Question

For a group of 200 candidates, the mean and standard deviations 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 standard deviation.

Answer 1

We have:
$n=200,\overline{X}=40,\sigma =15$

$$\begin{gathered} \frac{1}{n}\sum x_{\mathrm{i}}=\overline{\mathrm{X}} \\ \therefore \frac{1}{200}\sum _{}{x}_i=40 \\ \Rightarrow \sum _{}{x}_i=40\times 200=8000 \\ \mathrm{Since}\mathrm{the}\mathrm{score}\mathrm{was}\mathrm{misread},\mathrm{this}\mathrm{sum}\mathrm{is}\mathrm{incorrect}. \\ \Rightarrow \mathrm{Corrected}\sum _{}^{}{x}_i=8000-34-53+43+35 \\ =8000-7 \\ =7993 \\ \therefore \mathrm{Corrected}\mathrm{mean}=\frac{\mathrm{Corrected}\sum _{}^{}{x}_i}{200} \\ =\frac{7993}{200} \\ =39.955\mathbf{} \end{gathered}$$


$$\begin{gathered} \mathrm{SD}=\sigma \mathit{}=15\mathbf{}\mathbf{}\mathbf{} \\ \Rightarrow \mathrm{Variance}={15}^2=225 \\ \mathrm{According}\mathrm{to}\mathrm{the}\mathrm{formula}, \\ \mathrm{Variance}=\left(\frac{1}{n}\sum _{}{x_{\mathrm{i}}}^2\right)-{\left(\frac{1}{n}\sum _{}{\mathrm{x}}_{\mathrm{i}}\right)}^2 \\ \therefore \frac{1}{200}\sum _{}{x_{\mathrm{i}}}^2-{\left(40\right)}^2=225 \\ \Rightarrow \frac{1}{200}\sum _{}{\left(x_{\mathrm{i}}\right)}^2-1600=225 \\ \underset{}{\Rightarrow \sum }{\left(x_{\mathrm{i}}\right)}^2=200\times 1825=365000 \\ \mathrm{This}\mathrm{is}\mathrm{an}\mathrm{incorrect}\mathbf{}reading. \\ \therefore \mathrm{Corrected}\sum _{}{\left(x_i\right)}^{\mathbf{2}}=365000-{34}^2-{53}^2+{43}^2+{35}^2 \\ =365000-1156-2809+1849+1225 \\ =364109 \end{gathered}$$


$$\begin{gathered} \mathrm{Corrected}\mathrm{variance}=\left(\frac{1}{n}\times \mathrm{Corrected}\sum _{}{x}_i\right)-{\left(\mathrm{Corrected}\mathrm{mean}\right)}^2 \\ =\left(\frac{1}{200}\times 364109\right)-{\left(39.955\right)}^2 \\ =1820.545-1596.402 \\ =224.14 \\ \mathrm{Corrected}\mathrm{SD}=\sqrt{\mathrm{Corrected}\mathrm{variance}} \\ =\sqrt{224.14} \\ =14.97 \end{gathered}$$