If the median of the following data is 32.5,

Solution:

Given: Median = 32.5

We prepare the cumulative frequency table, as given below.

Now, we have

$N=40$

$31+f_{1}+f_{2}=40$

$f_{2}=9-f_{1}$.....(1)

Also, $\frac{N}{2}=20$

Since median $=32.5$ so the median class is $30-40$.

Here, $l=30, f=12, F=14+f_{1}$ and $h=10$

We know that

Median $=I+\left\{\frac{\frac{N}{2}-F}{f}\right\} \times h$

$32.5=30+\left\{\frac{20-\left(14+f_{1}\right)}{12}\right\} \times 10$

$2.5=\frac{\left(6-f_{1}\right) \times 10}{12}$

$2.5 \times 12=60-10 f_{1}$

$10 f_{1}=60+30$

$f_{1}=\frac{30}{10}$

$=3$

Putting the value of $f_{1}$ in (1), we get

$f_{2}=9-3$

$=6$

Hence, the missing frequencies are 3 and 6.