The mean of the following table will be 
Step 1: Find midpoints (class marks).
For each class interval:
0–5 → 2.5, 5–10 → 7.5, 10–15 → 12.5, 15–20 → 17.5
Step 2: Multiply each midpoint by its frequency.
\[ f_i x_i = (2)(2.5) + (4)(7.5) + (6)(12.5) + (10)(17.5) \] \[ = 5 + 30 + 75 + 175 = 285 \] Step 3: Find total frequency.
\[ \sum f_i = 2 + 4 + 6 + 10 = 22 \]
Step 4: Apply mean formula.
\[ \bar{x} = \frac{\sum f_i x_i}{\sum f_i} = \frac{285}{22} = 12.95 \]
Step 5: Conclusion.
Mean ≈ 13.05 (nearest hundredth).
The coefficient of correlation of the above two data series will be equal to \(\underline{\hspace{1cm}}\)
\[\begin{array}{|c|c|} \hline X & Y \\ \hline -3 & 9 \\ -2 & 4 \\ -1 & 1 \\ 0 & 0 \\ 1 & 1 \\ 2 & 4 \\ 3 & 9 \\ \hline \end{array}\]
Identify the median class for the following grouped data:
\[\begin{array}{|c|c|} \hline \textbf{Class interval} & \textbf{Frequency} \\ \hline 5-10 & 5 \\ 10-15 & 15 \\ 15-20 & 22 \\ 20-25 & 25 \\ 25-30 & 10 \\ 30-35 & 3 \\ \hline \end{array}\]
Find the unknown frequency if 24 is the median of the following frequency distribution:
\[\begin{array}{|c|c|c|c|c|c|} \hline \text{Class-interval} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency} & 5 & 25 & 25 & \text{$p$} & 7 \\ \hline \end{array}\]