Question

Mean and standard deviation of $20$ observations are found to be $10$ and $2$ respectively , On rechecking it was found that an observation $8$ was incorrect. Calculate the correct mean and standard deviation if $8$ is replaced by $12$.

Answer 1

Given mean $\left(\overline{x}\right)=10$

$\Rightarrow$ $\frac{\sum x_i}{20}=10$

$\Rightarrow$ $\sum x_i=10\times 20$

$\therefore$ $\sum x_i=200$

If wrong item $8$ is replaced by $12$

$\Rightarrow$ $\sum x_i=20-8+12$

$\therefore$ $\sum x_i=204$

Now, correct mean $\left(\overline{x}\right)=\frac{\sum x_i}{20}=\frac{204}{20}=10.2$

$\Rightarrow$ Given standard deviation $=2$

So variance $=(SD{)}^2=(2{)}^2=4$

$\Rightarrow$ $\frac{\sum x_i^2}{20}-{\left(\frac{\sum x_i}{20}\right)}^2=4$

$\Rightarrow$ $\frac{\sum x_i^2}{20}-(10{)}^2=4$

$\Rightarrow$ $\frac{\sum x_i^2}{20}-100=4$

$\Rightarrow$ $\frac{\sum x_i^2}{20}=104$

$\Rightarrow$ $\sum x_i^2=2080$

Now, replacing observation $8$ by $12$

$\Rightarrow$ $\sum x_i^2=2080-8^2+{12}^2$

$\Rightarrow$ $\sum x_i^2=2080-64+144$

$\therefore$ $\sum x_i^2=2160$

Now variance of remaining observations are

$\Rightarrow$ $\frac{\sum x_i^2}{20}-{\left(\frac{\sum x_i}{20}\right)}^2$

$\Rightarrow$ $\frac{2160}{20}-(10.2{)}^2$

$\Rightarrow$ $108-104.04$

$\Rightarrow$ $3.96$

$\Rightarrow$ Correct standard deviation $=\sqrt{3.96}=1.99$

$\therefore$ Correct mean and standard deviation are $10.2$ and $1.99$