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 it is replaced by $12$
When $8$ is replaced by $12$
Incorrect sum of observations $=200$
$\therefore$ Correct sum of observations $=200-8+12=204$
$\therefore$ Correct mean $=\frac{\text { Correct sum }}{20}=\frac{204}{20}=10.2$
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}} $
$=2080-64+144$
$=2160$
$\therefore$ Correct standard deviation $=\sqrt{\frac{\text { Correct } \sum x_{i}^{2}}{n}-(\text { Correct mean })^{2}}$
$=\sqrt{\frac{2160}{20}-(10.2)^{2}}$
$=\sqrt{108-104.04}$
$=\sqrt{3.96}$
$=1.98$
Statement $1$ : The variance of first $n$ odd natural numbers is $\frac{{{n^2} - 1}}{3}$
Statement $2$ : The sum of first $n$ odd natural number is $n^2$ and the sum of square of first $n$ odd natural numbers is $\frac{{n\left( {4{n^2} + 1} \right)}}{3}$
The frequency distribution:
$\begin{array}{|l|l|l|l|l|l|l|} \hline X & 2 & 3 & 4 & 5 & 6 & 7 \\ f & 4 & 9 & 16 & 14 & 11 & 6 \\ \hline \end{array}$
Find the standard deviation.
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If the standard deviation of $0, 1, 2, 3, …..,9$ is $K$, then the standard deviation of $10, 11, 12, 13 …..19$ is
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