During the medical check-up of $35$ students of a class,their weights were recorded as follows:

Weight (in kg)	Number of students
Less than $38$	$0$
Less than $40$	$3$
Less than $42$	$5$
Less than $44$	$9$
Less than $46$	$14$
Less than $48$	$28$
Less than $50$	$32$
Less than $52$	$35$

Draw a less than type ogive for the given data. Hence,obtain the median weight from the graph and verify the result by using the formula.

(N/A) The given cumulative frequency distributions of less than type are:

Weight (in kg) (upper class limits)	Number of students (cumulative frequency)
Less than $38$	$0$
Less than $40$	$3$
Less than $42$	$5$
Less than $44$	$9$
Less than $46$	$14$
Less than $48$	$28$
Less than $50$	$32$
Less than $52$	$35$

Taking upper class limits on the $x$-axis and their respective cumulative frequencies on the $y$-axis,the ogive is drawn as shown in the figure.
Here,$n = 35$.
So,$\frac{n}{2} = 17.5$.
Mark the point $A$ on the curve whose ordinate is $17.5$. The $x$-coordinate corresponding to this point is $46.5$. Therefore,the median of this data is $46.5$.
To verify using the formula,we first determine the class intervals and their frequencies:

Weight (in kg)	Frequency $(f)$	Cumulative frequency $(cf)$
$38-40$	$3-0=3$	$3$
$40-42$	$5-3=2$	$5$
$42-44$	$9-5=4$	$9$
$44-46$	$14-9=5$	$14$
$46-48$	$28-14=14$	$28$
$48-50$	$32-28=4$	$32$
$50-52$	$35-32=3$	$35$

The cumulative frequency just greater than $\frac{n}{2} = 17.5$ is $28$,which belongs to the class interval $46-48$.
Median class $= 46-48$.
Lower class limit $(l) = 46$.
Frequency $(f) = 14$.
Cumulative frequency of the class preceding the median class $(cf) = 14$.
Class size $(h) = 2$.
Using the median formula:
$\text{Median} = l + \left( \frac{\frac{n}{2} - cf}{f} \right) \times h$
$= 46 + \left( \frac{17.5 - 14}{14} \right) \times 2$
$= 46 + \left( \frac{3.5}{14} \right) \times 2$
$= 46 + \frac{7}{14} = 46 + 0.5 = 46.5$.
The result is verified.