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	The number of states/U.T.
$15 - 20$	$3$
$20 - 25$	$8$
$25 - 30$	$9$
$30 - 35$	$10$
$35 - 40$	$3$
$40 - 45$	$0$
$45 - 50$	$0$
$50 - 55$	$2$

Hint: As for finding the mode of the grouped data we use formula $L + \dfrac{{{f_m} - {f_{m - 1}}}}{{({f_m} - {f_{m - 1}}) + ({f_m} - {f_{m + 1}})}} \times w$where L is the lower class boundary of the modal group ,${f_{m - 1}}$ is the frequency of the group before the modal group ,${f_m}$ is the frequency of the modal group , ${f_{m + 1}}$ is the frequency of the group after the modal group , w is the group width and for the mean of grouped data Mean = $\dfrac{{\sum {f \times } X}}{{\sum f }}$ where $X$ is the midpoint of group and f is frequency of that .

Complete step by step answer:
In the first case we have to find the mode of the given data ,
Mode is the number that appears most frequently in a data set . In the grouped data we use the formula for mode that is ,
Estimated mode = $L + \dfrac{{{f_m} - {f_{m - 1}}}}{{({f_m} - {f_{m - 1}}) + ({f_m} - {f_{m + 1}})}} \times w$
Modal group is the group which have maximum frequency .
where L is the lower class boundary of the modal group
${f_{m - 1}}$ is the frequency of the group before the modal group
${f_m}$ is the frequency of the modal group
${f_{m + 1}}$ is the frequency of the group after the modal group
w is the group width .

For the given question
Modal Group is $30 - 35$
L = $30$ (is the lower class boundary of the modal group)
${f_{m - 1}} = 9$ (is the frequency of the group before the modal group)
${f_m} = 10$ (is the frequency of the modal group)
${f_{m + 1}} = 3$ (is the frequency of the group after the modal group)
w = $5$ (is the group width)
Now putting these values in the given Estimated mode equation ,
Mode = $30 + \dfrac{{10 - 9}}{{(10 - 9) + (10 - 3)}} \times 5$
Mode = $30 + \dfrac{1}{{1 + 7}} \times 5$
Mode = $30 + \dfrac{5}{8}$
When we divide $5$ by $8$ we get $0.625$
Hence the mode = $30 + 0.625 = 30.625$
Now for the mean of the given data ,
first we have to find the midpoint of the group data as ,
Midpoint of $15 - 20$is $17.5$
Midpoint of $20 - 25$ is $22.5$
The midpoint of $25 - 30$ is $27.5$ similarly for other as,
After finding the Midpoint of the group, multiple by these to the frequency of the data or Number of states.

Hence for the mean of the group data the formula is
Mean = $\dfrac{{\sum {f \times } X}}{{\sum f }}$
therefore $\sum {f \times } X = 52.5 + 180 + 247.5 + 325 + 112.5 + 0 + 0 + 105$ = $1022.5$
$\sum {f = 3 + 8 + 9 + 10 + 3 + 0 + 0 + 2 = 35} $
By putting the value in equation the ,
Mean = $\dfrac{{1022.5}}{{35}}$ on dividing we get
Mean $ = 29.21$

$\therefore $ The mode of the given data is 30.625 and the mean is $29.21$

Note:
We find the midpoint of the grouped data as we have to think like $17.5$ Number of students per teacher have $3$ number of states so from this the grouped data is converted into simple data or in another word we will say that
$17.5$ is three times, $22.5$ is eight times, $27.5$ is nine times, $32.5$ is ten times, $37.5$ is three times, $52.5$ two times now find the mean from simply.
For Median of the group data we use formula Median = $ = L + \dfrac{{\dfrac{n}{2} - B}}{G} \times w$ where:
$L$ is the lower class boundary of the group containing the median, $n$ is the total number of values, $B$ is the cumulative frequency of the groups before the median group, $G$ is the frequency of the median group, $w$ is the group width.