**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 $=\frac{2080-81+121}{20}-\left(\frac{202}{20}\right)^{2}$

$=\frac{2120}{20}-(10.1)^{2}=106-102.01=3.99$

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