Applied Maths Case study (4 Marks)

Mean Deviation $=\frac{1}{n} \times \sum\left|d_{i}\right|=\frac{86}{10}=8.6$
(iii) Sum of new scores $=564$
New mean $=\frac{564}{10}=56.4$
OR
It is called range of the data.

The girder of a railway bridge is a parabola with its vertex at the highest point, 10 metres above the ends and span is 100 metres.

The equation of the parabola in standard from is $x^{2}=-4 a y$.
Coordinates of one end of the girder is $(50,-10)$.
As it lies on parabola, $\therefore(50)^{2}=-4 \mathrm{a}(-10)$
$\Rightarrow a=\frac{125}{2}$.
$\therefore$ Parabola is $\mathrm{x}^{2}=-4 \times \frac{125}{2} y \Rightarrow x^{2}=-250 \mathrm{y}$
(ii) Let Q be a point as the girder which is 20 m from the middle. Let the coordinates of Q be $(20,-\mathrm{y})$.
As it lies on the parabola,
$(20)^{2}=-250(\mathrm{y})$
$\Rightarrow y=-\frac{20^{2}}{250}=-\frac{400}{250}=-\frac{8}{5}=-1.6$
$\therefore$ Height $=10-\mathrm{Iyl}=10-1.6=8.4 \mathrm{~m}$
(iii) Here, $a=\frac{125}{2}$ and focus of $\mathrm{x}^{2}=-4 \mathrm{ay}$ is ( $0,-\mathrm{a}$ ),
$\therefore$ Focus is $\left(0,-\frac{125}{2}\right)$.
The length of latus-rectum $=4 \mathrm{a}$ units
$
=4 \times \frac{125}{2}=250 \mathrm{~m}
$