Question:
The mean and variance of 20 observations are found to be 10 and 4 , respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11 . Then the correct variance is:
Correct Option: 1
Solution:
Let $x_{1}, x_{2}, \ldots . ., x_{20}$ be 20 observations, then
Mean $=\frac{x_{1}+x_{2}+\ldots . .+x_{20}}{20}=10$
$\Rightarrow \frac{\sum_{i=1}^{20} x_{i}}{20}=10$...(i)
Variance $=\frac{\Sigma x_{i}^{2}}{n}-(\bar{x})^{2}$
$\Rightarrow \quad \frac{\Sigma x_{i}^{2}}{20}-100=4$ .....(ii)
$\Sigma x_{i}^{2}=104 \times 20=2080$
Actual mean $=\frac{200-9+11}{20}=\frac{202}{20}$
Variance $=\frac{2080-81+121}{20}-\left(\frac{202}{20}\right)^{2}$
$=\frac{2120}{20}-(10.1)^{2}=106-102.01=3.99$