For a group of 200 candidates the mean and S.D. were found to be 40 and 15 respectively. Later on it was found that the score 43 was misread as 34. Find the correct mean and correct S.D.

Solution: Incorrect mean is $40$. There are $200$ candidates.
$\therefore \overline{X}=40$
So, $\sum x_i=40\ast 200=8000$
This is incorrect $\sum x_i$ as $43$ was misread as $34$.
$\therefore$ Corect $\sum x_i=8000-34+43=8009$
$\therefore$ Correct mean $=\frac{8009}{200}=40.045$
Now, we will find the correct standard deviation.
We know,
$\sigma =\sqrt{\frac{1}{N}\sum {\left(x_i\right)}^2-{\left(\overline{x}\right)}^2}$
When incorrect observations was present, then standard deviation was $15$.
$\therefore 15=\sqrt{\frac{1}{200}\sum {\left(x_i\right)}^2-{\left(40\right)}^2}$
$\Rightarrow 225=\frac{1}{200}\sum {\left(x_i\right)}^2-1600$
$\Rightarrow 365000=\sum {\left(x_i\right)}^2$
This is the sum when observations were incorrect.
$\therefore$ Correct sum$=365000-{\left(34\right)}^2+{\left(43\right)}^2=365693$
$\therefore$ Correct $\sum {\left(x_i\right)}^2=365693$
$\therefore \sigma =\sqrt{\frac{1}{200}\left(365693\right)-{\left(40.045\right)}^2}=14.995$
So, the correct standard deviation is $14.995$.