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

Consider the following distribution of daily wages of $50$ workers of a factory.
$\begin{array}{|l|c|c|c|c|c|} \hline \text{Daily wages (in Rs.)} & 100-120 & 120-140 & 140-160 & 160-180 & 180-200 \\ \hline \text{Number of workers} & 12 & 14 & 8 & 6 & 10 \\ \hline \end{array}$
Find the mean daily wages of the workers of the factory by using an appropriate method.

For a given frequency distribution,$\Sigma f_{i} x_{i} = 1790$ and $\Sigma f_{i} = 50$. Then,the mean $\bar{x} = $ ..........

The mean of the following frequency distribution is $65$ and the total frequency is $100$. Find the missing frequencies $f_{1}$ and $f_{2}$.

Class	$15-35$	$35-55$	$55-75$	$75-95$	$95-115$
Frequency	$17$	$f_1$	$32$	$f_2$	$19$

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

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.