Question

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

Answer 1

Less than method, It is given that,

On x-axis upper class limits. Y-axis cumulative frequency

We plot the points $(38,0)(40,3)(42,5)(44,9)(46,4)(48,28)(50,32)(52,35)$

Here, $N=35$ so, $\frac{N}{2}=17.5$

Mark the point $A$ whose coordinate is $17.5$ and its x-coordinate is $46.5$

$Weight$	$Frequency$	$cf$
Less than $38$	$0$	$0$
$38-40$	$3$	$3$
$40-42$	$2$	$5$
$42-44$	$4$	$9$
$44-46$	$5$	$14$
$46-48$	$14$	$28$
$48-50$	$4$	$32$
$50-52$	$3$	$35$
	$N=35$

$\Rightarrow$ $N=35$ and $\frac{N}{2}=17.5$

$\Rightarrow$ Class interval is $46-48$ then median class $=46-48$

$l=$ lower limit of the modal class

$h=$ size of the class intervals

$f=$ frequency of the modal class

$\Rightarrow$ $l=46,f=14,cf=14,h=2$

$\Rightarrow$ $Median=l+\frac{\frac{N}{2}-cf}{f}\times h$

$\Rightarrow$ $Median=46+\frac{17.5-14}{14}\times 2$

$\Rightarrow$ $Median=46+\frac{3.5}{7}=46.5$

$\therefore$ Median of this data is $46.5$.

Hence, the value of median is verified.