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

Mean of five observation =9

Sum of observation =45

      S.D=0

one observation changed

New Mean =10

Let the observations 9, 9, 9, 9, 9

   New Mean =10

$$\begin{gathered} \frac{9+9+9+9+(9+\mathrm{a})}{5}=10 \\ \mathrm{a}=5 \end{gathered}$$

New observations 9, 9, 9, 9, 14

$\begin{array}{rcl}\mathrm{S}.\mathrm{D}& =& \sqrt{\frac{\sum ({\mathrm{x}}_{\mathrm{i}}-\overline{\mathrm{x}})}{\mathrm{n}}}\\ & =& \sqrt{\frac{1+1+1+1+16}{5}}\\ & =& \sqrt{\frac{20}{5}}=\sqrt{4}=2\end{array}$