The mean and standard deviation of a group of 100 observations were found to be 20 and 3, respectively.

Number of observations (n) = 100

Incorrect mean $(\bar{x})=20$

Incorrect standard deviation $(\sigma)=3$

$\Rightarrow 20=\frac{1}{100} \sum_{i=1}^{100} x_{i}$

$\Rightarrow \sum_{i=1}^{100} x_{j}=20 \times 100=2000$

$\therefore$ Incorrect sum of observations $=2000$

$\Rightarrow$ Correct sum of observations $=2000-21-21-18=2000-60=1940$

$\therefore$ Correct mean $=\frac{\text { Correct sum }}{100-3}=\frac{1940}{97}=20$

Standard deviation $(\sigma)=\sqrt{\frac{1}{n} \sum_{i=1}^{n} x_{i}-\frac{1}{n^{2}}\left(\sum_{i=1}^{n} x_{i}\right)^{2}}=\sqrt{\frac{1}{n} \sum_{j=1}^{n} x_{i}^{2}-(\bar{x})^{2}}$

$\Rightarrow 3=\sqrt{\frac{1}{100} \times \operatorname{Incorrect} \sum x_{j}{ }^{2}-(20)^{2}}$

$\Rightarrow$ Incorrect $\sum x_{i}{ }^{2}=100(9+400)=40900$

Correct $\sum_{i=1}^{n} x_{i}^{2}=\operatorname{Incorrect} \sum_{i=1}^{n} x_{i}^{2}-(21)^{2}-(21)^{2}-(18)^{2}$

$=40900-441-441-324$

$=39694$

$\therefore$ Correct standard deviation $=\sqrt{\frac{\text { Correct } \sum x_{i}^{2}}{n}-(\text { Correct mean })^{2}}$

$=\sqrt{\frac{39694}{97}-(20)^{2}}$

$=\sqrt{409.216-400}$

$=\sqrt{9.216}$

$=3.036$