JEE Main 2020 (Online) 8th January Evening ShiftQuestion 1
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 :
Statistics
Solution
$\sum x=200$, $\sum x^2=20(4+100)=2080$. Corrected: $\sum x=202$, $\sum x^2=2080-81+121=2120$. Variance $=\dfrac{2120}{20}-(10.1)^2=106-102.01=3.99$. Option (B).