Question

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

Answer 1

We have

$n=200.\overline{x}40,\sigma =15$

Now,

$\overline{x}=\frac{1}{n}\sum xi$

$\therefore \frac{1}{200}\sum xi=40$

$\Rightarrow \sum xi=40\times 200=8000$

Since, the score was miss read this sum is incorrect

$\Rightarrow$ corrected $\sum xi=8000-34-53+43+35$

$=8000-7$

$=7993$

$\therefore$ corrected mean $=\frac{\sum xi}{200}$

$=\frac{7993}{200}$

$=39.955$

$S.D.=\sigma =15$

$\Rightarrow$ variance $={15}^2=225$

Now,

Variance $=(\frac{1}{n}\sum x{i}^2)-{(\frac{1}{n}\sum xi)}^2$

$\therefore \frac{1}{200}\sum x{i}^2–(40{)}^2=225$

$\Rightarrow \sum (xi{)}^2=200\times 1825=365000$

Now,

This is an incorrect reading.

$\therefore$ corrected $\sum x{o}^2=365000-{34}^2-{53}^2+{43}^2+{35}^2$

$=365000-1156-2809+1849+1225$

$=364109$

Corrected variance $=(\frac{1}{n}corrected\sum xi)–(correctedm{)}^2$

$=(\frac{1}{200}\times 364109)–(39.955{)}^2$

$=1820.545-1596.402$

Corrected variance $=224.14$

Corrected $S.D=\sqrt{correctedvariance}$

$=\sqrt{224.14}$

Corrected $S.D=14.97$

Hence, this is the answer.