The given table can be written as:
$$\begin{array}{ccccccc}\text{Height (in cm)}& \text{Less than 140}& \text{Less than 145}& \text{Less than 150}& \text{Less than 155}& \text{Less than 160}& \text{Less than 165}\\ \text{No. of girls}& 4& 11& 29& 40& 46& 51\end{array}$$
$$\begin{array}{ccccccc}\text{Height (in cm)}& \text{Less than 140}& 140-145& 145-150& 150-155& 155-160& 160-165\\ \text{Cumulative Frequency}& 4& 11& 29& 40& 46& 51\end{array}$$
n = 51
$\therefore \frac{n}{2}=\frac{51}{2}$
                                     = 25.5
Since there are a total of 51 observations, the median class is the one whose cumulative frequency is closest to and greater than 25.5. The cumulative frequency for the class '145 - 150' is 29 which is greater than 25.5. The median class is '140 - 145'. Thus, the class interval '145 - 150' is the required answer.

Mode is the most frequently observed value.

$\begin{array}{ccc}\text{Class Interval}& \text{Frequency}& \text{Cumulative Frequency}\\ 60-70& 2& 2\\ 70-80& 3& 5\\ 80-90& 5& 10\\ 90-100& 16& 26\\ 100-110& 14& 40\\ 110-120& 13& 53\\ 120-130& 7& 60\end{array}$

$\frac{n}{2}=30$
$\therefore$ the median class is 100-110 
Median $=l+\left(\frac{\frac{n}{2}-cf}{f}\right)$$\times h$  
                    $=100+\frac{30-26}{14}$$\times 10$
$=102.86$ 

The mean of the data is the sum of the values of all the observations divided by the total number of observations.

For calculating the mean of grouped data, the midpoint of each class interval is chosen to represent all the observations from that class. The midpoint is called the class midpoint or class mark.

Modal class is 4000-5000
$l=4000;D_1=27-21=6$
$D_2=27-25=2;h=1000$

$Mode=l+\left(\frac{D_1}{D_1+D_2}\right)\times h4000+\frac{6}{8}\times 1000=750$

$Mean=\frac{\sum FX}{\sum F}=\frac{391000}{100}=3910$

Difference between mode and mean
= 4750 – 3910 = 840
840 lies between 800 and 900.

The given table can be written as:
 $$\begin{array}{cccccc}\text{Marks}& 0-10& 10-20& 20-30& 30-40& 40-50\\ \text{No. of students}& 5& 5& 13& 8& 9\\ \text{Cumulative Frequency}& 5& 10& 23& 31& 40\end{array}$$
No. of students can be calculated as follows:
5
10 - 5 = 5
23 - 10 = 13
31 - 23 = 8
40 - 31 = 9
$\therefore \frac{n}{2}=\frac{40}{2}=20$
Since there are a total of 40 observations, the median class is the one whose cumulative frequency is closest to and greater than 20.
Here the cumulative frequency for the class interval '10 - 20' is 23 which is greater than 20.
The class interval 20 - 30 is the required answer.

Range is the difference between the largest and the smallest observation.
Range = 34-(-10)=44

$\overline{x}=A+\frac{\sum FD}{\sum F}$

In the formula, the A stands for the assumed mean.