The data for marks in Physics and History obtained by $10$ students is as given:
| Marks in Physics | $15$ | $12$ | $8$ | $8$ | $7$ | $7$ | $7$ | $6$ | $5$ | $3$ |
|---|---|---|---|---|---|---|---|---|---|---|
| Marks in History | $10$ | $25$ | $17$ | $11$ | $13$ | $17$ | $20$ | $13$ | $9$ | $15$ |
Using this data, compute the Karl Pearson's coefficient of correlation between the marks in Physics and History obtained by the $10$ students.
Number of students $= n = 10$
| Marks in Physics $\left(x\right)$ | Marks in History $\left(y\right)$ |
|---|---|
| $15$ | $10$ |
| $12$ | $25$ |
| $8$ | $17$ |
| $8$ | $11$ |
| $7$ | $13$ |
| $7$ | $17$ |
| $7$ | $20$ |
| $6$ | $13$ |
| $5$ | $9$ |
| $3$ | $15$ |
| $\Sigma x = 78$ | $\Sigma y = 150$ |
$\overline{x} = \dfrac{\Sigma x}{n} = \dfrac{78}{10} = 7.8$, $\;$ $\overline{y} = \dfrac{\Sigma y}{n} = \dfrac{150}{10} = 15$
| Marks in Physics $\left(x\right)$ | Marks in History $\left(y\right)$ | $d_x = x - \overline{x}$ | $d_y = y - \overline{y}$ | $d_x \cdot d_y$ | $d_x^2$ | $d_y^2$ |
|---|---|---|---|---|---|---|
| $15$ | $10$ | $7.2$ | $- 5$ | $- 36$ | $51.84$ | $25$ |
| $12$ | $25$ | $4.2$ | $10$ | $42$ | $17.64$ | $100$ |
| $8$ | $17$ | $0.2$ | $2$ | $0.4$ | $0.04$ | $4$ |
| $8$ | $11$ | $0.2$ | $- 4$ | $- 0.8$ | $0.04$ | $16$ |
| $7$ | $13$ | $- 0.8$ | $- 2$ | $1.6$ | $0.64$ | $4$ |
| $7$ | $17$ | $- 0.8$ | $2$ | $- 1.6$ | $0.64$ | $4$ |
| $7$ | $20$ | $- 0.8$ | $5$ | $- 4$ | $0.64$ | $25$ |
| $6$ | $13$ | $- 1.8$ | $- 2$ | $3.6$ | $3.24$ | $4$ |
| $5$ | $9$ | $- 2.8$ | $- 6$ | $16.8$ | $7.84$ | $36$ |
| $3$ | $15$ | $- 4.8$ | $0$ | $0$ | $23.04$ | $0$ |
| $\Sigma x = 78$ | $\Sigma y = 150$ | $\Sigma d_x d_y = 22$ | $\Sigma d_x^2 = 82.56$ | $\Sigma d_y^2 = 218$ |
Karl Pearson coefficient $= r = \dfrac{\Sigma d_x d_y}{\sqrt{\Sigma d_x^2 \cdot \Sigma d_y^2}}$
i.e. $\;$ $r = \dfrac{22}{\sqrt{82.56 \times 218}} = \dfrac{22}{\sqrt{17998.08}} = \dfrac{22}{134.16} = 0.1640$ (approx)