Consider the following data with frequency distribution:
\begin{tabular}{|l|l|l|l|l|l|}
\hline\( x_{i} \) & 3 & 4 & 2 & 7 & 11 \\
\hline\( f_{i} \) & 3 & 4 & 2 & 5 & 1 \\
\hline
\end{tabular}
\( \mathrm{P} \)
Match column-I with column-II
\begin{tabular}{|c|c|c|c|}
\hline & Column I & & Column II \\
\hline (A) & \begin{tabular}{l}
The mean of the given \\
data is
\end{tabular} & P. & 4 \\
\hline (B) & \begin{tabular}{l}
The median of the given \\
data is
\end{tabular} & Q. & 5 \\
\hline (C) & \begin{tabular}{l}
The mean deviation \\
about the mean of the \\
given data is
\end{tabular} & \( \mathrm{R} \). & 1.93 \\
\hline (D) & \begin{tabular}{l}
The mean deviation \\
about the median of the \\
above data is
\end{tabular} & \( \mathrm{S} \). & 2.13 \\
\hline A & B & \multicolumn{2}{|c|}{ D } \\
\hline (1) & Q & \multicolumn{2}{|c|}{ R } \\
\hline (2) & \( \mathrm{S} \) & \multicolumn{2}{|c|}{ R } \\
\hline (3) & \( \mathrm{R} \) & \multicolumn{2}{|c|}{ S } \\
\hline \multicolumn{4}{|c|}{ (4) \( \mathrm{P} \)} \\
\hline
\end{tabular}
.
0