Question

For a group of 200 candidates, the mean and standard deviation were found to be 40 and 15, respectively. Later on, it was found that the score 43 was misread as 34. Find the correct mean and correct standard deviation.

Answer 1

We have, n = 200, $\overline{x}$ = 40 and $\sigma$ = 15

$\therefore$ $\overline{x}=\frac{1}{n}\sum x_i\Rightarrow =n\overline{x}=200\times 40=8000$

Corrected$\sum x_i$ = Incorrect $\sum x_i$ - Incorrect value + Correct value = 8000 - 34 + 43 = 8009

Corrected mean = $\frac{\text{Corrected}\sum x_i}{n}=\frac{8009}{200}=40.045$

Now, $\sigma =15\Rightarrow {\sigma }^2=\frac{1}{n}(\sum x_i^2)-{\left(\frac{\sum x_i}{n}\right)}^2$

$\Rightarrow {15}^2=\frac{1}{200}(\sum x_i^2)-{\left(\frac{8000}{200}\right)}^2\Rightarrow 225=\frac{\sum x_i^2}{200}-1600$

$\Rightarrow \sum x_i^2=200\times 1825=365000$

Now, corrected $\sum x_i^2$ = Incorrect $\sum x_i^2$ - (Square of incorrect value) + (Square of correct value )

= $365000-(34{)}^2+(43{)}^2=365000-1156+1849=365693$

So, corrected $\sigma =\sqrt{\frac{1}{n}\text{corrected}\sum x_i^2-{(\frac{1}{n}\text{corrected}\sum x_i)}^2}=\sqrt{\frac{365693}{200}-{\left(\frac{8009}{200}\right)}^2}$

= $\sqrt{1828.465-1603.602}=14.995$