Prepare a continuous grouped frequency distribution table from the following data:

Mid-point	5	15	25	35	45
Frequency	4	8	13	12	6

Also find the size of class intervals.

Hint: We have given a table for mid – point and frequency corresponding to them and we are asked to construct a continuous grouped frequency distribution table. As you can see, the difference between consecutive mid – points is 10 so we are taking the size of the class interval as 10 and name it as “h”. Now, we are going to find the lower and upper limit of the class interval corresponding to each mid - point. The formula for the lower limit of class interval is $m-\dfrac{h}{2}$ where ‘m’ is the mid – point and “h” is the size of class interval. And the formula for the upper limit of class interval is $m+\dfrac{h}{2}$ where ‘m’ is the mid – point and “h” is the size of class interval.

Complete step by step answer:
The table given in the above problem is as follows:


We have to construct a continuous grouped frequency distribution table from the above table and which we are going to do by using the following formulae for lower limit and upper limit for each midpoint.
Formula for lower limit of the class interval having mid – point m and class size as h:
$m-\dfrac{h}{2}$
Formula for upper limit of the class interval having mid – point m and class size as h:
$m+\dfrac{h}{2}$
Now, in the above problem, we have given different mid – points corresponding to different frequencies so those mid – points will be m and the value of “h” is 10 which is the difference of two consecutive mid – points.
Let us take any two consecutive mid – points from the given set of mid – points and we are picking 15 and 25. The difference of these two mid – points is:
$\begin{align}
  & 25-15 \\
 & =10 \\
\end{align}$
Now, finding the continuous class interval for mid – point 5 we get,
Lower limit for 5 is equal to:
$\begin{align}
  & m-\dfrac{h}{2} \\
 & =5-\dfrac{10}{2} \\
 & =5-5=0 \\
\end{align}$
Upper limit for 5 is equal to:
$\begin{align}
  & 5+\dfrac{10}{2} \\
 & =5+5 \\
 & =10 \\
\end{align}$
Hence, we got the class interval for mid – point 5 as $\left( 0-10 \right)$.
Similarly, we can calculate for other mid – points also.
Class interval for mid - point 15 is equal to:
$\begin{align}
  & \left( \left( m-\dfrac{h}{2} \right)-\left( m+\dfrac{h}{2} \right) \right) \\
 & =\left( \left( 15-\dfrac{10}{2} \right)-\left( 15+\dfrac{10}{2} \right) \right) \\
 & =\left( \left( 15-5 \right)-\left( 15+5 \right) \right) \\
 & =\left( 10-20 \right) \\
\end{align}$
Class interval for mid – point 25 is equal to:
$\begin{align}
  & \left( \left( m-\dfrac{h}{2} \right)-\left( m+\dfrac{h}{2} \right) \right) \\
 & =\left( \left( 25-\dfrac{10}{2} \right)-\left( 25+\dfrac{10}{2} \right) \right) \\
 & =\left( \left( 25-5 \right)-\left( 25+5 \right) \right) \\
 & =\left( 20-30 \right) \\
\end{align}$
Class interval for mid – point 35 is equal to:
$\begin{align}
  & \left( \left( m-\dfrac{h}{2} \right)-\left( m+\dfrac{h}{2} \right) \right) \\
 & =\left( \left( 35-\dfrac{10}{2} \right)-\left( 35+\dfrac{10}{2} \right) \right) \\
 & =\left( \left( 35-5 \right)-\left( 35+5 \right) \right) \\
 & =\left( 30-40 \right) \\
\end{align}$
Class interval for mid – point 45 is equal to:
$\begin{align}
  & \left( \left( m-\dfrac{h}{2} \right)-\left( m+\dfrac{h}{2} \right) \right) \\
 & =\left( \left( 45-\dfrac{10}{2} \right)-\left( 45+\dfrac{10}{2} \right) \right) \\
 & =\left( \left( 45-5 \right)-\left( 45+5 \right) \right) \\
 & =\left( 40-50 \right) \\
\end{align}$
Now, we are making a table that constitutes the mid – point, class interval and frequencies corresponding to them.


Hence, we have constructed the continuous grouped frequency distribution table with the size of the class intervals as 10.

Note: In the above problem, you might have been wondering the role of frequencies in finding the class interval. The answer is there is no use of the frequencies corresponding to each mid – point in finding the class interval. Sometimes, we have given information in the question which is not useful in solving the problems so don’t over think about the extra information which is of no use.