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 option(s) is/are

• A. new mean $=10.2$
• B. new mean $=10.1$
• C. new variance $=3.99$
• D. new variance $=4.01$

Answer 1

Answer: (B), (C)

The correct options are
B new mean $=10.1$
C new variance $=3.99$

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

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

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

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