Question

The mean and variance of $10$ observations are found to be $10$ and $4$ respectively. On rechecking it was found that an observation $8$ was incorrect. If it is replaced by $18,$ then the correct variance is

• A. $7$
• B. $8$
• C. $9$
• D. $\frac{55}{6}$

Answer 1

Answer: (C)

${\bar{x}}_{(\text{old })}=10=\frac{\Sigma x_i}{10}\Rightarrow \Sigma x_{i(\text{ old })}=100$
$\Sigma x_{i(\text{ new })}=100-8+18=110$
$x_{(\text{new })}=\frac{110}{10}=11$
${\mathrm{Var}}_{(\text{old })}=4=\frac{\Sigma x_{i(\text{ old })}^2}{10}-{\left({\bar{x}}_{(\text{old })}\right)}^2$
$\Sigma x_{i(\text{ old })}^2=1040$
$\Sigma x_{i(\text{ new })}^2=1040-64+324$
$=1300$
${\mathrm{Var}}_{(\text{new })}=\frac{1300}{10}-(11{)}^2=130-121=9$