Topic: Maths 4

Solution:

$ \sum (x-10)^{2} = 180=\sum dx^{2}=180 $

$ \sum (y-15)^{2} = 215=\sum dy^{2}=215 $

$ \sum (x-10) (y-15) = 60=\sum dx dy=60 $

Now,

$\begin{aligned} \bar{x} &= A+ \cfrac{\sum dx}{N} \\ \therefore 14 &= 10+\cfrac{\sum dx}{10} \\ \therefore \sum dx &= 40 \end{aligned}$

Similarly,

$\begin{aligned} \bar{y} &= B+\cfrac{\sum dy}{N} \\ \therefore 15 &= 15+\cfrac{\sum dy}{10} \\ \therefore \sum dy &= 0 \end{aligned}$

$\gamma = \cfrac{\sum dx dy - \cfrac{\sum dx dy}{N}}{\sqrt{\sum dx^{2}-\cfrac{(\sum dx)^{2}}{N}} \sqrt{\sum dy^{2}- \cfrac{(\sum dy)^{2}}{N}}}$

$\gamma = \cfrac{60-\cfrac{40 \times 0}{10}}{\sqrt{180-\cfrac{(40)^{2}}{10}}\sqrt{215-\cfrac{(0)^{2}}{10}}}$

$= \cfrac{60}{\sqrt{20}\sqrt{215}}=0.915 $