The given set M is $\(\{-14,-11,-7,9,10,13\}\)$ whose range \((= Highest value - Smallest value)\) is \(13\)\( -(-14)=13+14=27\)$

Median of the set $\(M\)$, having 6 values is the average of the $\(3^{\text {rd \&$ the $4^{\text {th \)$ term $\(=\frac{-7+9}{2}=\frac{2}{2}=1 \&\)$ the arithmetic mean (average) of the set $\(M\)$ is

$\(=\frac{-14+(-11)+(-7)+9+10+13}{6}=\frac{0}{6}=0\)$


Finally we can find the standard deviation of set M about the mean $\((=\overline{\mathrm{X)\)$ i.e. about 0 so we get

\begin{tabular}{|l|l|l|}
\hline x_i$ & x_i- Mean $=x_i-0 & \left(x_i-\text { Mean }\right)^2 \\
\hline-14 & -14 & 196 \\
\hline-11 & -11 & 121 \\
\hline-7 & -7 & 49 \\
\hline 9 & 9 & 81 \\
\hline 10 & 10 & 100 \\
\hline 13 & 13 & 169 \\
\hline & {$\sum\left(x_i-\bar{X}\right)^2=716$} \\
\hline
\end{tabular}

So, the standard deviation is $\(\sqrt{\frac{\sum\left(x_i-\bar{X}\right)^2}{n}}=\sqrt{\frac{716}{6}}=\sqrt{119.3} \approx 10.92\)$

Now, checking from the statements

I. Median is greater than mean - which is true as Median $\(=1>\)$ Mean $\(=0\)$

II. Standard deviation is greater than range - which is false as standard deviation $\(=10.92<\) Range \(=27\)$

III. Mean is greater than median - which is false as Mean $\(=0\)< Median \(=1\)$

Hence only statement I is true, so the answer is (A).