Find the median.

Intelligence quotient (IQ)	60-79	80-99	100-119	120-139	140-159	Total
Number of students	2	9	15	11	3	40

Median $=\mathrm{l}+\frac{\left(\frac{\mathrm{N}}{2}-\mathrm{cf}\right)}{{\mathrm{f}}_{\mathrm{m}}}\times \mathrm{c}$

To change the discontinuous frequency distribution to continuous we will use the following approach:

New lower limit = Lower limit − Half of the difference 

For example: 

The difference between 79 and 80 is 1, 

Similarly, the difference between 99 and 100 is 1.

Therefore, the new lower limit is $60-\frac{1}{2}\left(1\right)$  or $60-0.5=59.5$

Similarly, New Upper limit = Upper limit + Half of the difference 

Therefore, the new upper limit is $79+\frac{1}{2}\left(1\right)$  or $79+0.5=79.5$

Therefore, for all the class intervals, we get the following continuous distribution :

Median $=\mathrm{l}+\frac{\left(\frac{\mathrm{N}}{2}-\mathrm{cf}\right)}{{\mathrm{f}}_{\mathrm{m}}}\times \mathrm{c}$

Here, Class Interval, $\mathrm{c}=20$

$\text{N}=\Sigma {\text{f}}_{\text{i}}=40$

$\frac{\text{N}}{2}=20$

$\Rightarrow \text{ Median class }=99.5-119.5$

 Frequency of median class, ${\mathrm{f}}_{\text{m}}=15$

The lower limit of the median class, $\mathrm{l}=99.5$

Cumulative frequency of the class before the median class $\mathrm{cf}=11$

Median $=99.5+\frac{\left(\frac{40}{2}-11\right)}{15}\times 20$

Median $=99.5+\frac{\left(9\right)}{15}\times 20$

 Median = 111.5

Thus, the median is 111.5.