**Question:**

**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.**

**Solution:**

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.

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