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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 :
$(11,26)$
$(10.5,25)$
$(11,25)$
$(10.5,26)$
Solution
Given :
Mean $(\bar{x})=\frac{\Sigma x_{i}}{20}=10$
or $\Sigma \mathrm{x}_{\mathrm{i}}=200$ (incorrect)
or $200-25+35=210=\Sigma \mathrm{x}_{\mathrm{i}}$ (Correct)
Now correct $\bar{x}=\frac{210}{20}=10.5$
again given $S . D=2.5(\sigma)$
$\sigma^{2}=\frac{\Sigma \mathrm{x}_{\mathrm{i}}^{2}}{20}-(10)^{2}=(2.5)^{2}$
or $\Sigma \mathrm{x}_{\mathrm{i}}^{2}=2125$ (incorrect)
or $\Sigma \mathrm{x}_{\mathrm{i}}^{2}=2125-25^{2}+35^{2}$ $=2725$ (Correct)
$\therefore$ correct $\sigma^{2}=\frac{2725}{20}-(10.5)^{2}$
$\underline{\underline{\sigma}}^{2}=26$
or $\sigma=26$
$\therefore \underline{\alpha}=10.5, \beta=26$
