"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 :"

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Accepted Answer

"Given :

Mean $(\overline x ) = {{\sum {{x_i}} } \over {20}} = 10$

or $\Sigma$x_i = 200 (incorrect)

or 200 $-$ 25 + 35 = 210 = $\Sigma$x_i (Correct)

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

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

${\sigma ^2} = {{\sum {{x_i}^2} } \over {20}} - {(10)^2} = {(2.5)^2}$

or $\sum {{x_i}^2} = 2125$ (incorrect)

or $\sum {{x_i}^2} = 2125 - {25^2} + {35^2}$

= 2725 (correct)

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

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

or $\sigma$ = 26

$\therefore$ $\alpha$ = 10.5, $\beta$ = 26"

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 :

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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 :

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