Following table shows the absentees record of $\displaystyle 40$ students in an academic year :

Number of Days	Number of Students
$\displaystyle 2$-$\displaystyle 6$	$\displaystyle 11$
$\displaystyle 6$-$\displaystyle 10$	$\displaystyle 10$
$\displaystyle 10$-$\displaystyle 14$	$\displaystyle 7$
$\displaystyle 14$-$\displaystyle 18$	$\displaystyle 4$
$\displaystyle 18$-$\displaystyle 22$	$\displaystyle 4$
$\displaystyle 22$-$\displaystyle 26$	$\displaystyle 3$
$\displaystyle 26$-$\displaystyle 30$	$\displaystyle 1$

Find the 'mean' and the 'mode' of the above data.

Age (in years)	Number of patients \(\displaystyle \left(f_{i}\right)\)	Class Mark \(\displaystyle \left(x_{i}\right)\)	\(\displaystyle f_{i} x_{i}\)
$\displaystyle 2$-$\displaystyle 6$	$\displaystyle 11$	$\displaystyle 4$	$\displaystyle 44$
$\displaystyle 6$-$\displaystyle 10$	$\displaystyle 10$	$\displaystyle 8$	$\displaystyle 80$
$\displaystyle 10$-$\displaystyle 14$	$\displaystyle 7$	$\displaystyle 12$	$\displaystyle 84$
$\displaystyle 14$-$\displaystyle 18$	$\displaystyle 4$	$\displaystyle 16$	$\displaystyle 64$
$\displaystyle 18$-$\displaystyle 22$	$\displaystyle 4$	$\displaystyle 20$	$\displaystyle 80$
$\displaystyle 22$-$\displaystyle 26$	$\displaystyle 3$	$\displaystyle 24$	$\displaystyle 72$
$\displaystyle 26$-$\displaystyle 30$	$\displaystyle 1$	$\displaystyle 28$	$\displaystyle 28$
Total	\(\displaystyle \sum f_{i}=40\)		\(\displaystyle \sum f_{i} x_{i}=452\)

\[\begin{aligned} \text { Mean } & =\frac{452}{40} \\ & =\frac{113}{10} \text { or } 11.3 \text { years } \end{aligned} \] Modal class is $\displaystyle 2$-$\displaystyle 6$ \[\begin{aligned} \text { Mode } & =2+\frac{11-0}{2 \times 11-0-10} \times 4 \\ & =\frac{17}{3} \text { or } 5.67 \text { years (approx.) } \end{aligned} \] Let the length of shortest side be \(\displaystyle x \mathrm{~m}\) \[\begin{aligned} & \therefore \text { length of longest side }=(x+4) \mathrm{m} \\ & \text { and length of third side }=(x+2) \mathrm{m} \\ & \text { Now, }(x+4)^{2}=x^{2}+(x+2)^{2} \\ & \Rightarrow x^{2}-4 x-12=0 \\ & \Rightarrow(x-6)(x+2)=0 \\ & \Rightarrow x=6 \\ & \therefore \text { sides are } 6 \mathrm{~m}, 8 \mathrm{~m} \text { and } 10 \mathrm{~m} \\ & \text { Area }=\frac{1}{2} \times 6 \times 8=24 \mathrm{~m}^{2} \\ & \text { Perimeter }=6+8+10=24 \mathrm{~m} \\ & \text { Difference }=0 \end{aligned} \] \[\begin{aligned} & \frac{x-2}{x-3}+\frac{x-4}{x-5}=\frac{10}{3} \\ & \Rightarrow \frac{(x-2)(x-5)+(x-4)(x-3)}{(x-3)(x-5)}=\frac{10}{3} \end{aligned} \] Simplifying, we get \(\displaystyle 2 x^{2}-19 x+42=0\) \[\begin{aligned} & \Rightarrow(x-6)(2 x-7)=0 \\ & \Rightarrow x=6 \text { or } x=\frac{7}{2} \end{aligned} \]