& y-12=0.8(-5) \\
& y-12=-4 \\
& y=8
\end{aligned}
$

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Question 31 Mark

The regression equation of y on $x$ is given by $3x + 2y – 26 = 0$. Find $b_{yx}.$

Answer

Given, regression equation of $\mathrm{Y}$ on $\mathrm{X}$ is
$
\begin{aligned}
& 3 x+2 y-26=0 \\
& \therefore 2 y=-3 x+26 \\
& \therefore y=\frac{-3}{2} x+13 \\
& \therefore b_{y x}=\frac{-3}{2}
\end{aligned}
$

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Question 41 Mark

If $b_{yx} = -0.6$ and $b_{xy} = -0.216$ then find correlation coefficient between $X$ and $Y$ comment on it.

Answer

$
\begin{aligned}
& r^2=b_{y x} \cdot b_{x y} \\
& r^2=-0.6 \times-0.216 \\
& r^2=0.1296 \\
& r= \pm \sqrt{ } 0.1296 \\
& r= \pm 0.36
\end{aligned}
$
Since $b_{y x}$ and $b_{x y}$ are negative
$
r=-0.36
$

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Question 51 Mark

\begin{align}
\text { For a bivariate data } \bar{x}=53, \bar{y}=28, \mathrm{~b}_{\mathrm{yx}}=-1.5 \text { and } \mathrm{b}_{\mathrm{xy}}=-0.2 \text {. Estimate } \mathrm{Y} \text { when } \mathrm{X}=50 \text {. }
\end{align}

Answer

Regression equation of $\mathrm{Y}$ on $\mathrm{X}$ is
$
\begin{aligned}
& (Y-\bar{y})=b_{y x}(X-\bar{x}) \\
& (Y-28)=-1.5(50-53) \\
& Y-28=-1.5(-3) \\
& Y-28=4.5 \\
& Y=32.5
\end{aligned}
$

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Question 61 Mark

Two sample from bivariate populations have 15 observation each. The sample means of X and Y are 25 and 18 respectively. The corresponding sum of square of deviations from respective means are 136 and 150. The sum of product of deviations from respective means is 123. Obtain the equation of line of regression of X on Y.

Answer

$\begin{aligned} & \text { Given, } \mathrm{n}=15, \bar{x}=25, y=18, \Sigma(\mathrm{x}-\bar{x})=136, \Sigma(\mathrm{y}-\bar{y})=150, \Sigma(\mathrm{x}-\bar{x})(\mathrm{y}-\bar{y})=123 \\ & \text { Regression equation of } \mathrm{X} \text { on } \mathrm{Y} \text { is }(X-\bar{x})=\mathrm{bxy}(\mathrm{Y}-\bar{y}) \\ & (\mathrm{X}-25)=0.82(\mathrm{y}-18) \\ & (\mathrm{X}-25)=082 \mathrm{y}-14.76 \\ & \mathrm{X}=0.82 \mathrm{y}+10.24\end{aligned}$

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Maths (commerce) Solve the following Question.(1 Marks)

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