Given mid-values and frequencies, construct a continuous series by calculating class interval width, lower limits, and upper limits.

The class interval width is the difference between any two consecutive mid-values.

$$i = 25 - 15 = 10$$

So, the class interval width is 10.

The lower limit ($L$) and upper limit ($U$) for a class can be found using the mid-value ($M$) and half of the class interval width ($\frac{i}{2}$).
Here, $\frac{i}{2} = \frac{10}{2} = 5$.
The formulas are:

$$L = M - \frac{i}{2}$$ $$U = M + \frac{i}{2}$$

For each mid-value:

• For $M = 15$: $L = 15 - 5 = 10$ $U = 15 + 5 = 20$ Class interval: $10 - 20$

• For $M = 25$: $L = 25 - 5 = 20$ $U = 25 + 5 = 30$ Class interval: $20 - 30$

• For $M = 35$: $L = 35 - 5 = 30$ $U = 35 + 5 = 40$ Class interval: $30 - 40$

• For $M = 45$: $L = 45 - 5 = 40$ $U = 45 + 5 = 50$ Class interval: $40 - 50$

• For $M = 55$: $L = 55 - 5 = 50$ $U = 55 + 5 = 60$ Class interval: $50 - 60$

• For $M = 65$: $L = 65 - 5 = 60$ $U = 65 + 5 = 70$ Class interval: $60 - 70$

Combine the calculated class intervals with their corresponding frequencies.

$$\boxed{ \begin{array}{|c|c|} \hline \text{Class Interval} & \text{Frequency} \\ \hline 10 - 20 & 20 \\ 20 - 30 & 19 \\ 30 - 40 & 17 \\ 40 - 50 & 62 \\ 50 - 60 & 41 \\ 60 - 70 & 73 \\ \hline \end{array} }$$