The mean and standard deviation of $100$ observations were calculated as $40$ and $5.1$ , respectively by a student who took by mistake $50$ instead of $40$ for one observation. What are the correct mean and standard deviation?

Given that number of observations $(n)=100$

$\text { Incorrect mean }(\bar{x})=40$

Incorrect standard deviation $(\sigma)=5.1$

We know that   $\bar x = \frac{1}{n}\sum\limits_{i = 1}^n {{x_i}} $

i.e.   $40 = \frac{1}{{100}}\sum\limits_{i = 1}^{100} {{x_i}} $  or  $\sum\limits_{i = 1}^{100} {{x_i}}  = 4000$

i.e.,     Incorrect sum of observations $=4000$

Thus    the correct sum of observations $=$ Incorrect sum $-50+40$

$=4000-50+40=3990$

Hence      Correct mean $=\frac{\text { correct sum }}{100}=\frac{3990}{100}=39.9$

Also     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}} } $

i.e.     $5.1 = \sqrt {\frac{1}{{100}} \times Incorrect\sum\limits_{i = 1}^n {x_i^2 - {{\left( {40} \right)}^2}} } $

or     $26.01 = \frac{1}{{100}} \times Incorrect\sum\limits_{i = 1}^n {x_i^2 - 1600} $

Therefore   $Incorrect\sum\limits_{i = 1}^n {x_i^2 = 100\left( {26.01 + 1600} \right) = 162601} $

Now   $Correct\sum\limits_{i = 1}^n {x_i^2}  = Incorrect\sum\limits_{i = 1}^n {x_i^2 - {{\left( {50} \right)}^2} + {{\left( {40} \right)}^2}} $

$=162601-2500+1600=161701$

Therefore Correct standard deviation

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

$=\sqrt{\frac{161701}{100}-(39.9)^{2}}$

$=\sqrt{1617.01-1592.01}=\sqrt{25}=5$