Question

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

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

Find the median length of the leaves.

Answer 1

The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to $(117.5-126.5,126.5-135.5,...,171.5-\mathrm{180.5.})$

Converting the given table into exclusive form and preparing the cumulative frequency table, we get

We have, $n=40$

$\Rightarrow \frac{n}{2}=20$

The cumulative frequency just greater than $\frac{n}{2}$ is $29$ and the corresponding class is $144.5-153.5$.

Thus, $144.5-153.5$ is the median class such that

$\frac{n}{2}=20,l=144.5,cf=17,f=12$, and $h=9$

Substituting these values in the formula

Median, $M=l+\left(\frac{\frac{n}{2}-cf}{f}\right)\times h$

$M=144.5+\left(\frac{20-17}{12}\right)\times 9$

$M=144.5+\frac{3}{12}\times 3=144.5+2.25=146.75$

Hence, median length $=146.75$ hours