The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.

Number of students per teacher	Number of states/U.T $\left( {{f_i}} \right)$
$15 - 20$	$3$
$20 - 25$	$8$
$25 - 30$	$9$
$30 - 35$	$10$
$35 - 40$	$3$
$40 - 45$	$0$
$45 - 50$	$0$
$50 - 55$	$2$

Hint: Here, we are asked to calculate the mode and median for the given frequency distribution table. To find the mode, we need to obtain the modal class. Then we shall substitute the calculated values in the formula of mode. To find the mean, we need to create a table containing the values that we need to apply in the formula.

Formula to be used:
$Mode = l + \left[ {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right] \times h$
Here $l$ is the lower limit, $h$ is the class size, ${f_1}$ is the frequency of the modal class, ${f_0}$ is the frequency preceding the modal class, and ${f_2}$ is the frequency succeeding the modal class.
\[{x_i} = \dfrac{{Upper{\text{ }}limit + Lower{\text{ }}limit}}{2}\]
$Mean,\overline x = A + \left( {\dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }} \times h} \right)$
Here $A$ is the assumed mean.
$\sum {{f_i}} $ is the sum of the frequency and $\sum {{f_i}} {d_i}$ is the sum of ${f_i}{d_i}$

Complete step-by-step answer:
Here we are asked to find the mode and mean of the given distribution.
First, we shall calculate the mode.
a) The modal class is the class that contains the highest frequency.
From the given data, we can note that the highest frequency is $10$ that belongs to the class interval $30 - 35$
Hence, the modal class is $30 - 35$
Let \[h\] be the size of the class.
Here, for the given frequency distribution, $h = 5$
Let $l$be the lower limit of the median class.
Hence, we have $l = 30$.
Let ${f_1}$be the frequency of the modal class.
Therefore, we have ${f_1} = 10$
Let us assume that ${f_0}$be the frequency of class above the calculated modal class.
Then, ${f_0} = 9$ is the required frequency of the class above the calculated modal class.
Let us assume that ${f_2}$be the frequency of the class below the obtained modal class.
Then, ${f_2} = 3$ is the required frequency of the class below the obtained modal class.
Now, we shall apply all the values in the formula $Mode = l + \left[ {\dfrac{{{f_1} - {f_0}}}{{2{f_1} - {f_0} - {f_2}}}} \right] \times h$
Thus, we have \[Mode = 30 + \left[ {\dfrac{{10 - 9}}{{2 \times 10 - 9 - 3}}} \right] \times 5\]
                                        \[ = 30 + \dfrac{5}{{20 - 12}}\]
      \[ = 30 + \dfrac{5}{8}\]
     $ = 30.6$
Thus, we have $Mode = 30.6$
Therefore, we can able to conclude that most of the states/U.T have a teacher-student ratio as $30.6$
b) Now we need to calculate the mean.
We need to create a frequency distribution table containing assumed mean $A$, classmark ${x_i}$, ${d_i}$ and${f_i}{d_i}$
To find the classmark, we shall apply \[{x_i} = \dfrac{{Upper{\text{ }}limit + Lower{\text{ }}limit}}{2}\]

Here the assumed mean is $A = 32.5$
Now, we shall apply the formula $Mean,\overline x = A + \left( {\dfrac{{\sum {{f_i}{d_i}} }}{{\sum {{f_i}} }} \times h} \right)$
$Mean,\overline x = 32.5 + \left( {\dfrac{{ - 23}}{{35}} \times 5} \right)$ (Here $\sum {{f_i}} $ is the sum of the frequency and $\sum {{f_i}} {d_i}$ is the sum of ${f_i}{d_i}$ )
$ \Rightarrow Mean,\overline x = 32.5 + \dfrac{{ - 23}}{7}$
$ \Rightarrow Mean,\overline x = 32.5 - 3.28$
$ \Rightarrow Mean,\overline x = 29.22$
Hence, the required mean is $29.2$
Therefore, we can conclude that the teacher-student ratio on average was $29.2$

Note: We can able to conclude that most of the states/U.T have a teacher-student ratio as $30.6$ and we can conclude that the teacher-student ratio on an average was $29.2$To find the mean and mode, we need to know the required formula. Also, we know how to create a frequency distribution table for calculating the mean.