Mean and standard deviation of 100 observations

Given 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

Now we have to find the correct standard deviation.

As per given criteria,

Number of observations, $\mathrm{n}=100$

Mean of the given observations before correction, $\overline{\mathrm{x}}=40$

But we know,

$\overline{\mathrm{X}}=\frac{\sum \mathrm{x}_{\mathrm{i}}}{\mathrm{n}}$

Substituting the corresponding values, we get

$40=\frac{\sum x_{i}}{100}$

$\Rightarrow \sum x_{1}=40 \times 100=4000$

It is said two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively,

So $\sum x_{i}=4000-30-70+3+27=3930$

So the correct mean after correction is

$\bar{x}=\frac{\sum x_{i}}{n}=\frac{3930}{100}=39.3$

Also given the standard deviation of the 100 observations is 10 before correctior i.e., $\sigma=10$

But we know

$\sigma=\sqrt{\frac{\sum x_{i}^{2}}{n}-\left(\frac{\sum x_{i}}{n}\right)^{2}}$

Substituting the corresponding values, we get

$10=\sqrt{\frac{\sum x_{i}^{2}}{100}-\left(\frac{4000}{100}\right)^{2}}$

Now taking square on both sides, we get

$10^{2}=\frac{\sum x_{i}^{2}}{100}-(40)^{2}$

$\Rightarrow 100=\frac{\sum x_{i}^{2}}{100}-1600$

$\Rightarrow 100+1600=\frac{\sum x_{i}^{2}}{100}$

$\Rightarrow \frac{\sum x_{i}^{2}}{100}=1700$

$\Rightarrow \Sigma x_{1}^{2}=170000$

It is said two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, so correction is

$\Rightarrow \Sigma x_{i}^{2}=170000-(30)^{2}-(70)^{2}+3^{2}+(27)^{2}$

$\Rightarrow \Sigma x_{i}^{2}=170000-900-4900+9+729=164938$

$\Rightarrow \Sigma x_{i}^{2}=164938$

So the correct standard deviation after correction is

$\sigma=\sqrt{\frac{164938}{100}}-\left(\frac{3930}{100}\right)^{2}$

$\sigma=\sqrt{1649.38-(39.3)^{2}}$

$\sigma=\sqrt{1649.38-1544.49}=\sqrt{104.89}$

σ=10.24

Hence the corrected standard deviation is 10.24.