If the mean of the frequency distribution Class : 0 - 10 10 - 20 20...

Given that the mean for the frequency distribution is $28$.

We know that Mean \left(x\right) = \dfrac{∑ {f}_{i}{x}_{i}}{∑ {f}_{i}}$</p><p>Now for the class interval$0 - 10$we get${x}_{1} = \dfrac{0 + 10}{2} = 5$. </p><p>Similarly we can calculate${x}_{i}$for other class intervals.</p><p>Now we know that mean$= 28$</p><p>$⇒ \dfrac{2 \times 5 + 3 \times 15 + x \times 25 + 5 \times 35 + 4 \times 45}{14 + x} = 28$</p><p>$⇒ x = 6$</p><p>Also we know that variance \left({\sigma}^{2}\right) = \dfrac{∑ {f}_{i}{\left(\overset{¯}{x} - {x}_{i}\right)}^{2}}{∑ {f}_{i}}$

${\sigma}^{2} = \dfrac{2 \times {\left(28 - 5\right)}^{2} + 3 \times {\left(28 - 15\right)}^{2} + 6 \times {\left(28 - 25\right)}^{2} + 5 \times {\left(28 - 35\right)}^{2} + 4 \times {\left(28 - 45\right)}^{2}}{20}$

${\sigma}^{2} = \dfrac{2 \times {\left(23\right)}^{2} + 3 \times {\left(13\right)}^{2} + 6 \times {\left(3\right)}^{2} + 5 \times {\left(7\right)}^{2} + 4 \times {\left(17\right)}^{2}}{20}$

${\sigma}^{2} = 151$

Hence this is the required answer.