Question

The mean and variance of $20$ observations are found to be $10$ and $4$ respectively. On rechecking, it was found that an observation which was $11$ was incorrectly recorded as $9$. If the correct variance is $V$, then the value of $\left[V\right]$ is

Answer 1

$\overline{x}=\frac{\sum x_i}{20}=10\Rightarrow \sum x_i=200\frac{\sum x_i^2}{20}-(\overline{x}{)}^2=4\Rightarrow \sum x_i^2=\left(104\right)\left(20\right)=2080$

Now, the actual mean
$\frac{\sum x_i-9+11}{20}=10.1$

The actual variance
$V=\frac{\sum x_i^2-81+121}{20}-(10.1{)}^2\Rightarrow V=106-102.01=3.99\therefore [V]=3$