For a group of 100 students, the mean x ¯ 1 and the standard deviation σ 1 of their marks were…

${\overline{x}}_1=40$

$\sigma =15$

Now, the mean is,

${\overline{x}}_1=\frac{\sum x_i}{n}$

$\Rightarrow 40=\frac{\sum x_i}{100}$

$\Rightarrow \sum x_i=4000$

Corrected summation,

$\sum x_i^{\text{'}}=\sum x_i-30-60+40+50=4000$

The corrected mean,

${\overline{x}}_2=\frac{\sum x_i^{\text{'}}}{100}$

$\Rightarrow {\overline{x}}_2=\frac{4000}{100}=40$

${\sigma }_1^2=\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$

$\Rightarrow 225=\frac{\sum x_i^2}{100}-{\left(\frac{4000}{100}\right)}^2$

$\Rightarrow \sum x_i^2=182500$

Corrected summation,

$\sum x_i^2=182500-{30}^2-{60}^2+{40}^2+{50}^2$

$\Rightarrow \sum x_i^2=182100$

The corrected standard deviation,

${\sigma }_2^2=\frac{\sum x_i^2}{n}-{\left(\frac{\sum x_i}{n}\right)}^2$

$\Rightarrow {\sigma }_2^2=\frac{182500}{100}-{\left(\frac{4000}{100}\right)}^2$

$\Rightarrow {\sigma }_2^2=1821-1600$

$\Rightarrow {\sigma }_2=14.86$

Thus, ${\overline{\mathit{x}}}_{\mathit{1}}\mathit{=}{\overline{\mathit{x}}}_{\mathit{2}}$ and ${\sigma }_{\mathit{1}}\mathit{>}{\sigma }_{\mathit{2}}$.