Question

The marks obtained by $10$ candidates in English and Mathematics are given below:

Marks in English	$20$	$13$	$18$	$21$	$11$	$12$	$17$	$14$	$19$	$15$
Marks in Mathematics	$17$	$12$	$23$	$25$	$14$	$8$	$19$	$21$	$22$	$19$

Estimate the probable score for Mathematics if the marks obtained in English are $24$.

Answer 1

Let $x$ denote the marks in English and $y$ in Mathematics.

$x$	$y$	$dx=x-16$	$dy=y-18$	$dx.dy$	${\left(dx\right)}^2$	${\left(dy\right)}^2$
$20$	$17$	$4$	$-1$	$-4$	$16$	$1$
$13$	$12$	$-3$	$-6$	$18$	$9$	$36$
$18$	$23$	$2$	$5$	$10$	$4$	$25$
$21$	$25$	$5$	$7$	$35$	$25$	$49$
$11$	$14$	$-5$	$-4$	$20$	$25$	$16$
$12$	$8$	$-4$	$-10$	$40$	$16$	$100$
$17$	$19$	$1$	$1$	$1$	$1$	$1$
$14$	$21$	$-2$	$3$	$-6$	$4$	$9$
$19$	$22$	$3$	$4$	$12$	$9$	$16$
$15$	$19$	$-1$	$1$	$-1$	$1$	$1$
$\sum x=160$	$\sum y=180$			$\sum dxdy=125$	$\sum {\left(dx\right)}^2=110$	$\sum {\left(dy\right)}^2=254$

$\overline{x}=\frac{160}{10}=16$ and $\overline{y}=\frac{180}{10}=18$
$r=\frac{\sum dx\sum dy}{\sqrt{\sum {\left(dx\right)}^2}.\sum {\left(dy\right)}^2}=\frac{125}{\sqrt{110\times 254}}$
$=\frac{125}{27940}=\frac{125}{167.17}=0.75$ (approx.)
$\therefore$ A line of best fit is $y-\overline{y}=\frac{\sum dxdy}{\sum {\left(dx\right)}^2}(x-\overline{x})$
$y-18=\frac{125}{110}(x-16)$
when $x=24$, then $y-18=\frac{125}{110}(24-16)$
$y-18=\frac{25}{22}\times 8$
$22y-396=200$
$22y=396+200$
$y=\frac{596}{22}=27.09$ or $27$ marks (nearly)