Question

For a group of 200 candidates, the mean and the standard deviation of scores were found to be 40 and 15, respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53, respectively. Find the correct mean and the standard deviation.

Answer 1

We have, n = 200, incorrect mean = 40 and incorrect standard devition = 15.

Now, incorrect mean = 40

$\Rightarrow \frac{Incorrect\sum x_i}{200}=40$

$\Rightarrow$ Incorrect $\sum x_i=8000$

$\therefore$ Correct $\sum x_i=8000-(34+53)+(43+35)$

= $8000-87+78=7991$

So, correct mean = $\frac{7991}{200}=39.955$

And incorrect SD = 15

$\Rightarrow$ Incorrect variance = $(15{)}^2=225$

$\Rightarrow \frac{Incorrect\sum x_i^2}{200}-(Incorrectmean{)}^2=225$

$\Rightarrow \frac{Incorrect\sum x_i^2}{200}-(40{)}^2=225$

$\Rightarrow$ Incorrect $\sum x_i^2=200(1600+225)$

= $200\times 1825=365000$

$\Rightarrow$ Correct $\sum x_i^2$ = Incorrect $\sum x_i^2$

$-(34{)}^2+{53}^2)+({43}^2+{35}^2)$

= $365000-3965+3074=364109$

So, correct variance

= $\frac{1}{200}(correct\sum x_i^2)-(correctmean{)}^2$

= $\frac{1}{200}\left(364109\right)-{\left(\frac{7991}{200}\right)}^2$

= $1820.545-1596.402=224.143$

$\therefore$ Correct SD= $\sqrt{224.143}=14.971$