Question

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

• A. $(11,25)$
• B. $(11,26)$
• C. $(10.5,25)$
• D. $(10.5,26)$

Answer 1

The correct option is D $(10.5,26)$

$\overline{x}=10\sqrt{\frac{\sum x_i^2}{20}-(\overline{x}{)}^2}=2.5$
$\frac{\sum x_i^2}{20}-(10{)}^2=6.25\Rightarrow \sum x_i^2=20\times 106.25=2125$

$\sum x_{i\text{(actual)}}^2=2125-{25}^2+{35}^2$
$=2125+600=2725$

For ${\overline{x}}_{\text{(actual)}}\Rightarrow \frac{\sum x}{n}=10\Rightarrow \sum x=200$
$\sum x_{\text{(actual)}}=200-25+35=210$
${\overline{x}}_{\left(\text{actual}\right)}=\frac{210}{20}=10.5$

$S.D=\sqrt{\frac{2725}{20}-(10.5{)}^2}=\sqrt{136.25-110.25}=\sqrt{26}$