Question

The lengths of $40$ leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length(in mm)	Number of leaves
$118 - 126$	$3$
$127 - 135$	$5$
$136 - 144$	$9$
$145 - 153$	$12$
$154 - 162$	$5$
$163 - 171$	$4$
$172 - 180$	$2$

Find the median length of the leaves.

Answer 1

Hint: The data needs to be converted to continuous classes for finding the median. Therefore, $0.5$ has to be added and subtracted to upper class limits and lower class limits respectively. The classes then change to $117.5 - 126.5,126.5 - 135.5....171.5 - 180.5$.
 The formula used to find the median of a given data is as follows: $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
Where $l$ is the lower limit of median class, $n$ is the sum of all frequencies, $cf$ is the cumulative frequency before the median class, $f$ is the frequency of median class and $h$ is the size of median class.

Complete step-by-step answer:
The continuous class interval with respective cumulative frequencies can be represented as follows:

Length(in mm)	Number of leaves$\left( f \right)$	Cumulative frequency$\left( {cf} \right)$
$117.5 - 126.5$	$3$	$3$
$126.5 - 135.5$	$5$	$8$
$135.5 - 144.5$	$9$	$17$
$144.5 - 153.5$	$12$	$29$
$153.5 - 162.5$	$5$	$34$
$162.5 - 171.5$	$4$	$38$
$171.5 - 180.5$	$2$	$40$
	$n = \sum {f = 40} $

From the table, we obtain $n = 40 \Rightarrow \dfrac{n}{2} = 20$
Cumulative frequency $\left( {cf} \right)$just greater than $\dfrac{n}{2}$$\left( {i.e.,20} \right)$ is $29$,which lies in the interval $144.5 - 153.5$.
Therefore, median class=$144.5 - 153.5$
Lower limit of the median class, $l = 144.5$
Frequency of the median class, $f = 12$
Cumulative frequency of the class preceding the median class, $cf = 17$
Class size, $h = 9$
Therefore, $Median = l + \left( {\dfrac{{\dfrac{n}{2} - cf}}{f}} \right) \times h$
$Median = 144.5 + \left( {\dfrac{{20 - 17}}{{12}}} \right) \times 9$
$ \Rightarrow Median = 144.5 + \dfrac{{27}}{{12}}$
$ \Rightarrow Median = 146.75$
Therefore, the median length of leaves is $146.75$.

Note: The cumulative frequency is calculated by adding each frequency from a frequency distribution table to the sum of its predecessors. The last value will always be equal to the sum of all frequencies, since all frequencies will already have been added to the previous total.