The mean and S.D. of the marks of $$200$$  candidates were found to be $$40$$  and $$15$$  respectively. Later, it was discovered that a score of $$40$$  was wrongly read as $$50.$$  The correct mean and S.D. respectively are :

Solution :
$$\eqalign{ & {\text{Corrected }}\sum x = 40 \times 200 - 50 + 40 = 7990 \cr & \therefore \,{\text{Corrected}}\,\,\overline x = \frac{{7990}}{{200}} = 39.95 \cr & {\text{Incorrect}}\,\sum {{x^2}} = n\left[ {{\sigma ^2} + {{\overline x }^2}} \right] \cr & = 200\left[ {{{15}^2} + {{40}^2}} \right] \cr & = 365000 \cr & {\text{Corrected}}\,\sum {{x^2}} = 365000 - 2500 + 1600 = 364100 \cr & \therefore \,{\text{Corrected}}\,\,\sigma = \sqrt {\frac{{364100}}{{200}} - {{\left( {39.95} \right)}^2}} \cr & = \sqrt {\left( {1820.5 - 1596} \right)} \cr & = \sqrt {224.5} \cr & = 14.98 \cr} $$