The mean and standard deviation of 20 observations were calculated as 10 and $2.5$ respectively. It was found that by mistake one data value was taken as 25 instead of $35 .$ If $\alpha$ and $\sqrt{\beta}$ are the mean and standard deviation respectively for correct data, then $(\alpha, \beta)$ is :
Correct Option: 4,
Given :
$\operatorname{Mean}(\overline{\mathrm{x}})=\frac{\sum \mathrm{x}_{\mathrm{i}}}{20}=10$
or $\Sigma x_{i}=200$ (incorrect)
or $200-25+35=210=\Sigma \mathrm{x}_{\mathrm{i}}($ Correct $)$
Now correct $\overline{\mathrm{x}}=\frac{210}{20}=10.5$
again given S.D $=2.5(\sigma)$
$\sigma^{2}=\frac{\sum x_{i}^{2}}{20}-(10)^{2}=(2.5)^{2}$
or $\Sigma x_{i}^{2}=2125$ (incorrect)
or $\sum x_{i}^{2}=2125-25^{2}+35^{2}$
$=2725($ Correct $)$
$\therefore$ correct $\sigma^{2}=\frac{2725}{20}-\left(10.5^{2}\right)$
$\underline{\underline{\sigma}}^{2}=26$
or $\sigma=26$
$\therefore \underline{\alpha}=10.5, \beta=26$