A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below.…

Number of cars	\(\displaystyle \mathrm{x}_{\mathrm{i}}\)	\(\displaystyle \mathrm{f}_{\mathrm{i}}\)	\(\displaystyle \mathrm{x}_{\mathrm{i}} \mathrm{f}_{\mathrm{i}}\)	c.f.
$\displaystyle 0$-$\displaystyle 10$	$\displaystyle 5$	$\displaystyle 7$	$\displaystyle 35$	$\displaystyle 7$
$\displaystyle 10$-$\displaystyle 20$	$\displaystyle 15$	$\displaystyle 14$	$\displaystyle 210$	$\displaystyle 21$
$\displaystyle 20$-$\displaystyle 30$	$\displaystyle 25$	$\displaystyle 13$	$\displaystyle 325$	$\displaystyle 34$
$\displaystyle 30$-$\displaystyle 40$	$\displaystyle 35$	$\displaystyle 12$	$\displaystyle 420$	$\displaystyle 46$
$\displaystyle 40$-$\displaystyle 50$	$\displaystyle 45$	$\displaystyle 20$	$\displaystyle 900$	$\displaystyle 66$
$\displaystyle 50$-$\displaystyle 60$	$\displaystyle 55$	$\displaystyle 11$	$\displaystyle 605$	$\displaystyle 77$
$\displaystyle 60$-$\displaystyle 70$	$\displaystyle 65$	$\displaystyle 15$	$\displaystyle 975$	$\displaystyle 92$
$\displaystyle 70$-$\displaystyle 80$	$\displaystyle 75$	$\displaystyle 8$	$\displaystyle 600$	$\displaystyle 100$
Total	$\displaystyle 100$	$\displaystyle 4070$

\[\text { Mean }=\frac{\sum \mathrm{x}_{\mathrm{i}} \mathrm{f}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=\frac{4070}{100}=40 \cdot 7 \] Median class : $\displaystyle 40$-$\displaystyle 50$ \[\text { Median }=40+\frac{50-46}{20} \times 10=42 \]