Question

The mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively. Later, it was discovered that a score of $40$ was wrongly read as $50.$ If the correct standard deviation is $d$ then the value of $\left[d^{2}\right]$ is ( where $\left[.\right]$ denotes greatest integer function )

Answer 1

We have, mean and $S.D.$ of the marks of $200$ candidates were found to be $40$ and $15$ respectively
Also, given that a score of $40$ was wrongly read as $50.$
Therefore, corrected mean $=\frac{\left(200 \times 40\right) - 50 + 40}{200}$
$=\frac{7990}{200}=39.95$
Now, $\frac{\displaystyle \sum \left(x_{i}\right)^{2}}{200}-\left(40\right)^{2}=225$
$\Rightarrow \displaystyle \sum \left(x_{i}\right)^{2}=\left(225 + 1600\right)\times 200=365000$
New standard deviation $=\sqrt{\frac{365000 + 1600 - 2500}{200} - \left(39 . 95\right)^{2}}$
$=\sqrt{1820 . 50 - 1596 . 0025}$
$=\sqrt{224 . 4975}$
Therefore, $d=\sqrt{224 . 4975}\Rightarrow d^{2}=\left[224 . 4975\right]=224$