The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively.

Given:

$\operatorname{Mean}(\bar{x})=\frac{\Sigma x_{i}}{20}=10$

or $\Sigma x_{i}=200$ (incorrect)

or $200-25+35=210=\Sigma x_{i}($ Correct $)$

Now correct $\overline{\mathrm{x}}=\frac{210}{20}=10.5$

again given S.D $=2.5(\sigma)$

$\sigma^{2}=\frac{\Sigma x_{i}^{2}}{20}-(10)^{2}=(2.5)^{2}$

or $\Sigma x_{1}^{2}=2125$ (incorrect)

or $\Sigma \mathrm{x}_{\mathrm{i}}^{2}=2125-25^{2}+35^{2}$

$=2725$ (Correct)

$\therefore$ correct $\sigma^{2}=\frac{2725}{20}-(10.5)^{2}$

$\underline{\sigma}^{2}=26$

or $\sigma=26$

$\therefore \underline{\alpha}=10.5, \beta=26$