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
$=\sqrt{\frac{\text { Correct } \sum x_{i}^{2}}{n}-(\text { Correct mean })^{2}}$
$=\sqrt{\frac{161701}{100}-(39.9)^{2}}$
$=\sqrt{1617.01-1592.01}=\sqrt{25}=5$
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