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

Answer: 224

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, $\frac{\sum {\left(x_i\right)}^2}{200}-{\left(40\right)}^2=225$
$\Rightarrow \sum {\left(x_i\right)}^2=\left(225+1600\right)\times 200=365000$
New standard deviation $=\sqrt{\frac{365000+1600-2500}{200}-{\left(39.95\right)}^2}$
$=\sqrt{1820.50-1596.0025}$
$=\sqrt{224.4975}$
Therefore, $d=\sqrt{224.4975}\Rightarrow d^2=\left[224.4975\right]=224$