Find the mean of the following frequency distribution:

Class	$10-20$	$20-30$	$30-40$	$40-50$	$50-60$	$60-70$
Frequency	$2$	$10$	$40$	$25$	$13$	$10$

(in $.7$)

In the formula $\bar{x} = A + \frac{\Sigma f_{i} d_{i}}{\Sigma f_{i}}$ for the mean,$d_{i} = \dots$

In calculating the mean of grouped data,grouped in classes of equal width,we may use the formula $\bar{x} = a + \frac{\sum f_i d_i}{\sum f_i}$,where $a$ is the assumed mean. $a$ must be one of the mid-points of the classes. Is the last statement correct? Justify your answer.

For a given frequency distribution,$\bar{x}=20, \Sigma f_{i} u_{i}=-50, n=100$ and $c=10$. Then,assumed mean $A = \ldots \ldots \ldots \ldots$

For a given frequency distribution,the total frequency is $25$ and $\Sigma f_{i} x_{i} = 120$. Then,the mean is $\ldots$

$A$ class teacher has the following absentee record of $40$ students of a class for the whole term. Find the mean number of $days$ a student was absent.
$\begin{array}{|l|c|c|c|c|c|c|c|} \hline \text{Number of days} & 0-6 & 6-10 & 10-14 & 14-20 & 20-28 & 28-38 & 38-40 \\ \hline \text{Number of students} & 11 & 10 & 7 & 4 & 4 & 3 & 1 \\ \hline \end{array}$