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

The correct option is B $3.99$

Mean, $\overline{x}=10$
$\Rightarrow \frac{\sum x_i}{20}=10\Rightarrow \sum x_i=200$

Variance, ${\sigma }^2=4$
$\Rightarrow \frac{\sum x_i^2}{20}-{\left(10\right)}^2=4\Rightarrow \sum x_i^2=2080$

$\text{Correct mean}=\frac{200-9+11}{20}=\frac{202}{20}=10.1$

$\therefore \text{Correct variance}=\frac{2080-9^2+{11}^2}{20}-(10.1{)}^2$
$=106-102.01=3.99$