Mumbai University > Mechanical Engineering > Sem 4 > Applied Mathematics IV

Marks: 5M

Year: Dec 2015

$$\overline{x}=5 , \overline{y}=10$$

The line of regression of y on x is parallel to the line 20y = 9x +40

y = $\frac{9}{20}\ x+\ \frac{40}{20}$

Slope = $\frac{9}{20}$ i.e. $b_{yx}=9/20$

The equation of line of regression of y on x is ,

$$y - \overline{y\ }=b_{yx}\ \left(x-\overline{x}\right)$$ $$y-10=\frac{9}{20}(\ x-5)\ $$ $$y = \frac{9}{20}\ x-\ \frac{45}{20}+10$$ Put x = 30 $$y = \frac{9}{20}*30-\ \frac{45}{20}+10$$

y = 21.25