Question

A student obtained the mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50 instead of 40. Find the correct mean and standard deviation.

Answer 1

Calculating correct Mean, by using the following observations:

$$\begin{gathered} {\overline{X}}_{wrong}=\frac{\Sigma X_{wrong}}{N} \\ \Sigma X_{wrong}={\overline{X}}_{wrong}\times N \\ \Sigma X_{wrong}=40\times 100 \\ \Rightarrow \Sigma X_{wrong}=4000 \\ {\mathrm{\Sigma X}}_{correct}=\Sigma X_{wrong}-\mathrm{incorrect}\mathrm{item}+\mathrm{correct}\mathrm{item} \\ {\mathrm{\Sigma X}}_{correct}=4000-50+40 \\ \Rightarrow {\mathrm{\Sigma X}}_{correct}=3990 \\ {\overline{\mathrm{X}}}_{\mathrm{correct}}=\frac{{\mathrm{\Sigma X}}_{correct}}{N}=\frac{3990}{100}=39.9 \\ \mathrm{Calculation}\mathrm{of}\mathrm{Correct}\mathrm{S}.\mathrm{D}. \\ \sigma =\sqrt{\frac{\Sigma x^2}{N}-{\left(\overline{X}\right)}^2} \\ 5.1=\sqrt{\frac{\Sigma x^2}{20}-{\left(20\right)}^2} \end{gathered}$$

Squaring both sides

$26.01=\frac{\mathrm{\Sigma }{x}^2}{100}-1600$

$$\begin{gathered} 1626.01\times 100=\Sigma x^2 \\ \Sigma {x^2}_{wrong}=162601 \\ \Sigma {x^2}_{correct}=\Sigma {x^2}_{wrong}-{\left(\mathrm{Incorrect}\mathrm{value}\right)}^2+{\left(\mathrm{Correct}\mathrm{value}\right)}^2 \\ \Sigma {x^2}_{correct}=162601-{\left(50\right)}^2+{\left(40\right)}^2 \\ \Sigma {x^2}_{correct}=162601-2500+1600=161701 \\ {\mathrm{\sigma }}_{\mathrm{Correct}}=\sqrt{\frac{\Sigma {x^2}_{correct}}{\mathrm{N}}-\left({\overline{\mathrm{X}}}_{\mathrm{correct}}\right)} \\ or,{\mathrm{\sigma }}_{\mathrm{Correct}}=\sqrt{\frac{161701}{100}-{\left(39.9\right)}^2} \\ or,{\mathrm{\sigma }}_{\mathrm{Correct}}=\sqrt{1617.01-1592.01} \\ or,{\mathrm{\sigma }}_{\mathrm{Correct}}=\sqrt{25} \\ \Rightarrow {\mathrm{\sigma }}_{\mathrm{Correct}}=5 \end{gathered}$$


Hence, correct mean and standard deviation are 39.9 and 5 respectively.