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 :

• A. $4.01$
• B. $3.99$
• C. $3.98$
• D. $4.02$

Answer 1

Answer: (B)

$\frac{\sum x_{i}}{20} = 10 \quad\Rightarrow \quad\sum x_{i} = 200 \quad...\left(i\right)$
$\frac{\sum x_{i}^{2}}{20} = 100 = 4 \quad \Rightarrow \quad \sum x_{i} ^{2}= 2080 \quad ...\left(ii\right)$
Actual mean $= \frac{200 - 9 + 11}{20} = \frac{202}{20}$
Variance $= \frac{2080 - 81 + 121}{20} - \left(\frac{202}{20}\right)^{2} = 3.99$