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 in each of the following cases:
(i) If wrong item is omitted
(ii) if it is replaced by 12.

Answer 1

$$\begin{gathered} n=20 \\ \mathrm{Mean}=\overline{X} \\ \mathrm{SD}=\sigma =2\mathbf{} \\ \frac{1}{n}\underset{}{\sum x_i=}\overline{X} \\ \therefore \frac{1}{20}\underset{}{\sum x_i=10} \\ \Rightarrow \sum x_{\mathrm{i}}=200\left[\mathrm{This}\mathrm{is}\mathrm{incorrect}\mathrm{due}\mathrm{to}\mathrm{misread}\mathrm{values}.\right]...\left(1\right) \\ \Rightarrow \mathrm{Variance}={\sigma }^2=4 \\ \frac{1}{n}\sum _{}{x_i}^2-\left(\overline{X}\right){}^2=4 \\ \Rightarrow \frac{1}{20}\sum _{}{x_i}^2-{10}^2=4 \\ \Rightarrow \frac{1}{20}\sum _{}{x_i}^2=104 \\ \Rightarrow \sum _{}{x_{\mathrm{i}}}^2=104\times 20=2080\left[\mathrm{This}\mathrm{is}\mathrm{incorrect}\mathrm{due}\mathrm{to}\mathrm{misread}\mathrm{values}.\right]...\left(2\right) \end{gathered}$$


(i) If observation 8 is omitted, then total 19 observations are left.

Incorrected $\sum _{}{x}_i=200$

$\mathrm{Corrected}\sum _{}{x}_i+8=200$

$$\begin{gathered} \mathrm{Corrected}\sum _{}^{}{x}_i=192 \\ \Rightarrow \left(\mathrm{Corrected}\mathrm{mean}\right)=\frac{\mathrm{Corrected}\sum _{}^{}{x}_i}{19} \\ =\frac{192}{19} \\ =10.10 \\ \mathrm{Using}\mathrm{equation}\left(2\right),\mathrm{we}\mathrm{get}: \\ \mathrm{Corrected}\sum _{}^{}{x_i}^2+8^2=2080 \\ \Rightarrow \mathrm{Corrected}\sum _{}^{}{x_i}^2=2080-64 \\ =2016 \\ \therefore \frac{1}{19}\mathrm{Corrected}\sum _{}^{}{x_i}^2-{\left(\mathrm{Corrected}\mathrm{mean}\right)}^2=\mathrm{Corrected}\mathrm{variance} \\ \Rightarrow \mathrm{Corrected}\mathrm{variance}=\frac{1}{19}\times 2016-{\left(\frac{192}{19}\right)}^2 \\ \Rightarrow \mathrm{Corrected}\mathrm{variance}=\frac{\left(2016\times 19\right)-{\left(192\right)}^2}{{19}^2} \\ \Rightarrow \mathrm{Corrected}\mathrm{variance}=\frac{38304-36864}{{19}^2} \\ \Rightarrow \mathrm{Corrected}\mathrm{variance}=\frac{1440}{{19}^2} \\ \mathrm{Corrected}\mathrm{SD}=\sqrt{\mathrm{Corrected}\mathrm{variance}} \\ =\sqrt{\frac{1440}{{19}^2}} \\ =\frac{12\sqrt{10}}{19} \\ =1.997 \end{gathered}$$

Thus, if 8 is omitted, then the mean is 10.10 and SD is 1.997.



(ii) When incorrect observation 8 is replaced by 12:


$$\begin{gathered} \text{From}\mathrm{eq}uation\left(1\right): \\ \mathrm{Incorrected}\sum _{}^{}{x}_i=200 \\ \mathrm{Corrected}\sum _{}^{}{x}_i=200-8+12=204 \\ \mathrm{Corrected}\overline{X}=\frac{204}{20}=10.2 \\ \mathrm{Incorrected}\sum _{}^{}{x_i}^2=2080\left[\mathrm{from}\left(2\right)\right] \\ \mathrm{Corrected}\sum _{}^{}{{\mathrm{x}}_{\mathrm{i}}}^2=2080-8^2+{12}^2 \\ =2160 \\ \mathrm{Corrected}\mathrm{variance}=\frac{1}{20}\times \mathrm{Corrected}\sum _{}^{}{{\mathrm{x}}_{\mathrm{i}}}^2-{\left(\mathrm{Corrected}\overline{\mathrm{X}}\right)}^2 \\ =\frac{1}{20}\times 2160-{\left(\frac{204}{20}\right)}^2 \\ =\frac{\left(2160\times 20\right)-{\left(204\right)}^2}{{20}^2} \\ =\frac{43200-41616}{400} \\ =\frac{1584}{400} \\ \mathrm{Corrected}\mathrm{SD}=\sqrt{\mathrm{Corrected}\mathrm{variance}} \\ =\sqrt{\frac{1584}{400}} \\ =\frac{\sqrt{396}}{10} \\ =\frac{19.899}{10} \\ =1.9899 \end{gathered}$$

If 8 is replaced by 12, then the mean is 10.2 and SD is 1.9899.