Question

The mean and the standard deviation of a group of $20$ items was found to be $40$ and $15$ respectively. While checking it was found that an item $43$ was wrongly written as $53$. Calculate the correct mean and standard deviation.

Answer 1

Let us find the correct mean.
Mean of $20$ items, $\overline{x}=\frac{\sum x}{n}=40$
$\Rightarrow \frac{\sum x}{20}=40$
$\Rightarrow \sum x=20\times 40=800$
Corrected $\sum x=800-\left(wrongvalue\right)+\left(correctvalue\right)$
Now, corrected $\sum x=800-53+43=790$.
$\therefore$ The corrected Mean$=\frac{790}{20}=39.5$
Variance, ${\sigma }^2=\frac{\sum x^2}{n}-{\left(\frac{\sum x}{n}\right)}^2=225$ (given)
$\Rightarrow \frac{\sum x^2}{20}-{40}^2=225$
$\Rightarrow \sum x^2-32000=225\times 20=4500$
$\therefore \sum x^2=32000+4500=36500$
corrected $\sum x^2=36500-(wrongvalue{)}^2+(correctvalue{)}^2$
corrected $\sum x^2=36500-{53}^2+{43}^2=36500-2809+1849$
$=36500-960=35540$.
Now, the corrected ${\sigma }^2=\frac{Corrected\sum x^2}{n}-(Correctedmean{)}^2$
$=\frac{35540}{20}-(39.5{)}^2$
$=1777-1560.25=216.75$
Corrected $\sigma =\sqrt{216.75}\simeq 14.72$
$\therefore$ The corrected Mean$=39.5$ and the corrected S.D. $\simeq 14.72$.