The mean and variance of 10 observations were calculated as 15 and...

Given,

$n = 10$ and mean $\overset{¯}{x} = \dfrac{∑ {x}_{i}}{10} = 15$

$⇒ \mathop{∑}_{i = 1}^{10}{x}_{i} = 150$

$⇒ \mathop{∑}_{i = 1}^{9}{x}_{i} + 25 = 150$

$⇒ \mathop{∑}_{i = 1}^{9}{x}_{i} = 125$

Now new sum will be $\mathop{∑}_{i = 1}^{9}{x}_{i} + 15 = 140$

So, actual mean $= \dfrac{140}{10} = 14 = \overset{¯}{x}new$

Now finding variance

$\dfrac{∑ {x}_{i}^{2}}{10} - {\left(\overset{¯}{x}\right)}^{2} = 15$

$⇒ \dfrac{∑ {x}_{i}^{2}}{10} - {15}^{2} = 15$

$⇒ \mathop{∑}_{i = 1}^{10}{x}_{i}^{2} = 2400$

Removing observation of $25$ we get,

$⇒ \mathop{∑}_{i = 1}^{9}{x}_{i}^{2} + 625 = 2400$

$⇒ \mathop{∑}_{i = 1}^{9}{x}_{i}^{2} = 1775$

Now adding observation $15$ we get,

$\mathop{∑}_{i = 1}^{9}{x}_{i}^{2} + {15}^{2} = 2000 = {\left(∑ {x}_{i}^{2}\right)}_{actual}$

Now calculating actual standard deviation we get,

${\sigma}_{actual}^{2} = \dfrac{{\left(∑ {x}_{i}^{2}\right)}_{actual}}{10} - {\left(\overset{¯}{x}new\right)}^{2}$

$= \dfrac{2000}{10} - {14}^{2}$

$= 200 - 196 = 4$

Calculating standard deviation, ${\left(S . D\right)}_{actual} = \sqrt{4} = 2$