The 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 in each of the following cases:

If wrong item is omitted.

Number of observations (n) $=20$

Incorrect mean $=10$

Incorrect standard deviation $=2$

$\bar x = \frac{1}{n}\sum\limits_{i = 1}^{20} {{x_i}} $

$10 = \frac{1}{{20}}\sum\limits_{i = 1}^{20} {{x_i}} $

$ \Rightarrow \sum\limits_{i = 1}^{20} {{x_i}}  = 200$

That is, incorrect sum of observations $=200$

Correct sum of observations $=200-8=192$

$\therefore$ Correct mean $=\frac{\text { correct sum }}{19}=\frac{192}{19}=10.1$

Standard deviation $\sigma  = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {{x_i}^2 - \frac{1}{{{n^2}}}{{\left( {\sum\limits_{i = 1}^n {{x_i}} } \right)}^2}} } $

$ = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {x_i^2 - {{\left( {\bar x} \right)}^2}} } $

$ \Rightarrow 2 = \sqrt {\frac{1}{{20}}Incorrect\sum\limits_{i = 1}^n {x_i^2 - {{\left( {10} \right)}^2}} } $

$ \Rightarrow 4 = \frac{1}{{20}}Incorrect\sum\limits_{i = 1}^n {x_i^2 - 100} $

$ \Rightarrow Incorrect\sum\limits_{i = 1}^n {x_i^2 = 2080} $

$\therefore Correct\,\,\sum\limits_{i = 1}^n {x_i^2 = \,} Incorrect\,\,\sum\limits_{i = 1}^n {x_i^2 - {{\left( 8 \right)}^2}} $

$=2080-64$

$=2016$

$\therefore$ Correct standard deviation $=\sqrt{\frac{\text { Correct } \sum x_{i}^{2}}{n}-(\text { Correct mean })^{2}}$

$=\sqrt{\frac{2016}{19}-(10.1)^{2}}$

$=\sqrt{1061 \cdot 1-102 \cdot 1}$

$=\sqrt{4.09}$

$=2.02$