The mean and the standard deviation(s.d.)of five observations are 9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :

Let a, b, c, d, e be five observations.

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\text{ S.D. }=\sqrt{\frac{\mathrm{\Sigma }(a-\overline{x}{)}^2}{5}}=0\Rightarrow \mathrm{\Sigma }(a-\overline{x}{)}^2=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}(a-\overline{x}{)}^2+(b-\overline{x}{)}^2+(c-\overline{x}{)}^2+(d-\overline{x}{)}^2+(e-\overline{x}{)}^2=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a-\overline{x}=b-\overline{x}=c-\overline{x}=d-\overline{x}=e-\overline{x}=0\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a=b=c=d=e=\overline{x}=9\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}a+b+c+d+e=9\times 5=45\end{array}$

Let $x$ be the change in term. Then

New sum  $a+b+c+(d+x)+e=5\times 10=50$

$\Rightarrow x=5$

$\therefore$ New observations becomes 9, 9, 9, 14, 9

New S.D.= $\sqrt{\frac{\mathrm{\Sigma }(a-\overline{x}{)}^2}{5}}=\sqrt{\frac{1^2+1^2+1^2+4^2+1^2}{5}}=\sqrt{\frac{20}{5}}=2$