Let the regression equation of Y on X be
3X + 2Y = 26
2Y = –3X + 26
Y = \(\frac{1}{2}\)(-3X + 26)
Y = -1.5X + 13
r\(\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } \) = -1.5
Implies by x = r\(\frac { { \sigma }_{ y } }{ { \sigma }_{ x } } \) = -1.5
Let the regression equation of X on Y be
6X + 3Y = 31
X = \(\frac{1}{6}\)(-3Y + 31) = -0.5Y + 5.17
r\(\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } \) = -0.5
Implies b_xy = r\(\frac { { \sigma }_{ x } }{ { \sigma }_{ y } } \) = -0.5
r = ±\(\sqrt { { b }_{ xy }.{ b }_{ yx } } \)
=\(-\sqrt { (-1.5).(-0.5) } \) (Since both the regression coefficient are negative r is negative)
\(\therefore\) r = -8.66

Let X, Y represent the expenditure on accommodation and food.
\(\bar{X} =178, \bar{Y}=47.8 \)
\(\sigma_x =63.15, \sigma_y=22.98, r=0.43 \)
\(b_{y x} =r \frac{\sigma_y}{\sigma_x}=0.43\left(\frac{22.98}{63.15}\right)=0.1565\)
Regression equation of Y on X is
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 47.8 = 0.1565(X - 178)
Y - 47.8 = 0.1565 X - 27.86
Y = 0.1565 X + 19.94
If  X = 200
Y = 31.3 + 19.94 = 51.24

\(r =\frac{N \Sigma X Y-(\Sigma X)(\Sigma Y)}{\sqrt{N \Sigma X^2-(\Sigma X)^2} \sqrt{N \Sigma Y^2-(\Sigma Y)^2}} \)
\(=\frac{6(13852)-(392)(194)}{\sqrt{6(28580)-(392)^2} \sqrt{6(6842)-(194)^2}} \)
\(=\frac{83112-76048}{\sqrt{171480-153664} \sqrt{41052-37636}} \)
\(=\frac{7064}{\sqrt{17816 \times 3416}}=\frac{7064}{7801.25} \)
= 0.906

\(\bar{X} =\frac{\Sigma X}{N}=\frac{55}{10}=5.5 \)
\(\bar{Y} =\frac{\Sigma Y}{N}=\frac{55}{10}=5.5 \)
\(b_{y x} =\frac{N \Sigma X Y-\Sigma X \Sigma Y}{N \Sigma X^2-(\Sigma X)^2} \)
\(=\frac{10(350)-(55)(55)}{10(385)-55^2}=\frac{3500-3025}{3850-3025} \)
\(=\frac{475}{825}=0.576\)
Regression line of Y on X is.
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 5.5 = 0.576(X - 5.5)
Y = 0.576 X - 3.168 + 5.5
Y = 0.576 X + 2.332
At  X = 6, Y = 0.576(6) + 2.332
= 3.456 + 2.332
= 5.788

\(\bar{X}\) = 6, \(\bar{Y}\) = 8, \(\sigma _x=5\)
\(\sigma_y=\frac{40}{3}, \quad r=\frac{8}{15} \)
\(b_{y x}=r \frac{\sigma_x}{\sigma_y}=\frac{8}{15}\left(\frac{40}{3 \times 5}\right)=\frac{64}{45}=1.422\)
Regression line of Y on X is
\(Y-\bar{Y}=b_{y x}(X-\bar{X}) \)
Y - 8 = 1.422(X - 6)
Y = 1.422 X - 8.532 + 8
Y = 1.422 X - 0.532 .
If  X = Rs 100,
Y = 142.2 - 0.532 = Rs 141.67