Question

For a frequency distribution consisting of 18 observations, the mean and the standard deviation were found to be 7 and 4 respectively. But on comparison with the original data, it was found that a figure 12 was miscopied as 21 in calculations. Find the correct mean and standard deviation.___

Answer 1

Given $\frac{\sum x_i}{18}=7(\because$ mean =7 and n=18)
$\Rightarrow \sum x_i=18\times 7=126$

Since an observation 12 was miscopied as 21, therefore,

$correct\sum x_i=126-21+12=117$

Hence true mean $=\frac{correct\sum x_i}{18}=\frac{117}{18}=6.5$

Also variance is given to be $4^2=16,$

therefore, $\frac{\sum x_i^2}{18}-(mean{)}^2=16$

$\Rightarrow \frac{\sum x_i^2}{18}=4^2+(mean{)}^2=16+7^2$

$\Rightarrow \sum x_i^2=18(16+49)=1170$

But, in the summation on R.H.S., one observation (=12) was miscopied as 21, therefore,

$correct\sum x_i^2=1170-{21}^2+{12}^2=1170-441+144=873.$

Hence true variance $=\frac{correct\sum x_i^2}{18}-(truemean{)}^2$

$=\frac{873}{18}-(6.5{)}^2=48.5-42.25=6.25$

$\therefore$ True $S.D=\sqrt{truevariance}=\sqrt{6.25}=2.5$

$\therefore$ Correct mean =6.5 and
correct standard Deviation =2.5