Ans: (a) Mean $=3.5$, inter-quartile range $=2$, standard deviation $=1.5$ (b) $0.048\ 5$
- The mean
$\begin{array}{cl}
= & 3.5
\end{array}$The inter-quartile range
$\begin{array}{cl}
= & 4-2 \\
= & 2
\end{array} $The standard deviation
$\begin{array}{cl}
= & 1.5
\end{array}$ - The new standard deviation
$\begin{array}{cl}
= & 1.451~456~116
\end{array}$Therefore, the change in the standard deviation
$\begin{array}{cl}
= & 1.5 – 1.451~456~116 \\
= & 0.048~543~884
\end{array}$
