Question

Mean and variance of $20$ observations are $10$ and $4$, respectively. It was found, that in place of $11,9$ was taken by mistake, then correct variance is

• A. $3.99$
• B. $3.98$
• C. $4.01$
• D. $4.02$

Answer 1

The correct option is A $3.99$

Let $20$ observations be $x_1,x_2,\cdots \cdots x_{19},s$ (wrong series)
Here,$s$ is the incorrect observation

$\text{Mean}=\frac{\sum _{i=1}^{19}{x}_i+9}{20}=10$
$\Rightarrow \sum _{i=1}^{19}{x}_i=191\cdots \left(1\right)$

Let series with correct observations be
$x_1,x_2,x_3,\cdots \cdots ,x_{19},x_{20}$

Now, $\overline{x}$(Corrected mean)$=\frac{\sum _{i=1}^{19}{x}_i+x_{20}}{20}=\frac{191+11}{20}=\frac{202}{20}=10.1$

Variance of wrong series$=4$

$\Rightarrow \text{Variance}=\frac{\sum _{i=1}^{19}(x_i{)}^2+s^2}{20}-(\overline{x}{)}^2=4$
$\Rightarrow \sum _{i=1}^{19}(x_i{)}^2=(104\left)\right(20)-81$
$\Rightarrow \sum _{i=1}^{19}(x_i{)}^2=1999\cdots (2)$



Correct variance
$=\frac{\sum _{i=1}^{19}(x_i{)}^2+(x_{20}{)}^2}{20}-(10.1{)}^2$
$=\frac{1999+(11{)}^2}{20}-(10.1{)}^2$
$=106-102.01=3.99$