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  • Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

Mean and standard deviation of 100 observations were found to be 40 and 10, respectively. If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

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Mean and standard deviation of 100 observations were found to be 40 and 10, respectively.

If at the time of calculation two observations were wrongly taken as 30 and 70 in place of 3 and 27 

respectively.

We have to find the correct standard deviation

As per the given criteria, Number of observations, n=100

Mean of the given observations before correction,\\\\ \overline{X~}=40 \\\\

But we know \\\overline{X~}=\frac{ \Sigma x_{i}}{n}~~ \\\\

 

Substituting the corresponding values, we get 40=\frac{ \Sigma x_{i}}{100}~~ \\\\


\\ \\ ~~ \Sigma x_{i}=40 * 100=4000~ \\\\ \\ \text{It is given in the question that 2 observations were wrongly taken as 30 and 70 in place of 3 and 27} \\\\ \\ ~respectively~~~~ \\\\ \\ So~~ \Sigma x_{i}=4000-30-70+3+27=3930~ \\\\ \\ \text{ So the correct mean after correction is }\overline{X~}=\frac{ \Sigma x_{i}}{n}=\frac{3930}{100}=39.3~ \\\\

Also given that standard deviation of the 100 observations is 10 before correction \sigma =10 \\\\


\\ \text{But~we know that } \\ \sigma = \sqrt {\frac{ \Sigma x_{i}^{2}}{n}- \left( \frac{ \Sigma x_{i}}{n} \right) ^{2}}~~ \\ ~\text{ Substituting the corresponding values we get }10=\sqrt {\frac{ \Sigma x_{i}^{2}}{100}- \left( \frac{4000}{100} \right) ^{2}}~~ \\ \text{~Now taking square on both sides, we get }10^{2}=\frac{ \Sigma x_{i}^{2}}{100}- \left( 40 \right) ^{2}~~ \\ ~ 100=\frac{ \Sigma x_{i}^{2}}{100}-1600~ \\ ~ 100+1600=\frac{ \Sigma x_{i}^{2}}{100}~~ \\ ~~ \Sigma x_{i}^{2}=170000~ \\ \text{It is said that 2 observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively. }


 \\ \text{So,~correction is } \Sigma x_{i}^{2}=170000- \left( 30 \right) ^{2}- \left( 70 \right) ^{2}+ \left( 3 \right) ^{3}+ \left( 27 \right) ^{2}~~ \\\\ \\ ~~~ \Sigma x_{i}^{2}=170000-900-4900+9+729=164938 \\\\ \\ \text{~~So, the correct standard deviation after correction is } \\\\ \\ \sigma = \sqrt {\frac{164938}{100}- \left( \frac{3930}{100} \right) ^{2}} =\sqrt {1649.38- \left( 39.3 \right) ^{2}} = \sqrt {104.89}=10.24 \\\\

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