It is given that the mean and standard deviation of a group of 100 observations are
\[ \bar{x} = 20, \quad \sigma = 3. \]
Step 1: Find the sum of the 100 observations
We know that
\[ \bar{x} = \frac{\sum x}{n}. \]
Hence,
\[ \sum x = n \bar{x} = 100 \times 20 = 2000. \]
Step 2: Find the sum of squares of the observations
We use the formula
\[ \sigma^2 = \frac{\sum x^2}{n} - \bar{x}^2. \]
Substituting the given values:
\[ 3^2 = \frac{\sum x^2}{100} - 20^2. \]
\[ 9 = \frac{\sum x^2}{100} - 400. \]
\[ \frac{\sum x^2}{100} = 409. \]
\[ \sum x^2 = 409 \times 100 = 40900. \]
Step 3: Subtract the incorrect observations
The incorrect observations are \[ 21,\; 21,\; 18. \]
Their sum is
\[ 21 + 21 + 18 = 60. \]
Their sum of squares is
\[ 21^2 + 21^2 + 18^2 = 441 + 441 + 324 = 1206. \]
Step 4: Find the corrected sum and sum of squares
Corrected number of observations:
\[ n' = 100 - 3 = 97. \]
Corrected sum:
\[ \sum x' = 2000 - 60 = 1940. \]
Corrected sum of squares:
\[ \sum x'^2 = 40900 - 1206 = 39694. \]
Step 5: Find the corrected mean
\[ \bar{x}' = \frac{1940}{97} = 20. \]
Step 6: Find the corrected standard deviation
Using the formula
\[ \sigma'^2 = \frac{\sum x'^2}{n'} - (\bar{x}')^2. \]
\[ \sigma'^2 = \frac{39694}{97} - 20^2. \]
\[ \sigma'^2 = 409.216 - 400. \]
\[ \sigma'^2 = 9.216. \]
\[ \sigma' = \sqrt{9.216} \approx 3.04. \]
Final Answer:
\[ \boxed{ \text{Corrected Mean} = 20, \quad \text{Corrected Standard Deviation} \approx 3.04 } \]