The mean and standard deviation of 100 observations were calculated...

Standard deviation $\left(\sigma\right) = \sqrt{\dfrac{1}{n}\mathop{∑}_{i = 1}^{n}{x}_i^{2} - \dfrac{1}{n^{2}}{\left(\mathop{∑}_{i = 1}^{n}{x}_{i}\right)}^{2}}$
$= \sqrt{\dfrac{1}{n}\mathop{∑}_{i = 1}^{n}{x}_i^{2} - {\left(\overset{-}{x}\right)}^{2}}$

~$⇒ 5.1 = \sqrt{\dfrac{1}{100} \times Incorrect\mathop{∑}_{i = 1}^{n}{x}_i^{2} - \left(40\right)2}$
or, $26.01 = \dfrac{1}{100} \times$ Incorrect $\mathop{∑}_{i = 1}^{n}{x}_i^{2} - 1600$
Therefore,
Incorrect $\mathop{∑}_{i = 1}^{n}{x}_i^{2} = 100 ( 26 .01 + 1600 ) = 162601$

Correct mean,~$\overset{¯}{x} = \dfrac{3990}{100} = 39 . 9$
Now, Correct $\mathop{∑}_{i = 1}^{n}{x}_i^{2} =$ Incorrect $\mathop{∑}_{i = 1}^{n}{x}_i^{2} - {\left(50\right)}^{2} + {\left(40\right)}^{2}$
$= 162601 - 2500 + 1600 = 161701$
Therefore correct standard deviation
$= \sqrt{\dfrac{Correct∑ {x}_i^{2}}{n} - {\left(Correctmean\right)}^{2}}$
$= \sqrt{\dfrac{161701}{100} - {\left(39.9\right)}^{2}} = \sqrt{1617.01 - 1592.01} = \sqrt{25} = 5$