Question

The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation when

(i) the wrong item is omitted,

(ii) the wrong item is replaced by 12.

Answer 1

Mean $=10\Rightarrow \frac{{\sum }_{i=1}^{20}{x}_i}{20}=10$

$\Rightarrow {\sum }_{i=1}^{20}{x}_i=200$

$SD=2\Rightarrow {\sigma }^2=4$

$\Rightarrow \frac{{\sum }_{i=1}^{20}{x}_i^2}{20}-(10{)}^2=4$

$\Rightarrow {\sum }_{i=1}^{20}{x}_i^2=2080$

Thus, incorrect $\left({\sum }_{i=1}^{20}{x}_i^2\right)=200$ and incorrect $\left({\sum }_{i=1}^{20}{x}_i^2\right)=2080$

CASE (i) When the wrong item is omitted

On omitting 8, we are left with 19 observations.

$\therefore \text{correct}\left({\sum }_{i=1}^{19}{x}_i\right)=\text{incorrect}\left({\sum }_{i=1}^{20}{x}_i\right)-8$

= (200 - 8) =192.

Thus, correct $\left({\sum }_{i=1}^{19}{x}_i\right)=192$

$\therefore$ correct mean $=\frac{192}{19}=10.105$ . . . (i)

Also, correct $\left({\sum }_{i=1}^{19}{x}_i^2\right)$ = incorrect $\left({\sum }_{i=1}^{20}{x}_i^2\right)-64$

= (2080 - 64) = 2016.

$\therefore$ correct variance $=\frac{1}{19}$ $\left(\text{correct}{\sum }_{i=1}^{19}{x}_i^2\right)-(\text{correct mean}{)}^2$

$(\frac{1}{19}\times 2016)-{\left(\frac{192}{19}\right)}^2=(\frac{2016}{19}-\frac{36864}{361})$

$=\frac{(38304-36864)}{361}=\frac{1440}{361}$

$\therefore \text{correct SD}=\sqrt{\frac{1440}{361}}=\frac{12\sqrt{10}}{19}=\frac{(12\times 3.162)}{19}$

$=\frac{37.944}{19}=1.997$

$\therefore$ mean = 10.105 and SD = 1.997

CASE (II) When incorrect observation 8 is replaced by 12.

Incorrect $\left({\sum }_{i=1}^{20}{x}_i\right)=200$

$\Rightarrow \text{correct}\left({\sum }_{i=1}^{20}{x}_i\right)=(200-8+12)=204$

$\Rightarrow$ correct mean $=\frac{204}{20}=10.2$

Also, incorrect $\left({\sum }_{i=1}^{20}{x}_i^2\right)=2080$

$\Rightarrow \text{correct}\left({\sum }_{i=1}^{20}{x}_i^2\right)=(2080-8^2+{12}^2)=2160$

$\Rightarrow$ correct variance $=\frac{\text{correct}\left({\sum }_{i=1}^{20}{x}_i^2\right)}{20}-(\text{correct mean}{)}^2$

$=\{\frac{2160}{20}-(10.2{)}^2\}=(108-104.04)=3.96$

$\Rightarrow \text{correct SD}=\sqrt{3.96}=1.989$

Thus, in this case, we have mean = 10.2 and SD = 1.989